1. The solution of the quadratic equation x2-x-2 = 0 is ....................................... ().
A.x 1= 1，x2=2 B.x 1= 1，x2=-2 C.x 1=- 1，x2=-2 D.x 1=- 1，x2=2
2. It is known that point A is within ⊙O with a radius of R, and the distance from point A to point O is 6, so the range of R is ................................................................................................................................
A.r > 6 B.r ≥ 6 C.r < 6 D.r ≤ 6
3. As shown in the figure, a seagoing vessel is located at A, 60 nautical miles away from Lighthouse P, and after sailing due south for a period of time, it reaches B, 45 east of Lighthouse P. At this time, the distance between B where the seagoing vessel is located and Lighthouse P is ....................................................
302 nautical miles 303 nautical miles 60 nautical miles 306 nautical miles
4. A machinery factory produced 500,000 pieces in July and 1.96 million pieces in the third quarter. Let the average monthly growth rate of this factory be X, then the equation that X satisfies is .......................... ().
a . 50( 1+x)2 = 196 b . 50+50( 1+x)2 = 196
c . 50+50( 1+x)+50( 1+x)2 = 196d . 50+50( 1+x)+50( 1+2x)= 196
The school organized a talent show competition, and the top six won prizes. There are 65,438+03 students taking part in the competition, and their scores are different. After a student knows his score, he only needs to know one quantity in the score statistics of these 65,438+03 students, namely
A. Mode B. Variance C. Median D. Mean
6. As the picture shows, there is a pond between A and B. Xiao Ming measures the distance between A and B in the following way: first, choose a point C outside AB, then measure the midpoint M and N between AC and BC, and measure the length of MN as 6m, thus knowing the distance between A and B.. The description of his inquiry activities is wrong. ...................................................................
A.AB = 12m B. MN∨AB c .△CMN∽△ cab d.cm: ma =1∶ 2
7. As shown in the figure, the image of quadratic function Y = AX2+BX+C (A ≠ 0) is known, and there are four conclusions as follows: ① B2-4ac > 0; ②ABC < 0； ③b < a+c； ④ 4a+b = 1, in which the correct conclusion is ...................... ().
A.①② B.①②③ C.①②④ D.①③④
8. As shown in the figure, the radius ⊙O is 1, the delta △ABC is an equilateral triangle inscribed by ⊙O, the points D and E are on a circle, and the quadrilateral BCDE is a rectangle, and the area of this rectangle is ………………………………………………………………………………………………….
A.2 B. 3 C. 32 D. 32
9. As shown in the figure, point A(a, b) is the moving point of the second quadrant on the parabola Y = 12x2, OB⊥OA.
The parabola intersects with point B(c, d) ... When point A moves on the parabola, the following conclusions are drawn:
①ac is a constant value; ②AC =-BD； ③ The area of δ ③△AOB is constant; ④ straight AB
Absolutely. Among them, the correct conclusion is ............................ ().
A.4 B.3 C.2 D. 1
10. Now, define a transformation: for a sequence S0 consisting of any five numbers, replace each number with the number of times that the number appears in S0, and a new sequence S 1 can be obtained. For example, the sequence S0: (4, 2, 3, 4, 2) can be transformed to generate a new sequence s 1:(2, 2.
A.( 1，2， 1，2，2) B.(2，2，2，3，3)
C.( 1， 1，2，2，3) D.( 1，2， 1， 1，2)
Fill in the blanks (this topic is entitled ***8 small questions, 2 points for each small question, *** 16 points. )
The vertex coordinates of 1 1. parabola y = x2-2x+3 are.
12. Write the seven letters in the English word theorem on seven identical cards, clean them word by word and put them on the table. Choose any one, and the probability of getting the letter e is.
13. The known proposition "The quadratic equation of X is x2+BX+ 14 = 0, and when b < 0, there must be a real number solution" can explain that this proposition is false.
14. As shown in the figure, the surface development diagram of the cone consists of a sector and a circle. It is known that the area of a circle is 100π, the central angle of a sector is 120, and the area of this sector is.
15. As shown in the figure, add a condition: make △ ade ∽△ ACB.
16. It is known that y is a function about x, and the function image is shown in the figure, so when y > 0, the value range of the independent variable x is.
17. As shown in the figure, in Rt△ABC, ∠ C = 90, AC = 3, BC = 4, ⊙O is the inscribed circle of △ABC, and point D is the midpoint of the oblique side of AB, then tan∠ODA is equal to.
18. as shown in the figure, in Rt△ABC, ∠ b = 90, SIN ∠ BAC = 13, point d is a point on AC, and BC = BD = 2. rotate Rt△ABC around point c to the position of Rt△FEC, so that point e is on ray BD.
Third, answer the question (this big question is * * 10 small question, ***84 points. )
19. (8 points in this question) Solve the equation: (1) (4x-1) 2-9 = 0 (2) x2-3x-2 = 0.
20. As shown in the figure, in △ABC, AB = AC = 5, BC = 6, P is a point above BC, and BP = 2. Put the vertex of the angle equal to ∠B at point P, and then rotate the angle around point P, so that both sides of the angle always intersect AB and AC respectively, and the intersection points are D and E.
(1) Verify △ BPD ∽△ CEP.
② Does PD ⊥ DE have such a position? If it exists, find out the length of BD;
If it does not exist, explain why.
2 1. (8 points in this question) As shown in the figure, AB is the diameter of ⊙O, BC is the tangent of ⊙O, D is a point on ⊙O, and CD = CB. Extend the extension line of CD to BA at point e.
(1) Verification: CD is the tangent of ⊙ o. 。
(2) If the distance from the center o to the chord DB is 1, ∠ Abd = 30, find the area of the shaded part in the figure. (Results keep π)
22. (8 points in this question) At about 23: 35 pm on February 3, 20 14, a crowded stampede accident occurred in Chen Yi Square on the Bund in Shanghai. In order to eliminate potential safety hazards, Wuxi municipal government decided to rebuild an observation deck in Lihu Park. As shown in the figure, the tilt angle of the platform is ∠ ABC = 62. (The results were kept to 0.0 1 m) (reference data: sin62 ≈0.88, cos62 ≈0.47, TAN 50 ≈ 1.20).
23. (8 points in this question) There are seven cards that are exactly the same except for the marked values. Put four cards with tag values of -2,-1, 3 and 4 into bag A, and put three cards with tag values of -3, 0 and 2 into bag B. Now, randomly draw one card from bag A and bag B, and use X respectively.
(1) Please write all the situations of point A (x, y) by tree diagram or list method.
(2) Find the probability that point A belongs to the first quadrant.
24. (8 points in this question) The school winter fun sports meeting has set up the project of "grabbing the harvest and grabbing the seeds", and both Group A and Group B of Class 8 (5) want to participate in the competition on behalf of the class. In order to select a better team, the Class Committee of Class 8 (5) organized a tryout, and the results of each group were as follows:
Group a 7 8 9 7 10
10 9 10 10 10
Group b 10 8 7 9 8 10
10 9 10 9
(1) The median score of group A is 0, and the mode score of group B is 0.
(2) Calculate the average score and variance of group B. 。
(3) In view of the variance of group A's score of 1.4, this group was selected to represent Class 8 (5) in the school competition.
25. (8 points in this question) In the activity of "beautifying the campus", an interest group wants to enclose a rectangular garden AB=x (m) with right angles (both sides of Yamato are long enough) as shown in the figure. The fence only surrounds both sides of AB and BC), and let AB = x (m).
(1) If the area of the garden is 192m2, find the value of x. 。
(2) If the distances between a tree at P and the wall at DA are 15m and 6m respectively, the tree should be enclosed in the garden (including the boundary, regardless of the thickness of the tree). Find the value of garden area s.
26. (8 points for this question) As shown in the figure, the right-angle OABC is in the plane right-angle coordinate system xoy, point A is on the positive semi-axis of the X axis, point C is on the positive semi-axis of the Y axis, OA = 4, OC = 3, if the vertex of the parabola is on the side of BC, the parabola passes through points O and A, and the straight line AC intersects the parabola at point D (1, n).
(1) Find the function expression of parabola.
(2) If point M is on a parabola and point N is on the X axis, is there a point?
Is a quadrilateral with vertices d, m and n a parallelogram? If it exists, find it.
Coordinates of point n; If it does not exist, please explain why.
27. (This question 10) As shown in the figure, in △ABC, ∠ A = 90, AB = 2 cm, AC = 4 cm. Moving points P and Q start from point A and point B, respectively, and move in opposite directions at the speed of 1 cm/s, making a square with AP as one side.
(1) When t = s, point P coincides with point Q. 。
(2) When t = s, point D is on QF.
(3) When point P is between Q and B (excluding Q and B), find the function expression between S and T. 。
28.( 10) Master Huang, a carpenter, made a round desktop as big as possible with rectangular boards with length AB = 3 and width BC = 2. He designed four schemes:
Scheme 1: directly saw a circle with radius;
Scheme 2: The center O 1 and O2 are on CD and AB respectively, and the radii are O 1C and O2A respectively. Saw two circumscribed semicircles to form a circle;
Scheme 3: AC saw the rectangle into two triangles along the diagonal, and translate the triangles appropriately to see a new circle;
Scheme 4: Saw a small rectangular BCEF and put it under the rectangular AFED, and saw a circle as big as possible with the assembled board.
(1) Write the radius of the circle in the scheme 1.
(2) Through calculation, which circle has a larger radius in Scheme 2 and Scheme 3?
(3) in the fourth scheme, let ce = x (0 < x < 1), and what is the radius of the circle when x is taken? And explain the radius of which round desktop in the four schemes.
I. Multiple-choice questions: (This topic is entitled *** 10, with 3 points for each question and 30 points for * * *. )
1.D 2。 A 3。 A 4。 C 5。 C 6。 D 7。 B 8。 B 9。 B 10。 D
Fill-in-the-blank question: (This big question is entitled ***8 small questions, with 2 points for each small question, *** 16 points. )
11.(12)12.2713. When b =- 12, the equation has no solution (no answer) 14.300 π.
15.∠ aed = ∠ b (no answer) 16.x
Answer: (This big question is * * 10, and the score is ***84. )
19.( 1)(4x- 1)2-9 = 0(2)x2―3x―2 = 0
4x- 1 =+/-3 ..........................................................................................................................................................
X 1 = 1, x2 =- 12...4 points, x 1 = 3+ 172, x2 = 3- 172...4 points.
20. solution: (1) ∵ AB = communication ∴∠ b = ∠ c ................................1min.
∠∠DPC =∠DPE+∠EPC =∠b+∠BDP...2 points.
∴∠
∴△ Abd ∽△ DCE ........................................... 4 points.
(2) AH⊥BC
In Rt△ABH and Rt△PDE.
∴ COS ∠ ABH = COS ∠ DPE = BHAB = PDPE = 35 .................................... 6 points.
∴ PDPE = BDPC = 35 and ∴ PC = 4 ∴ BD =125 .......................................................... 8 points.
2 1.( 1) proves that the connecting line od÷BC is a tangent of ÷ o ∴ ABC = 90. ...........................................................................................................
cd = cb，OB = OD ∴∠ CBD = ∠ CDB，∠ OBD = ∠ ODB ..........................................................................................................
∴∠ ODC =∠ ABC = 90, that is, OD ∴ CD ∴ CD is the tangent of ∵ O. ..........................................................................................................
(2) solution: OF⊥DB, in Rt△OBF,
∵∠ Abd = 30，of = 1，∴∠ BOF = 60，OB = 2，BF = 3 .....................................................................................................
∴∵of⊥bd BD = 2bf = 23，∠ BOD = 2 ∠ BOF = 120 ...................................................................................................
∴S shadow = 43 π-3 ............................................................................................... 8 points.
22. Solution: Make AE⊥CD in E after point A.
In Rt△ABE, ∠ Abe = 62. ∴ AE = AB？ Sin 62 = 25× 0.88 = 22 meters, ... 2 points.
BE=AB？ Cos 62 = 25× 0.47 =11.75m, ... 4 points.
In rtδade, ∠ ADB = 50,
∴ Germany = Etan 50 = 553 ........................... 6 points.
∴ DB = DC-BE ≈ 6.58 ........................... 7 points.
A: The outward width is about 6.58 meters, and .......................................................................... is 8 points.
23.( 1)
-2 - 1 3 4
-3 (-2, -3) (- 1, -3) (3, -3) (4, -3)
0 (-2, 0) (- 1, 0) (3, 0) (4, 0)
2 (-2, 2) (- 1, 2) (3, 2) (4, 2)
∴ As shown in the table, there are 12 kinds of * * * ..............................................................................................................................................
(2) Because the coordinates of the points belonging to the first quadrant are (3,2) and (4,2) * * *, so
So the probability p =16 ... 8 points.
24.( 1) 9.5 10 ...2 points (2) X-= 9, variance = 1...6 points (3) B...8 points.
25.( 1) According to the meaning of the question, x (28-x) = 192. ..........................................................................................................................
The solution is x = 12 or x =16 ... 3 points.
The value of ∴x is 12m or16m ........................................ 4 points.
(2)∵ According to the meaning of the question, get 6 ≤ x≤13 ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….
And ∵ s = x (28-x) =-(x-14) 2+196 ................................................................ 6 points.
When x≤ 14, s increases with the increase of x.
Therefore, when x = 13, the garden area s, with a value of195m2, .................................. is 8 points.
26. Solution: (1) Let the vertex of the parabola be E. According to the meaning of the question, OA = 4 and OC = 3, we get: E (2 2,3), ... 1.
Then the parabola function relation can be obtained as y =-34 (x-2) 2+3 =-34x2+3x; .............................., 3 points.
(2) The coordinate of available point D is (1, 94)4 points.
There are considered to be two situations:
① When point M is above the X axis, as shown in the answer sheet 1:
Quadrilateral ADMN is a parallelogram, DM∑AN, DM = an,
Dm = 2, ∴ An = 2, ∴ N1(2,0), N2 (6 6,0) ..............................................................................................................
② When the point M is below the X axis, as shown in Figure 2:
If the DQ⊥x axis is at point Q when passing through point D, and the MP⊥x axis is at point P when passing through point M, you can get △ adq △ NMP.
∴ MP = DQ = 94，NP = AQ = 3，∴ N3 (-7- 1，0)，N4 (7- 1，0) .......................................................................................
27. Solution: (1)1...1(2) 45 ...
(3) When 1 < t ≤ 43, as shown in Figure ②, let DE intersect FQ at point H, and the overlapping part is trapezoidal DHQP.
We can get: PQ = 2t-2, HD = 52t-2...3 points.
∴s= 12( pq+ HD)? DP= 12 ( 2t-2+52 t-2)？ T = 94t2-2t (1 < t ≤ 43)...5 points.
When 43 < t < 2, as shown in Figure ③, let DE pass BC at M point and DP pass BC at N point.
The overlapping part is hexagonal EFQPNM.
We can get: AQ = 2-t, AF = 4-2t.
∴S△FAQ = 12 AQ？ Af =(2-t)2 7 points.
Similarly, DN = 3t-4, DM = 12 (3t-4).
∴S△DMN = 12 DM？ DN= 12？ 12 (3t-4) (3t-4) =14 (3t-4) 2 .........................................., 8 points.
∴S=S = s squared apde-s △ FAQ-s △ DMN =-94t2+10t-8. .......................................................................................................................
To sum up, s = 94t2-2t (1< t ≤ 43)-94t2+10t-8 (43 < t < 2)-10.
28. Solution: (1) The radius in Scheme I is 1 ...........................................................................................................................................
(2) Let the radius be r,
Scheme 2: In Rt△O 1O2E, (2r) 2 = 22+(3-2r) 2, and the solution is r = 13 12 … 4.
Scheme 3: ∫△aom∽△ofn, ∴ R3-R = 2-RR, and the solution is R = 65 … 6 points.
∫ 13 12 < 65, ∴ the radius of scheme 3 is relatively large. ................................................................................................................................
(3) The horizontal span and vertical span of the graphics assembled in the fourth scheme are 3-x and 2+x respectively.
Therefore, the diameter of the cutting circle is the smaller of (3-x) or (2+x), which is ............................................................................................ 8 points.
When 3-x < 2+x, that is, x > 12, r =12 (3-x); At this time, r decreases with the increase of x, so r <12 (3-12) = 54;
When 3-X = 2+X, that is, X = 12, r =12 (3-12) = 54;
When 3-X > 2+X, that is, X < 12, R = 12 (2+X). At this time, r increases with the increase of x, so r <12 (2+12) = 54;
In the fourth scheme, when x = 12, r is 54. .............................................................................................................................................
∫1<1312 < 65 < 54, ∴ the radius of the circular desktop obtained in the fourth scheme. ...........................................................................................................