The first volume (multiple choice questions, ***36 points)
First, multiple-choice questions (* * 12 small questions, 3 points for each small question, ***36 points)
There are four alternative answers to the following questions, only one of which is correct. Please black out the code of the correct answer on the answer sheet.
1. The reciprocal of rational number -3 is
A.3. B.-3。 C. D。
2. The value range of the independent variable X in the function is
A.x≥0。 B.x≥-2。 C.x≥2。 D.x≤-2。
3. As shown in the figure, the solution set of an inequality group is expressed on the number axis, so this inequality group may be
a . x+ 1 & gt； 0，x-3 & gt； 0.b . x+ 1 & gt； 0，3-x & gt； 0.
c . x+ 1 & lt； 0，x-3 & gt； 0.d . x+ 1 & lt； 0，3-x & gt； 0.
4. In the following events, the inevitable events are
A. win the lottery.
B.turn on the TV, the advertisement is playing.
C. toss a coin, face up.
There are only five black balls in a bag, and one of them is a black ball.
5. If x 1 and x2 are two roots of the unary quadratic equation x2+4x+3=0, then the value of x 1x2 is
a4 . B3 . c .-4d-3。
6. It is reported that in 20 1 1 year, the enrollment plan of colleges and universities nationwide is about 6.75 million. The number 6,750,000 is expressed by scientific calculation method as follows
A.675× 104。 B.67.5× 105。 C.6.75× 106。 D.0.675× 107。
7. As shown in the figure, in the trapezoidal ABCD, AB∨DC, AD=DC=CB, and if ∠ Abd = 25, the size of ∠BAD is
.40 caliber. b45。
C.50D.60。
8. The picture on the right is a direct view of an object, and its top view is
9. In the rectangular coordinate system, we call the points whose abscissa and ordinate are integers the whole point, and stipulate that the interior of the square does not contain the points on the boundary. Observe the square whose center is at the origin and one side is parallel to the X axis, as shown in the figure: a square with a side length of 1 has 1 integer points, and a square with a side length of 2 and a side length of 3 has 1 integer points.
64. B.49. C.36. D.25
10. As shown in the figure, the railway MN and highway PQ intersect at point O, ∠ QON = 30. The distance from point O to point A on PQ highway is 240 meters. If the train is running, it will be affected by noise within 200 meters around. Then, when the train runs on the railway MN at a speed of 72 km/h, the time when the noise is affected at point A is
A.65438+C.20 seconds b. 16 seconds C.20 seconds d.24 seconds.
1 1. In order to extensively carry out sunshine fitness activities, Hongxing Middle School invested 380,000 yuan on 20 10 for projects such as site maintenance, installation of facilities and purchase of equipment. Figures 1 and 2 respectively reflect the allocation of funds invested in 20 10 and the annual growth rate of funds invested in purchasing equipment since 2008.
Based on the above information, make the following judgment:
① Among the total investment of 20 10, the funds for purchasing equipment are the most;
② The investment in purchasing equipment in 2009 increased by 8% compared with that in 20 10;
③ ③ If the investment growth rate of equipment purchased in 20 1 1 year is the same as that in 20 10, the investment of equipment purchased in 201+32% year is 38×38%×( 1+32%).
A.0. B. 1。 C.2. D.3
12. As shown in the figure, in diamond-shaped ABCD, AB=BD, points E and F are on AB and AD respectively, and AE=DF. Connecting BF and DE intersect at G point, and connecting CG and BD intersect at H point.
①△AED?△DFB；
②S quadrilateral bcdg = cg2
③ If AF=2DF, BG=6GF. The correct conclusion is.
① ③. Only 122.b only 133.c only 233.d only 123.
Volume 2 (multiple choice questions, ***84 points)
Fill in the blanks (***4 small questions, each with 3 points, *** 12 points).
The following questions do not need to be solved. Please fill in the results directly on the answer sheet.
The value of 13.SIN 30 is _ _ _ _.
14. In a math exam, the scores of five students were: 89, 9 1, 105, 105, 1 10. The median of this set of data is _ _ _ _ _ _ _.
15. For the container with water inlet pipe and water outlet pipe, from a certain moment on, only open the water inlet pipe to feed water, and then open the water outlet pipe to drain water. When the water inlet pipe is opened to 12 minutes, the water inlet pipe is closed. During the period from opening the water inlet pipe to closing the water inlet pipe, the functional relationship between the water quantity y (unit: liter) in the container and the time x (unit: minute) is shown in the figure. After closing the water inlet pipe,
16. As shown in the figure, the coordinates of vertices A and B of □ABCD are A (- 1, 0) and B (0 0,2) respectively, vertices C and D are on hyperbola y=, edge AD intersects Y axis at point E, and the area of quadrilateral BCDE is five times that of △ABE, so k = _ _.
Third, answer questions (***9 small questions, ***72 points)
The following questions need to be written in the designated position on the answer sheet to prove the process, calculation steps or draw a graph.
17. (The full mark of this question is 6) Solve the equation: x2+3x+ 1=0.
18. (The full mark of this question is 6) Simplify first, then evaluate:, where x=3.
19. (The full mark of this question is 6) As shown in the figure, D and E are points on AB and AC respectively, AB=AC and AD=AE. Verify that ∠ b = ∠ c.
20. (Full mark for this question is 7) When a car passes an intersection, it may continue straight, or turn left or right. If these three possibilities are the same, then there are two cars passing through this intersection.
(1) Try one of the tree diagrams or list methods to list all possible results of the driving directions of these two cars;
(2) Find the probability of at least one car turning left.
2 1. (The full mark of this question is 7) In the plane rectangular coordinate system, the vertex coordinates of △ABC are A (-7, 1), B (1, 1) and C (1, 7). The endpoint coordinate of the line segment DE is D A(-7).
(1) Try to explain how to translate the line segment AC to coincide with the line segment ED;
(2) Rotate △ABC counterclockwise around the coordinate origin o, so that the corresponding side of AC is de. Please directly write down the coordinates of point B corresponding to point F;
(3) Draw △DEF in (2), and at the same time use △ABC to rotate 90 counterclockwise around the coordinate origin O to draw the rotated figure.
22. (The full mark of this question is 8) As shown in the figure, PA is the tangent of ⊙O, and A is the tangent point. The vertical line AB passing through A is OP, the vertical foot is point C, the intersection point ⊙O is point B, the extension line between BO and ⊙O is point D, and the extension line of PA is point E. 。
(1) Verification: PB is the tangent of ⊙O;
(2) If tan∠ABE=, find the sine value.
23. (The full mark of this question is 10) The extracurricular activity group of Xingguang Middle School will build a rectangular biological nursery garden. One side of it is against the wall, and the other three sides are surrounded by a fence 30 meters long. It is known that the wall is18m (as shown in the figure), and the side length perpendicular to the wall of this nursery garden is x meters.
(1) If the length of the side parallel to the wall is y meters, directly write the functional relationship between y and x and the value range of its independent variable x;
(2) When the length of the side perpendicular to the wall is several meters, the area of this nursery garden is the largest, and the maximum value is found;
(3) When the area of this nursery is not less than 88 square meters, try to write the value range of X directly by combining the function image.
24. (The full mark of this question is 10)
(1) As shown in figure 1, in △ABC, points D, E and Q are on AB, AC and BC respectively, and DE∑BC and AQ intersect at point P, which verifies that:
(2) As shown in the figure, in △ABC, ∠ BAC = 90, the four vertices of the square deFG are on the side of △ABC, connecting AG and AF, and intersecting with DE at m and n respectively.
① As shown in Figure 2, if AB=AC= 1, write the length of MN directly;
② As shown in Figure 3, verify MN2=DM? Yeah.
25. (The full mark of this question is 12) as shown in figure 1, and the parabola y=ax2+bx+3 passes through two points: a (-3,0) and B (- 1 0).
(1) Find the analytical formula of parabola;
(2) Let the vertex of the parabola be M, and the straight line y=-2x+9 intersect with the Y axis at point C and with the straight line OM at point D. Now translate the parabola and keep the vertex on the straight line OD. If there is only one common point between the translated parabola and the ray CD (including the end point C), find the value or range of its vertex abscissa;
(3) As shown in Figure 2, parabolic translation. When the vertex reaches the origin, a straight line that passes through Q (0 0,3) and is not parallel to the X axis intersects with the parabola at two points, E and F. Find out whether there is a point P on the negative semi-axis of the Y axis, so that the center of △PEF is on the Y axis. If yes, find the coordinates of point p; If it does not exist, please explain why.
20 1 1 Mathematical Answers for Senior High School Entrance Examination in Wuhan, Hubei Province
First, multiple choice questions
1.A 2。 C 3。 B 4。 D 5。 B 6。 C 7。 C 8。 A nine. B 10。 B 1 1。 C 12。 D
Second, fill in the blanks
13. 1/2
14. 105; 105; 100
15.8
16. 12
Third, answer questions.
17. (6 points for this question) Solution:
∵a= 1,b=3,c= 1∴△=b2-4ac=9-4× 1× 1=5>0∴x=-3
∴x 1=-3+，x2=-3-
18. (6 points in this question) Solution: The original formula = x (x-2)/x ÷ (x+2)/x = x (x-2)/x? x/(x+2)(x-2)= x/(x+2)
When x=3, the original formula =3/5.
19. (6 points for this question) Solution:
Proof: in △ABE and △ACD, AB = AC ∠ A = ∠ A AE = AD.
∴△ABE≌△ACD
∴∠B=∠C
20.(7 points) Solution 1:
Left and right
Left (left, left) (left, straight) (left, right)
Straight (straight, left) (straight, straight) (straight, right)
Right (right, left) (right, straight) (right, right)
(1) According to the meaning of the question, you can draw the following "tree diagram":
∴∴, there are nine possible outcomes for the direction of these two cars.
(2) According to the "tree diagram" in (1), there are five outcomes for at least one car to turn left, and the possibility of all outcomes is equal.
∴P (at least one car turns left) = 5/9
Solution 2: According to the meaning of the problem, the following table can be listed:
The following solution is 1 (abbreviated)
2 1. (7 points for this question) (1) First move the line segment AC to the right by 6 units.
Then translate it down by 8 units. (Other translation methods are also possible)
(2)F(- 1，- 1)
(3) Draw the correct figure as shown in the figure.
22. (8 points for this question) (1) Proof: connect OA.
∵PA is the tangent of⊙ o,
∴∠PAO=90
Oa = ob, OP⊥AB is represented by C.
∴BC=CA,PB=PA
∴△PBO≌△PAO
∴∠PBO=∠PAO=90
∴PB is the tangent of⊙ O.
(2) Solution 1: Connect AD, ∫BD is the diameter, ∠ Bad = 90.
According to (1), ∠ BCO = 90.
∴AD∥OP
∴△ADE∽△POE
∴ ea/EP = ad/OP=5t from AD∥OC, ad = 2oc ∫ tan ∠ Abe =1/2 ∴ oc/BC =1/2, let oc = t, then BC =
∴EA/EP=AD/OP=2/5. Let EA = 2m, EP = 5m and PA=3m.
∵PA=PB∴PB=3m
∴sinE=PB/EP=3/5
(2) Solution 2: If AD is connected, then ∠ bad = 90 is known from (1) ∠ bco = 90: from ad∨oc, ∴ AD = 2oc: Tan ∠ Abe = 67.
Pa = Pb = 2 t If A is AF, Pb is in F, then AF? PB=AB？ personal computer
∴AF= t and then get pf = t from pythagorean theorem.
∴sinE=sin∠FAP=PF/PA=3/5
23.( 10) solution: (1) y = 30-2x (6 ≤ x
(2) If the area of rectangular nursery garden is S, then S = XY = X (30-2x) =-2x2+30x ∴ S =-2 (x-7.5) 2+1 2.5 can be known from (1).
That is, when the length of the rectangular nursery perpendicular to the wall is 7.5m, the area of the nursery is the largest, and the maximum value is112.5 (3) 6 ≤ x ≤11.
24. (The score of this question is 10) (1) is proved in △ABQ, DP∨bq, ∴△ADP∽△ABQ, ∴ DP/BQ = AP/AQ.
Similarly, at △ACQ, EP/CQ = AP/AQ.
∴ DP/BQ = EP/CQ。 (2) (3) Proof: ∫∠b+∠C = 90, ∠ CEF+∠ C = 90. ∴∠b =∞。 EF=CF？ Bulgaria
∵ DG = GF = EF，∴ GF2 = CF？ Bulgaria
From ( 1)，dm/BG = Mn/gf = en/cf∴(Mn/gf)2 =(DM/BG)？ (EN/CF)
∴MN2=DM？ (constituting a verb) means "to put in a state of ..."
25.( 1) parabola y=ax2+bx+3 passes through two points: a (-3,0) and B(- 1, 0).
∴ 9a-3b+3 = 0，A-b+3 = 0
The solution is a = 1
The analytical formula of b = 4 ∴ parabola is y=x2+4x+3(2). Y=(x+2)2- 1∴ The vertex m (-2, 1) ∴ line of parabola is obtained by the formula (1).
So let the coordinate of the vertex of the translation parabola be (h, h), and the analytical formula of the translation parabola be y=(x-h)2+ h.① When the parabola passes through point C, ∫ c (0,9), ∴h2+ h=9.
The solution is h=. Ⅶ when ≤ h
There is only one common point between the translational parabola and the ray CD.
② When there is only one common point between parabola and straight line CD,
Through the equation y=(x-h)2+ h, y=-2x+9.
X2+(-2h+2)x+h2+ h-9=0，∴△=(-2h+2)2-4(h2+ h-9)=0，
The solution is h=4.
At this time, the parabola y=(x-4)2+2 has only one thing in common with the ray CD, which is consistent with the meaning of the question.
To sum up, when there is only one common point between the translational parabola and the ray CD, the value or range of the vertex abscissa is h=4 or ≤ H.
(3) Method 1
When the parabola is translated, when the vertex reaches the origin, its analytical formula is y=x2.
Let the analytical formula of EF be y=kx+3(k≠0).
Suppose there is a point P(0, t) that satisfies the condition, as shown in the figure, if P passes through the gh∨x axis and E and F pass through the GH vertical line respectively, then the vertical foot center is G, H.∵△PEF is on the Y axis, GEP =∠EPQ =∞.
∴-xe/xf=(ye-t)/(yf-t)=(kxe+3-t)/(kxf+3-t)
∴2kxE？ xF=(t-3)(xE+xF)
From y=x2, y=-kx+3, x2-kx-3=0.
∴xE+xF=k,xE？ Xf = -3。 ∴ 2k (-3) = (t-3) k，∫k≠0，∴ t =-3。 ∴ There is a point P(0, -3) on the negative semi-axis of the Y axis, so that the center of △PEF is on the Y axis.
In the second method, let the analytical formula of EF be y=kx+3(k≠0), and the coordinates of points E and F are (m, m2)(n, n2) respectively. It is known by the method 1 that mn=-3. Let point E be the symmetrical point R(-m, m2) about the Y axis, and make a straight line FR at this point to intersect with the Y axis. The analytical formula of the straight line FR is y=(n-m)x+mn. When x=0, y=mn=-3,∴P(0,-3). There is a point P(0, -3) on the negative semi-axis of the ∴ Y axis, so that the heart of △PEF is on the y axis.
Ran Ruihong's Arrangement in Wuhan Optical Valley Phase III