Mathematics Test
I. Multiple-choice questions (This big question consists of * *12 small questions, with 2 points for each small question and 24 points for each small question). Only one of the four options given in each small question meets the requirements of the topic.
1.(- 1) 3 equals ()
A.- 1b . 1C。 -3d . 3
Analyze this problem and examine the power of rational numbers. (-1) 3 =- 1, so choose a.
A: A.
2. In the real number range, X is meaningful, so the value range of X is ().
a . x≥0b . x≤0c . x > 0d . x < 0
Analyze this problem and investigate the meaningful conditions of quadratic root. From the meaningful condition of quadratic root, we can know that x ≥0, so we choose a.
A: A.
3. As shown in figure 1, ABCD in the diamond, AB = 5, ∠ BCD = 120, diagonal AC = ().
A.20
B 15
C. 10
D.5
This paper analyzes the properties of rhombus and the judgment of equilateral triangle. According to the nature of diamonds, AB = BC, ∠ B+∠ BCD = 180, while ∠ BCD = 120, ∴ B = 60, so the triangle ABC is an equilateral triangle, so AC = AB.
Answer: d
4. In the following operations, the correct one is ().
a . 4m-m = 3B。 ―(m―n)=m+n
C.(m2)3 = m6d . m2÷m2 = m
Analyze this problem and examine the operation of algebraic expressions.
Answer: c
5. As shown in Figure 2, four small squares with a side length of 1 form a big square, where A, B and O are the vertices of the small square, the radius ⊙O is 1, and P is the point on ⊙O, and it is located in the upper right small square, then ∠APB is equal to ().
A.30
b45
C.60
Cao 90
Analyze this problem and examine the knowledge about central angle and central angle. According to the fillet theorem: the fillet of an arc is equal to half the central angle of an arc, so the answer to this question is 90× 12 = 45.
Answer: b
6. The image of inverse proportional function y = 1x (x > 0) is shown in Figure 3. With the increase of x value, y value ().
A. Improve
B. Reduce
C. unchanged
D. decrease first and then increase.
Analyze this problem to investigate the properties of inverse proportional function. When k > 0, the inverse proportional function is in each quadrant, and the value of y decreases with the increase of x.
Answer: b
7. The following events, belong to the impossible event is ()
A. the absolute value of a number is less than 0B. The reciprocal of a number is equal to itself.
C. the sum of two numbers is less than 0D. The product of two negative numbers is greater than 0.
According to the probability of occurrence, events can be divided into inevitable events, random events and impossible events. According to the absolute value of real numbers, the event in option A is impossible, so choose A. 。
A: A.
8. Figure 4 is a schematic diagram of the walking elevator between the first floor and the second floor of the shopping center. Where AB and CD represent the horizontal line ∠ ABC = 150 on the first floor and the second floor respectively, and the length of BC is 8 m, then the height h rising from point B to point C by elevator is ().
A.m
B.4 m
C.m
8 meters in diameter
The analysis of this question belongs to the basic question, which examines students' ability to use trigonometric function definition for simple calculation. In Rt△CBE, it can be seen from the definition of trigonometric function that CE=BC? sin 30 = 8×4m = 4m。 So B. Some students often make mistakes because the definition of memory trigonometric function is inaccurate.
Answer: b
9. The quadratic function y = 120x2 (x > 0) is satisfied between the braking distance y(m) and the speed x(m/s) at the start of braking. If the car braking distance is 5 m at a time, the speed at the beginning of braking is ().
40 m/s 20 m/s
C. 10/0m/standard deviation 5m/s
Analyze this problem and investigate the practical application of quadratic function. If the braking distance is 5m, that is, y=5m, it is 5= 120x2. Therefore, x= 10, (x=- 10), so the braking starting speed is10m/s. 。
Answer: c
10. Dig a small cube with a side length of 1 from one corner of a cube blank with a side length of 2 to obtain the part shown in Figure 5, and the surface area of this part is ().
A.20
b22
C.24
Cao 26
Analyze this problem and investigate the overall thinking and the calculation ability of simple geometric surface area. Digging a small cube from one corner of the cube blank results in a part surface area equal to that of the original cube, that is, the surface area of this part is 2×2×6=24, so c 。
Answer: c
1 1. In the calculation program shown in Figure 6, the image corresponding to the functional relationship between y and x should be ().
Analyze this problem and investigate the ability to determine the function image according to the calculation program. According to the calculation program, it is easy to get the functional relationship between y and x as y=-2x+4. From k =-2 < 0, y decreases with the increase of x, when x=0, y = 4；; When y=0, x=2. So the function image that meets the meaning of the question is d.
Answer: d
12. The famous Pythagorean school in ancient Greece called the numbers 1, 3,6, 10 … as "triangular numbers" and the numbers 1, 4,9, 16 … as "square numbers". As can be seen from fig. 7,
a . 13 = 3+ 10b . 25 = 9+ 16
c . 36 = 15+2 1d . 49 = 18+3 1
Analyze this problem and investigate the mathematical thinking method of inquiry and induction. It is clearly pointed out in the title that any "square number" greater than 1 can be regarded as the sum of two adjacent "triangular numbers". Obviously, 13 in option A is not a "square number"; The right side of options b and d is not the sum of two adjacent triangles, so the answer is C.
Answer: c
2. Fill in the blanks (there are 6 small questions in this big question, 3 points for each small question, *** 18 points. Write the answer on the horizontal line of the question)
13. Comparative size: -6-8. (Fill in "")
The analysis of this question is the basic question, and the comparison of real numbers is investigated. Two negative numbers are larger than the size, and the absolute value is larger, but smaller; Or just imagine that the number on the right is always greater than the number on the left.
Answer: >;
14. According to the statistics of Chinese Academy of Sciences, by May this year, China has become the fourth largest country in wind power generation in the world, with an annual power generation of about120,000 kilowatts. 120,000 kilowatts is expressed by scientific notation.
The analysis of this problem investigates the scientific notation. Any number whose absolute value is greater than 10 or less than 1 can be written in the form of a× 10n. Where 1 ≤| a | < 10. For numbers whose absolute value is greater than 10, the exponent n is equal to the integer digits of the original number minus 1. So12000000 =1.2×107.
Answer:1.2×107;
15. within a week, Xiao Ming insisted on self-testing his temperature three times a day. The statistics of the measurement results are as follows:
Body temperature (℃) 36.136.2 36.3 36.4 36.5 36.6 36.7
Degree 2 3 4 6 3 65 438+0 2
Then the median value of these body temperatures is degrees Celsius.
Analyze this problem and examine the concept of median. According to the information provided in the table, the median of a set of data is the number in the middle when this set of data is arranged from small to large (or from large to small). Accordingly, the median of this set of data should be 1 1, and the number is 36.4. Students who solve this kind of problems often choose the wrong one because they don't grasp the calculation method of median well.
Answer: 36.4;
16. If m and n are reciprocal, the value of Mn2-(n- 1) is.
Analyze this problem and examine the knowledge of reciprocal. There are absolute values, opposites and so on. It is related to this knowledge point, and such questions can only be answered according to their concepts. When m and n are reciprocal, Mn2-(n-1) = n-(n-1) =1.
Answer:1;
17. As shown in Figure 8, the side length of equilateral △ABC is 1 cm, and d and e are points on AB and AC respectively. Fold △ADE along the straight line DE, and point A falls at this point, and this point is outside △ABC, then the perimeter of the shadow figure is cm.
The essence of analytic folding problem is "axisymmetric", and the key to solve the problem is to find out the equivalent relationship obtained by axisymmetric transformation. Fold △ADE along a straight line DE, and point A falls at point A ′, so AD = A ′ D and AE = A ′ E, then the perimeter of the shadow figure is equal to BC+BD+CE+A ′ D+A ′ E = BC+BD+CE+.
Answer: 3;
18. As shown in Figure 9, two iron bars stand upright in a bucket with a horizontal bottom. After adding water into the barrel, the length of one exposed to the water is 13, and the length of the other exposed to the water is 15. The sum of the two iron bars is 55 cm, and the water depth in the barrel is cm.
Solving this problem is a problem of ability, and the idea of equation and the ability to extract information by observing graphics are investigated. Let the length of a long iron bar be xcm and the length of a short iron bar be ycm, and we can get the solution from the meaning of the problem, so the depth of water in the bucket is 30×23=20cm.
Answer: 20.
Third, the solution (this big question is ***8 small questions, ***78 points. The solution should be written in words, proof process or calculus steps)
19. (Full score for this small question)
As we all know, α= 2, the value of α.
Answer: Solution: Original formula =
= .
When a = 2 and b =- 1,
Original formula = 2.
Note: If you directly replace the evaluation with this question, the correct result will get the corresponding score.
20. (The full score for this short question is 8)
Figure 10 is a schematic cross-sectional view of a semi-circular bridge hole, with the center of O, the diameter AB as the bottom line of the river, the chord CD as the water level line, CD∥AB, CD = 24 m, OE⊥CD at point E, and the measured sin ∠ Doe =.
(1) Find the radius od;
(2) As required, if the water surface drops at a speed of 0.5 m per hour, how long will it take to drain?
Answer: solution: (e point 1)∵OE⊥CD, CD=24.
∴ED = = 12。
In Rt△DOE,
∫sin∠DOE = =，
∴OD = 13 (m).
(2)OE=
= .
∴ Drainage requirements:
5/0.5 =10 (hour).
2 1. (Full score for this small question)
During the four-month trial period, a store only sold two brands of TVs, A and B, and * * * sold 400 sets. Only one brand can be distributed after the trial sale. In order to make a decision, the dealer is drawing two statistical charts, as shown in figure11and figure 165438+.
(1) The percentage of sales in the fourth month to the total sales is:
(2) Complete the dotted line in figure 1 1-2, which represents the monthly sales of B brand TV sets;
(3) In order to track and investigate the use of TV sets, randomly select the TV sets 1 set sold in our store in the fourth month, and work out the winning probability of B brand TV sets;
(4) After calculation, the average monthly sales of TV sets of the two brands are the same. Please make a brief analysis according to the trend of broken line to determine which brand of TV set this store should distribute.
Answer: Solution: (1) 30%;
② As shown in figure 1;
(3) ;
(4) As the average level of monthly sales is the same, the monthly sales of brand A show a downward trend, while the monthly sales of brand B show an upward trend.
Therefore, the store should distribute B brand TV sets.
22. (The full score for this short question is 9)
It is known that the parabola y = ax2+bx passes through point A (-3, -3) and point P (t, 0), t ≠ 0. ..
(1) If the parabola axis of symmetry passes through point A, as shown in figure 12, please point out the minimum value of y at this time by observing the image and write the value of t;
(2) If yes, find the values of a and b, and point out the opening direction of the parabola at this time;
(3) Write directly the t value that makes the parabolic opening downward.
Answer: Solution: (1)-3.
t =-6。
(2) Substitute (-4,0) and (-3,3) into y = ax2+bx, respectively, to obtain
solve
Up.
(3)- 1 (the answer is not unique).
Note: writing t >-3, t≠0 or any of them will get extra points.
23. (Full score for this small question 10)
As shown in figure 13- 1 to figure 13-5, ⊙O all the rolls without sliding, and ⊙O 1, ⊙O2, ⊙O3 and ⊙O4 all represent .
Reading comprehension:
(1) As shown in figure 13- 1, ⊙O starts at ⊙O 1 and scrolls along AB to ⊙O2. When AB = c, ⊙O just rotates 1 cycle.
(2) As shown in figure 13-2, the complementary angle adjacent to ∠ABC is n, ⊙O rolls along A-B-C outside ∠ABC, and must be rotated from ⊙O 1 to ⊙O2 at point B.
Practical application:
(1) In reading comprehension (1), if AB = 2c, then ⊙O rotates and circulates; If AB = l, then ⊙O rotates. In reading comprehension (2), if ∠ ABC = 120, ⊙O rotates at point B; If ∠ ABC = 60, O rotates at point B. 。
(2) As shown in figure 13-3, ∠ABC = 90, AB = BC = C. ⊙O starts from ⊙O 1 and rolls along A-B-C to ∠ ABC ⊙O4.
Extended association:
(1) As shown in figure 13-4, the circumference of △ABC is l, and ⊙O starts from the position where point D is tangent to AB, and outside △ABC, it rolls clockwise along the triangle and returns to the position where point D is tangent to AB. How many times has ⊙O rotated? Please explain the reason.
(2) As shown in figure 13-5, the perimeter of the polygon is l, and ⊙O starts from the position where point D is tangent to an edge, then rolls clockwise along the polygon outside the polygon, and then returns to the position where point D is tangent to the edge, and directly writes the cycle number of ⊙O rotation.
Answer: Solution: Practical application
( 1)2; . ; .
(2) .
Extended association
The circumference of (1)∑△ABC is L, and ∴⊙O rotates in three planes.
The sum of the outer angles of a triangle is 360 degrees,
∴ At three vertices, ⊙O rotates (week).
∴⊙O*** rotated (+1) weeks.
(2) + 1.
24. (Full score for this small question 10)
In figure 14- 1 to 14-3, point b is the midpoint of AC line and point d is the midpoint of CE line. Quadrilateral BCGF and CDHN are both squares. The midpoint of AE is m.
(1) As shown in figure 14- 1, point E is on the extension line of AC, and when point N coincides with point G, point M coincides with point C,
Verification: FM = MH, FM ⊥ MH;
(2) Rotate the CE in figure 14- 1 clockwise by an acute angle around point C to get figure 14-2.
It is proved that △FMH is an isosceles right triangle;
(3) shorten the CE in figure 14-2 to the situation in figure 14-3,
△FMH or isosceles right triangle? (No need.
Explain the reason)
Answer: (1) Proof: ∵ Quadrilateral BCGF and CDHN are both squares,
In addition, point n coincides with point g, point m coincides with point c,
∴FB = BM = MG = MD = DH，∠FBM =∠MDH = 90。
∴△fbm≔△mdh。
∴FM = MH。
∠∠fmb =∠DMH = 45° ,∴∠fmh = 90°。 ∴FM⊥HM.
(2) Proof: Connect MB and MD, as shown in Figure 2, and let FM and AC intersect at point P. 。
∫B, D and M are the midpoint of AC, CE and AE respectively.
∴MD∥BC and md = bc = bf；; MB∑CD，
And MB = CD = DH.
Quadrilateral BCDM is a parallelogram.
∴ = Clean development mechanism.
And ∠FBP =∠HDC, ∴∠FBM =∠MDH. ..
∴△fbm≔△mdh。
∴FM = MH，
And ∠ MFB = ∠ HMD.
∴∠fmh =∠FMD-∠HMD =∠APM-∠mfb =∠FBP = 90。
∴△FMH is an isosceles right triangle.
(3) Yes.
25. (The full score of this short question is 12)
A company needs 240 A-type boards and 80 B-type boards/KLOC-0 for decoration. The specification of A-type plate is 60 cm×30 cm, and that of B-type plate is 40 cm×30 cm ... At present, only the standard plate with the specification of 150 cm×30 cm can be purchased. A standard plate can cut as many A-type plates and B-type plates as possible. * * There are the following three types.
Cut one, cut two, cut three
The number of blocks of type A plate is 1.20.
The number of blocks of B-plate is 2 m n.
We assume that all the purchased standard plates have been cut, including X-rays according to the first cutting method and Y-rays according to the second cutting method.
According to the cutting method, Z plate is cut, and A and B plates are just enough.
(1) In the above table, m =, n =；;
(2) The functional relationships between Y and X and between Z and X are obtained respectively;
(3) If Q is used to represent the number of purchased standard boards, find the functional relationship between Q and X,
It is also pointed out that when x takes what value, q is the smallest. At this time, the standard plate is cut according to three cutting methods.
How many/much?
Answer: Solution: (1) 0,3.
(2) From the meaning of the question, you can get
, ∴ .
,∴ .
(3) From the meaning of the question, it is concluded that.
Tidy it up and bring it here.
From the meaning of the question, get
The solution is x ≤ 90.
Note: In fact, 0≤x≤90, and x is an integer multiple of 6.
According to the properties of linear functions, when x = 90, q is the minimum.
At this time, 90 sheets, 75 sheets and 0 sheets are cut respectively according to three cutting methods.
26. (The full score of this short question is 12)
As shown in figure 16, in Rt△ABC, ∠ C = 90, AC = 3, AB = 5..P Point P starts from point C, moves along CA to point A at a constant speed of 1 unit per second, and immediately returns to point A at the original speed; Point Q starts from point A and moves along AB to point B at a constant speed of 1 unit per second. With the movement of P and Q, DE divides PQ vertically, and intersects PQ at point D. The intersection line QB-BC-CP starts from point E, and points P and Q move at the same time. When point Q reaches point B, it stops moving, and point P also stops moving. The moving time of points P and Q is set as t seconds.
(1) When t = 2, AP =, and the distance from Q to AC is;
(2) In the process of moving point P from C to A, find the functional relationship between the area s and t of △APQ; (Don't write the range of T)
(3) In the process of moving from B to C, can the quadrilateral QBED become a right-angled trapezoid? If yes, find the value of t, if not, please explain the reason.
(4) When DE passes through point C, please write the value of t directly.
Answer: Solution: (1) 1,;
(2) QF⊥AC at point F, as shown in Figure 3, AQ = CP= t, ∴.
By △AQF∽△ABC
Yes ∴ 。
∴ ,
Namely.
(3) Yes.
(1) When DE∑QB, as shown in Figure 4.
∴pq⊥qb ∵de⊥pq, quadrilateral QBED is a right trapezoid.
At this time ∠ aqp = 90.
From △APQ ∽△ABC
Which is the solution.
② as shown in fig. 5, when pq∑BC, DE⊥BC and quadrilateral QBED are right-angled trapezoid.
At this time ∠ apq = 90.
From △AQP ∽△ABC
Which is the solution.
(4) or.
Note: ① point p moves from c to a, and DE passes through point C.
Method 1: Connect QC and do QG⊥BC at G point, as shown in Figure 6.
, .
Gradually, gradually.
Method 2: from, from, and then from.
, um, ∴. ∴
② point p moves from a to c, and DE passes through point c, as shown in figure 7.
,