There are many on Baidu, and this one is from Guangdong.
20 13 Guangzhou junior high school graduates' mathematics examination paper
The first part of the multiple-choice questions (***30 points)
First,? Multiple choice question:
1. The number greater than 0 is ()
A.- 1 ? B? C.0 D. 1
2. The main view of the geometry as shown in the figure is (? )
A.？ BC? D.
3. In a 6×6 grid, the translated position of the number n in Figure ① is as shown in Figure ②, and the correct translation method of the number n is ().
A. Move down 1 grid? B. Move up 1 grid C. Move up 2 grids? Move down 2 squares.
4. Calculation: (M3N) The result of 2 is ()
A.m6n？ B.m6n2 C.m5n2？ D.m3n2
5. In order to know the main channels for middle school students to get information, set five options (A: newspaper, B: TV, C: network, D: people around them, E: others)? Questionnaire survey: First, 50 middle school students were randomly selected for questionnaire survey, and the questionnaire was drawn according to the survey results. The crossing chart is shown in the figure, and the survey method is (? ), the value of a in the figure is ()
A. comprehensive investigation, 26 b. comprehensive investigation, 24? C. sampling survey, 26? D. Sample survey, 24
6. It is known that the sum of two numbers x and y is 10, and x is 2 times larger than y, then the following equation is correct ().
A.B. C. D。
7. The position of the real number A on the number axis is shown in the figure, so = (? )
Canadian Broadcasting Corporation? D.
8. If the algebraic expression is meaningful, the value range of the real number x is (? )
A.x ≠ 1b.x ≥ 0c.x > 0d.x ≥ 0 and x≠ 1.
9. If 5k+20 < 0, the root of quadratic equation x2+4x-k = 0 is (? )
A. there is no real number root B. There are two equal real roots.
C. there are two unequal real roots. D. it is impossible to judge
10. As shown in the figure, the quadrilateral ABCD is trapezoidal, ad∨BC, CA is the bisector of ∠BCD, AB⊥AC, AB=4, AD=6, then tanB=? (? )
A. BC? d？
2. Fill in the blanks (6 small questions in this big question, 3 points for each small question, full score 18 points)
1 1. Point P is on the vertical line of line segment AB, and PA=7, then Pb = _ _ _ _ _ _ _ _ _ _
12. A charity in Guangzhou * * * raised 5.25 million yuan, which was expressed by scientific notation as _ _ _ _ _ _ _.
13. Decomposition factor: x2+xy = _ _ _ _ _ _ _ _ _ _ _ _
14. The linear function y = (m+2) x+ 1. If y increases with the increase of x, the range of m is _ _ _ _ _ _ _? .
15. As shown in the figure, the hypotenuse AB of Rt△ABC is equal to16, and Rt△ABC rotates clockwise around point O to get RT △ A ′ B ′ C ′, then the length of the median line C ′ D on the hypotenuse A ′ B ′ C ′ of RT △ A ′ B ′ is _ _ _ _ _ _ _ _
16. As shown in the figure, in the plane rectangular coordinate system, point O is the coordinate origin, point P is in the first quadrant, ⊙P and X axis intersect at points O and A, the coordinate of point A is (6,0), and the radius ⊙P is, then the coordinate of point P is _ _ _ _ _ _ _ _ _ _ _ _ _.
3. Solution (this big topic is ***9 small question, full mark 102, and the solution should be written as proof process or calculus steps)
17. (Full score for this small question)
Solve the equation: x2- 10x+9 = 0.
18. (Full score for this small question)
As shown in the figure, the quadrilateral ABCD is a diamond, and the diagonal AC and BD intersect at O, AB=5, AO=4. Find the length of BD.
19. (Full score for this small question 10)
Simplify first, then evaluate:, where
20. (Full score for this small question 10)
The known quadrilateral ABCD is a parallelogram (as shown in the figure). Fold △ABD along diagonal BD 180 to get △ a ′ BD.
(1) Use a ruler to make △ a ′ bd. (it is required to keep the traces from writing);
(2) Let Da ′ and BC meet at point E, and prove: △ Ba ′ e △ DCE.
2 1. (The full score of this small question is 12)
In a survey on the number of daily hair in Weibo for young people aged 18 ~ 35, it is now random to assume that a person's "average number of Weibo hair per day" is m, and it is stipulated that when m≥ 10, it is grade A, and when 5 ≤ m < 10, it is grade C.
1 1? 10? 6 15? 9? 16 13 120? eight
2 8 10? 17? 6? 13 7? 5? 7? three
12? 10? 7 1 1? 36 8? 14 15 12
(1) Find the frequency of class A in the sample data;
(2) Try to estimate the young people aged 1 10,000 18 ~ 35, and their "daily average number of articles in Weibo" is Grade A;
(3) Randomly select two people from the crowd whose sample data is Grade C, and get the probability that the average daily Weibo number of two people is 3 through enumeration.
22. (The full score of this short question is 12)
As shown in the figure, there are two ships, A and B, on the east-west coastline MN, both of which have received the distress signal of ship P, and ship P has run aground on the rocks. It is known that P ship is 58 northeast of A ship, P ship is 35 northwest of B ship, and AP distance is 30 nautical miles.
(1) Find the distance MN from the ship P to the coastline (accurate to 0. 1 nautical mile);
(2) If Ship A and Ship B set out at the same time at the speed of 20 knots and 15 knots respectively, and went to the rescue in a straight line at a uniform speed, try to determine which ship arrived at Ship P first by calculation.
23. (The full score of this short question is 12)
As shown in the figure, in the plane rectangular coordinate system, the point O is the coordinate origin, the sides OA and OC of the square OABC are on the X axis and the Y axis respectively, the coordinate of the point B is (2,2), and the image of the inverse proportional function (x > 0, k≠0) passes through the midpoint D of the line segment BC.
(1) Find the value of k;
(2) If the point P(x, y) moves on the image of the inverse proportional function (which does not coincide with the point d), the passing point p is defined as the PR⊥y axis at the point R, the PQ⊥BC is defined as the straight line at the point Q, and the area of the quadrilateral CQPR is defined as S, so as to obtain the analytical expression of S about X and write the value range of X. 。
24. (The full score of this short question is 14)
It is known that AB is the diameter of ⊙O, AB=4, point C moves on the extension line of AB, point D moves on ⊙O (not coincident with point B), and connects CD, and CD = OA.
(1) When OC= (as shown in the figure), it is proved that CD is tangent to ⊙O;
(2) When oc >, the straight line where CD is located intersects ⊙O, and the other intersection point is E, connecting AE.
① when d is the midpoint of CE, find the circumference of △ACE;
② Connect OD, there is a quadrangle. Is AODE a trapezoid? If yes, please specify the number of trapezoid, and then find AE. The value of ED; If it does not exist, please explain why.
25. (The full score of this short question is 14)
It is known that the parabola Y 1 = AX2+BX+C (A ≠ 0, a≠c) passes through point a (1, 0), and the vertex is b, and the parabola does not pass through the third quadrant.
(1) use a and c to represent b:
(2) Determine the quadrant where point B is located and explain the reasons;
(3) If the straight line Y2 = 2x+m passes through point B and the parabola intersects with another point C (), find the value range of y 1 when x≥ 1.
Reference answer
Comparison of rational numbers of 1. test sites.
The analysis of numbers greater than 0 must be positive.
Answer d
2. Simple combination of three views of test sites.
It is found that the graphics obtained from the front view can be analyzed. Note that all the edges you see should be displayed in the front view.
Answer a
Annotations on the main view are views seen from the front of the object.
3. Translation phenomenon in the life of test center.
Through the analysis and observation of the graph, we can know that the graph n can be moved down 2 squares from the graph 1 to the graph 2.
Answer d
Comment, observe and compare the positions of characters before and after translation, and get the translation rules.
4. The strength of the test center and the strength of the products.
The analysis is calculated according to the nature of power sum product, (M3N)2 = M6·N2.
Answer b
5. Bar chart of test sites; Comprehensive survey and sampling survey.
According to the key sentence "First, randomly select 50 middle school students to conduct a questionnaire survey", it can be concluded that the survey method is sampling survey, and the sample size is 50, so 6+ 10+6+A+4 = 50.
Answer d
Comments get the necessary information from different statistical charts to solve problems.
6. Test sites abstract binary linear equations from practical problems.
Analysis of equivalence relation: the sum of x and y is10; X is three times bigger than y, 2.
Answer c
7. Test the real number and axis of the center.
The analysis is as follows: A < 2.5, that is, A-2.5 < 0, then | a-2.5 | =-(a-2.5) = 2.5-a.
Answer b
Comment on any two numbers on the number axis. The number on the right is always greater than the number on the left.
8. The meaningful conditions for testing the secondary roots of the center; Conditions for meaningful scores.
According to the nature of quadratic roots and the meaning of fractions, if the number of roots is greater than or equal to 0 and the denominator is not equal to 0, the range of X can be found.
Answer x≥0 and x≠ 1
Annotation score is meaningful, and denominator is not 0; The square root of quadratic form is nonnegative.
9. The discriminant formula of the root of a quadratic equation with one variable.
If ∵ 5k+20 < 0, that is, K.
Answer a
10. Test center trapezoid; Determination and properties of isosceles triangle: Pythagorean theorem: triangle midline theorem.
First, judge that DA=DC, intersection D is DE∑AB, intersection AC is at point F, and intersection BC is at point E. From the properties of isosceles triangle, it can be concluded that point F is the midpoint of AC and the middle line of EF is △CAB, then the lengths of EF and DF can be obtained, then AF in Rt△ADF can be obtained, and then AC can be obtained, and the value of tanB can be calculated.
∵CA is the bisector of ∠BCD, ∴ DCA = ∠ ACB, and ∵ad∨BC, ∴ ACB = ∠ CAD, ∴∠ DAC =
∵AB⊥AC, ∴DE⊥AC (the nature of three lines in an isosceles triangle), ∴ Point F is the midpoint of AC, ∴AF=CF, ∴EF is the center line of △CAB, ∴EF=AB=2, ∨.
Answer b
The key to answer this question is to make an auxiliary line, and the judgment point F is the communication midpoint.
1 1. Test the properties of the vertical line in the center line segment.
According to the nature of the median vertical line, PA = Pb is obtained.
Solution 7
The distance between the point on the perpendicular of the line segment and the two endpoints of the line segment is equal.
12. Scientific notation of test sites (indicating large numbers).
The analysis shows that 5250000 is expressed as 5.25× 106 by scientific notation.
Solution 5.25× 106
Annotate scientific notation in the form of a× 10n, where 1 ≤| a | < 10, n is an integer. When determining the value of n, it depends on how many digits the decimal point moves when the original number becomes a, and the absolute value of n is the same as the number of digits the decimal point moves. When the absolute value of the original number is greater than 1, and when the absolute value of the original number is less than 1, n is negative.
13. factorization of test sites (common factor method).
Analysis x2+xy = x (x+y).
Answer x (x(x+y)
The steps of factorization are as follows: firstly, the common factor is extracted; Second, look at the formula. Generally speaking, if we can extract the common factor, we should first extract the common factor, and then see if the remaining factors can be decomposed.
14. Relationship between linear function image and coefficient.
Analytic ∵ linear function y = (m+2) x+ 1, if y increases with the increase of x, ∴ m+2 > 0, then the solution is m >-2.
Answer m >-2
Comment on the relationship between the image of a function and the coefficient: Does the function value y decrease with the increase of X? k < 0； The function value y increases with the increase of x? k>0。
15. The nature of the rotation of the test center; The centerline of the hypotenuse of a right triangle.
After analysis, ∫rt△ABC rotates clockwise around point O, RT △ A ′ B ′ C ′, ∴ A ′ B ′ = AB =16, ∫ C ′ D is the median line on the hypotenuse A ′ B ′ of RT △ A ′ B ′ C ′, ∴
Answer 8
Comments on the essence of rotation: the two figures are identical before and after rotation; The distance from the corresponding point to the rotation center is equal; The included angle between the corresponding point and the rotation center is equal to the rotation angle, and the properties of the midline on the hypotenuse of the right triangle are also investigated.
16. The vertical diameter theorem of the test center; Coordinate and graphic attributes; Pythagorean theorem
After analysis, point P is connected with OP as the PD⊥x axis of point D,
∫a(6，0)，PD⊥OA，∴ OD = OA = 3。 At Rt△OPD, op =, OD=3, ∴PD===2, ∴ P (3 3,2).
Solutions (3, 2)
Comment on the auxiliary line according to the meaning of the question and construct the right triangle solution.
17. Try to solve the quadratic equation of one variable (factorization method).
Through the analysis of factorization, two linear equations are obtained and their solutions are obtained.
Solution: ∫x2- 10x+9 = 0,
(x- 1)(x-9)=0，
X- 1 = 0 or X-9 = 0,
∴x 1= 1,x2=9.
Comment on factorization method to solve a quadratic equation with one variable, the key is to transform solving a quadratic equation with one variable into solving a quadratic equation with one variable.
18. properties of diamonds in the testing center; Pythagorean theorem
AC⊥BD is calculated according to the nature of diamond, and then the length of BO is calculated by Pythagorean theorem, and the answer is obtained.
Solution: ∫ quadrilateral ABCD is a diamond, diagonal AC and BD intersect at O,
∴AC⊥BD,DO=BO.
∵AB=5,AO=4,∴BO==3，
∴BD=2BO=2×3=6.
Comment on the nature of diamond and the application of Pythagorean theorem, and draw the conclusion that the length of BO is the key to solving problems.
19. Simplified evaluation of test center score: simplified calculation of quadratic root.
Denominator remains the same, numerator is subtracted, which simplifies the evaluation of future generations.
Solution: = = x+y,
The original formula at that time was = 1+2+ 1-2 = 2.
20. Test the properties of the central parallelogram; Congruent triangles's judgment; Stretch-axisymmetric transformation; Folding transformation (folding problem)
Analysis (1) firstly works as ∠ a ′ BD = ∠ △A'BD, and then draws an arc with B as the center and AB length as the radius, and intersects BA ′ at point A ′ to connect BA ′ and DA ′, and works as △ a ′ BD.
(2) From the fact that the quadrilateral A ′ b = CD is a parallelogram and folded, it is easy to prove that ∠ ba ′ d = ∠ c and A ′ b = CD, and then judge △ ba ′ e △ DCE by AAS.
Solution: (1) as shown in the figure: ① For ∠ a ′ BD = ∠ Abd,
(2) Draw an arc with B as the center and AB as the radius, and intersect Ba' at point A'.
③ Connect Ba' and Da',
Then △ a ′ BD is the demand;
(2)∵ quadrilateral ABCD is a parallelogram, ∴AB=CD, ∠BAD=∠C,
From the nature of folding, we can get: ∠BA'D=∠BAD, A'B=AB, ∴ ba'd = ∠ c, a 'b = cd.
In △ ba ′ e and △DCE, ∠ ba ′ e = ∠ c, ∠ ba ′ e = ∠ c, a ′ b = CD,
∴△ba′e≌△dce(aas).
Comments pay attention to the corresponding relationship before and after graphic folding and the application of the idea of combining numbers with shapes.
2 1. Test site list method and tree diagram method; Estimate the population with samples; Frequency and frequency.
Analysis (1): Among 30 young people who meet the age conditions, 15 is Grade A, and the frequency of Grade A in the sample data is obtained;
(2) According to the meaning of the question, among 100018-35-year-old young people, the number of Weibo articles per day is1000× = 500;
(3) Draw a tree diagram according to the meaning of the question, then draw all equally possible results from the tree diagram and the situation that the average number of Weibo sent by two people is 3, and then use the probability formula to get the answer.
Solution: (1)∵ Among the 30 young people who meet the age conditions, 15 is Grade A, and the frequency of Grade A in the sample data is =;
(2) Among the 65,438+0,000 young people aged 65,438+08 ~ 35, the number of Weibo articles per day is 65,438+0,000× = 500;
(3) There are four people in Grade C: 0, 2, 3, 3. Draw a tree diagram:
∫ * * There are 12 possible results, and there are two cases where the "daily average Weibo number" of two people is 3.
∴ The probability that the "daily average Weibo number" of two people is 3 is =.
Annotation list method or tree diagram method can list all possible results without repetition or omission. List method is suitable for events completed in two steps, and tree diagram method is suitable for events completed in two or more steps.
22. Try to solve the application of right triangle (direction angle problem).
Analyze PE⊥AB with (1) intersection point P as point E, and solve PE in Rt△APE;
(2) At Rt△BPF, calculate BP, calculate the time required for two ships respectively, and make a judgment.
Solution: (1) PE⊥AB whose intersection point P is E.
From the meaning of the question, PAE = 32, AP=30 nautical miles,
In Rt△APE, PE = apsin ∠ PAE = apsin32 ≈15.9 nautical mile;
(2) At Rt△PBE, PE = 15.9 nm, ∠ PBE = 55,
Then BP = ≈ 19.4,
∴ Time required for ship A = 1.5 hours, and time required for ship B = 1.3 hours.
So ship B arrived first.
23. Test center inverse proportional function synthesis questions.
Analysis (1) First, get the coordinates of point C according to the meaning of the question, then get the coordinates of point D according to the formula of midpoint coordinates. The image of inverse proportional function Y = (x > 0, k≠0) passes through the midpoint d of BC line, and then substitute the coordinates of point D into the analytical formula to get k;
(2) The solution is divided into two steps. (1) when d is above the straight line BC, that is, 0 < x < 1, as shown in figure 1, according to the s quadrilateral CQPR=CQ? PD lists the analytical formula of S about X. ② When D is below the straight line BC, that is, X > 1, as shown in Figure 2, it is still based on S quadrilateral CQPR=CQ? PD lists the analytical formula of s about x.
Solution: (1)∫ The sides OA and OC of the square OABC are on the X axis and Y axis respectively, and the coordinate of point B is (2,2).
∴ c (0 0,2), ∫d is the midpoint of BC, ∴ d (1, 2).
The image of inverse proportional function y = (x > 0, k≠0) passes through point d,
∴k=2；
(2) When d is above the straight line BC, it is 0 < x < 1.
As shown in figure 1, ∵ point P(x, y) moves on the image of inverse proportional function, ∴y=,
∴S quadrilateral CQPR=CQ? PD=x？ (-2)= 2-2x(0 < x < 1)；
As shown in Figure 2, similarly, we can find that the S quadrilateral CQPR=CQ? PD=x？ (2-)=2x-2(x> 1)。
To sum up, s =
Comments focus on solutions (2). Analytic function needs to be solved by segments.
24. Test center circle comprehensive questions.
The key of analysis (1) is to determine that △OCD is a right triangle by using the inverse theorem of Pythagorean theorem, as shown in Figure ①.
(2)① As shown in Figure ②, the key is to determine that △EOC is a right triangle with an angle of 30 degrees, so as to solve the right triangle and find the perimeter of △ACE;
② There are two trapezoids that meet the meaning of the question, and Figure ③ shows one of them. When calculating the ED value, similar triangles is skillfully used to draw a simple conclusion and avoid complicated operation.
Proof of solution (1): connected with OD, as shown in Figure ①.
According to the meaning of the question, CD=OD=OA=AB=2, OC=2, ∴ OD2+Cd2 = OC2,
According to the inverse theorem of Pythagorean theorem, if △OCD is a right triangle, then OD⊥CD,
And ∵ point d is on ⊙O,
∴CD is the tangent of⊙ O.
(2) Solution: ① Connect OE, OD,
Then CD=DE=OD=OE, ∴△ODE is an equilateral triangle, ∠ 1 = ∠ 2 = ∠ 3 = 60.
∵OD=CD,∴∠4=∠5，
∵∠3=∠4+∠5,∴∠4=∠5=30 ,
∴∠EOC=∠2+∠4=90，
Therefore, △EOC is a right triangle with an angle of 30 degrees, and △AOE is an isosceles right triangle.
In Rt△EOC, CE=2OA=4, OC=4cos30 =2 2,
In the isosceles right triangle AOE, AE=OA=2,
The circumference of ace is AE+CE+AC = AE+CE+(OA+OC) = 2+4+(2+2) = 6+2+2.
② There are two such trapezoids.
Figure ③ shows that point D is above AB. Similarly, there is a trapezoid below AB, and they are symmetrical about the straight line AB.
∵OA=OE,∴∠ 1=∠2，
∵CD=OA=OD,∴∠4=∠5.
∵ Quadrilateral OD is a trapezoid, ∴OD∥AE, ∴∠4=∠ 1, ∠3=∠2,
∴∠3=∠5=∠ 1.
In △ODE and △COE, ∠OEC=∠OEC, ∠3=∠5,
∴△ Ode to ∴△ Ke, and then =, ∴CE? DE=OE2=22=4。
∵∠ 1=∠5,∴AE=CE,∴AE？ DE=CE？ DE=4。
In a word, there are two kinds of trapezoidal AODE, AE? DE=4。
25. Test site quadratic function synthesis questions.
(1) parabola is analyzed by a (1, 0), and the points can be substituted into the function to get b =-a-c;
(2) To determine which quadrant the point is in, you need to draw it according to the meaning of the question. The conditions are as follows: the image can be pushed up without going through the third quadrant, a > 0, and it can be solved only by knowing how many intersections the parabola has with the X axis. Judging that there are two intersections with the X-axis, delta can be considered, so it can be judged that there are two intersections with the X-axis, so it is in the fourth quadrant; Or directly use the formula method (or cross multiplication) to calculate, and from two different solutions x 1= 1, x2=, (a≠c), get the quadrant where point B is located;
(3) When x≥ 1, the range of y 1 is clear as long as drawing. The difficulty lies in observing that c (,b+8) is another intersection of parabola and X axis, because x 1= 1, x2=, (a≦. It can be determined that the straight line passes through points B and C. Looking at the image, it can be concluded that when x≥ 1, y 1 is greater than or equal to the minimum value. At this time, the minimum value of quadratic function is calculated, that is, B =-8 and A+C b=-8 are found, and A and C are calculated. Then, find a formula related to A and C, solve the equations, and find A and C..
Solution: (1)∵ parabola Y 1 = AX2+BX+C (A ≠ 0, a≠c), after A (1, 0), substitute the point into the function to get b =-a-c;
(2)B is in the fourth quadrant. The reason for this is the following：
∵ parabola y 1 = ax2+bx+c (a ≠ 0, a≠c) passes through point a (1, 0), ∴x 1= 1, x2=
So the parabola has two intersections with the X axis, because the parabola does not pass through the third quadrant.
So a > 0, the vertex is in the fourth quadrant;
(3)∵C (,b+8), and on the parabola, ∴ b+8 = 0, ∴ b =-8,
∵a+c=-b,∴a+c=8.
Substituting B and C into the analytical formula of straight line c-a=4, it is easy to get C-A = 4, that is, to get the solution.
As shown in the picture, C is to the right of A,
When x≥ 1, y 1 ≥- 2.
Comment on the application of the idea of combining numbers with shapes.