1. In the fraction, the value range of x is ().
A.x≠ 1 b . x≠0 c . x & gt； 1d . x & lt； 1
2. In the following four signs: recycling, green food, energy saving and water saving, the axis symmetry is ().
A.B. C. D。
3. It is known that α and β are two roots of the unary quadratic equation x2-2x-3 = 0, so the value of α+β is ().
A.2b ﹣2 c 3d ﹣3
4. As shown in the figure, the image of inverse proportional function y= passes through point A, which is perpendicular to X axis and Y axis respectively, and the vertical feet are B and C. If the area of right-angle ABOC is 2, the value of k is ().
A.4 B. 2 C. 1 D。
5. As shown in the figure, in ABCD, diagonal AC and BD intersect at point O, and E is the midpoint of CD, connecting OE. If OE=3cm, the length of AD is ().
A.3cm wide 6cm high 9cm deep12cm
6. The equation x2+6x-5 = 0 is completely squared, and the equation is ().
A.(x+3)2 = 14 b .(x﹣3)2= 14)
7. Every inner corner of a polygon is 108, so this polygon is ().
A. pentagon hexagon heptagon octagon
8. The solution of the fractional equation is ()
A.x=﹣5 B. x=5 C. x=﹣3 D. x=3
9. As shown in the figure, in the diamond-shaped ABCD, it is known that ∠ D = 1 10, then ∠BAC has a degree of ().
A.30 B. 35 C. 40 D. 45
10. If the unary quadratic equation kx2-6x+9 = 0 about x has two unequal real roots, then the value range of k is ().
A.k< 1 and k≠0 B. k≠0 C. k 1
1 1. The following figures are all composed of squares with an area of 1 according to certain rules, in which the (1) th figure has 9 squares with an area of 1, and the (2 nd) figure has 14 squares with an area of/
A.72 B. 64 C. 54 D. 50
12. It is known that the quadrilateral OABC is a rectangle with the side OA on the X axis and the side OC on the Y axis. The hyperbola intersects the side BC at point D and the diagonal OB at midpoint E. If the area of △OBD is 10, the value of k is ().
A. 10 5th century BC.
2. Fill in patiently (this big question is 6 small questions, with 4 points for each small question and 24 points for each small question). Please fill in the correct answer to each question in the table below.
13. Decomposition factor: 2m2-2 =.
14. If the value of the score is zero, then x=.
15. As shown in the figure, in the rectangular ABCD, diagonal AC and BD intersect at point O, AB=4, ∠ AOD = 120, then the length of diagonal AC is.
16. It is known that x=2 is the root of the equation x2+mx+2=0, then the value of m is.
17. Due to the hot weather, a school fumigated the classroom to prevent mosquito bites according to the school hygiene regulations. It is known that the relationship between the drug content y (mg) per cubic meter in indoor air and the combustion time x (min) is as shown in the figure (that is, the point A in the figure and the part of the line segment OA and hyperbola on its right).
18. As shown in the figure, in the square ABCD, AB=2, rotate clockwise around point A ∠BADα(0
3. When answering questions (this big question ***4 small questions, 19 questions 10 points, 20 questions 8 points, 2/kloc-0 questions 8 points, 22 questions 8 points, ***34 points), each small question must give the necessary calculus process or reasoning steps.
19. Solve the equation:
( 1)x2﹣6x﹣2=0
(2)=+ 1.
20. As shown in the figure, in ABCD, the bisector BE of ∠ABD intersects with AD at point E, and the bisector DF of ∠CDB intersects with BC at point F to connect BD.
(1) verification: △ Abe △ CDF;
(2) If AB=DB, it is proved that the quadrilateral DFBE is a rectangle.
2 1. As shown in the figure, the image of linear function y=kx+b(k≠0) passes through point P (0) and intersects with the image of inverse proportional function y=(m≠0) at point A (2, 1).
(1) Find the analytical expressions of linear function and inverse proportional function;
(2) Find the coordinates of point B and answer according to the image: When the value of x is in what range, the function value of linear function is smaller than that of inverse proportional function?
22. In clothing sales, children's clothing stores found that a child's clothing with a purchase price of 100 yuan can sell an average of 20 pieces a day. In order to meet the "June 1", children's wear store decided to take appropriate measures to reduce prices, expand sales and increase profits. After investigation, it is found that if the price of each piece of children's clothing is reduced by 1 yuan, two pieces can be sold on average every day.
(1) What is the daily profit of the children's wear store before the price reduction?
(2) If the children's clothing store sells this kind of children's clothing every day, and the profit is 1200 yuan, and at the same time, it wants customers to get more benefits, how much should the price of each child's clothing be reduced?
Four, answer the question (this big question ***2 small questions, each small question 10 points, ***20 points), each small question must give the necessary calculus process or reasoning steps.
23. Simplify first, then evaluate: (﹣) ﹣ (﹣1), where A is the solution of the equation a2﹣4a+2=0.
24. in the plane rectangular coordinate system xOy, the "extraordinary distance" between any two points P 1(x 1, y 1) and P2(x2, y2) is defined as follows:
If it is | x1-x2 |≥| y1-y2 |, the "extraordinary distance" between point P 1 and point P2 is | x1-x2 |;
If | x1| x2 | <| y1| y2 |, then the "extraordinary distance" between point P 1 and point P2 is | y1| y2 |.
For example: point P 1 (1, 2) and point P 1 (3,5). Because | 1 | < | 2-5 |, the "extraordinary distance" between point p 1 and point P2 is |.
(1) Point A(﹣) is known, and B is the moving point on the Y axis. ① If the "extraordinary distance" between point A and point B is 2, write the coordinates of point B that meet the conditions; (2) Write the minimum value of "extraordinary distance" between point A and point B directly;
(2) As shown in Figure 2, it is known that C is the moving point on a straight line, and the coordinate of point D is (0, 1). Find the coordinates of point C when the "extraordinary distance" between point C and point D is minimum.
5. When answering questions (this big question ***2 small questions, 25 questions 12 points, 26 questions 12 points, ***24 points), each small question must give the necessary calculus process or reasoning steps.
25. As shown in the figure, in the diamond-shaped ABCD, ∠ ABC = 60, e is any point on the diagonal AC, f is a point on the extension line of line segment BC, and CF=AE connects BE and EF.
(1) As shown in figure 1, when E is the midpoint of line segment AC and AB=2, find the area of △ABC;
(2) As shown in Figure 2, when point E is not the midpoint of line segment AC, verify that BE = EF.
(3) As shown in Figure 3, does the conclusion in (2) hold when point E is any point on the extension line of line segment AC? If yes, please give proof; If not, please explain why.
26. As shown in the figure, it is known that point A is a straight line y=2x+ 1 and an inverse proportional function y =(x >；; 0) The intersection of images, and the abscissa of point A is 1.
(1) Find the value of k;
② As shown in figure 1, hyperbola y =(x >；; 0) point m, if S△AOM=4, find the coordinates of point m;
(3) As shown in Figure 2, if the inverse proportional function y =(x >；; 0) point b (3, 1) on the image, point p is the moving point on the straight line y=x, and point q is the inverse proportional function y = (x >; 0) Is there a parallelogram with vertices p, a, b and q at another point on the image? If it exists, please write the coordinates of Q point directly; If it does not exist, please explain why.
Reference answers and analysis of test questions
Choose carefully: (There are 12 small questions in this big question, with 4 points for each small question and * * 48 points) Four answers with code names A, B, C and D are given under each small question, and only one is correct. Please fill in the code of the correct answer in the form.
1. In the fraction, the value range of x is ().
A.x≠ 1 b . x≠0 c . x & gt； 1d . x & lt； 1
Test center: conditions for meaningful scores.
Analysis: If the score is meaningful and the denominator is not equal to 0, the solution can be obtained.
Answer: solution: from the meaning of the question, x- 1 ≠ 0,
The solution is x≠ 1.
So choose a.
Comments: This question examines the meaningful conditions of the score from the following three aspects, and thoroughly understands the concept of the score:
(1) The score is meaningless and the denominator is zero;
(2) The meaningful denominator of the score is not zero;
(3) The fractional value is zero, the numerator is zero, and the denominator is not zero.
2. In the following four signs: recycling, green food, energy saving and water saving, the axis symmetry is ().
A.B. C. D。
Examination center: Axisymmetric graphics.
Analysis: According to the concept of axisymmetric figure, the options are analyzed and judged by exclusion method.
Solution: Solution: A, it is not an axisymmetric figure, so this option is wrong;
B, is an axisymmetric figure, so this option is correct;
C, not axisymmetric graphics, so this option is wrong;
D is not an axisymmetric figure, so this option is wrong.
Therefore, choose; B.
Comments: This topic examines the concept of axisymmetric graphics. The key to an axisymmetric figure is to find the axis of symmetry, and the two parts of the figure can overlap after folding.
3. It is known that α and β are two roots of the unary quadratic equation x2-2x-3 = 0, so the value of α+β is ().
A.2b ﹣2 c 3d ﹣3
Test center: the relationship between root and coefficient.
Analysis: According to the relationship between root and coefficient, α+β =-= 2, the answer can be obtained.
Solution: ∵ α and β are two roots of the unary quadratic equation x2-2x-3 = 0.
∴α+β=﹣=2;
So choose a.
Comments: This question examines the relationship between roots and coefficients: if x 1 and x2 are two roots of the unary quadratic equation ax2+bx+c=0(a≠0), x 1+x2=﹣, X65438+X2 =.
4. As shown in the figure, the image of inverse proportional function y= passes through point A, which is perpendicular to X axis and Y axis respectively, and the vertical feet are B and C. If the area of right-angle ABOC is 2, the value of k is ().
A.4 B. 2 C. 1 D。
Test site: the geometric meaning of the inverse proportional function coefficient K.
Analysis: Let the coordinates of point A be (x, y), and use x and y to represent the lengths of OB and AB. According to the area of rectangle ABOC is 2, the value of k is listed by formula.
Solution: Solution: Let the coordinate of point A be (x, y).
Then OB=x, AB=y,
The area of the rectangular ABOC is 2,
∴k=xy=2，
Therefore, choose: B.
Comments: This question examines the geometric meaning of the inverse proportional function coefficient k. If any point on the hyperbola is perpendicular to the two coordinate axes, the rectangular area surrounded by the coordinate axes is equal to |k|.
5. As shown in the figure, in ABCD, diagonal AC and BD intersect at point O, and E is the midpoint of CD, connecting OE. If OE=3cm, the length of AD is ().
A.3cm wide 6cm high 9cm deep12cm
Test site: triangle midline theorem; Properties of parallelogram.
Analysis: From the properties of parallelogram, it is easy to prove that OE is the midline and solve it according to the midline theorem.
Solution: According to the basic properties of parallelogram, the diagonal of parallelogram is equally divided. It is known that point O is the midpoint of BD, so OE is the centerline of △BCD.
According to the midline theorem, AD=2OE=2×3=6(cm).
So choose B.
Comments: This paper mainly investigates the basic properties and midline properties of parallelogram, and uses properties to solve problems. The basic properties of parallelogram are as follows: ① two groups of opposite sides of parallelogram are parallel respectively; ② Two groups of opposite sides of parallelogram are equal respectively; ③ The two diagonal lines of parallelogram are equal respectively; Diagonal bisection of parallelogram.
6. The equation x2+6x-5 = 0 is completely squared, and the equation is ().
A.(x+3)2 = 14 b .(x﹣3)2= 14)
Test center: Solve quadratic equation with unary matching method.
Title: Matching method.
Analysis: General steps of matching method:
(1) Move the constant term to the right of the equal sign;
(2) Convert the coefficient of quadratic term into1;
(3) The square of half of the coefficient of the first term is added to both sides of the equation.
Solution: Solution: By moving the terms in the original equation, we get
x2+6x=5，
At the same time, add the square of half the coefficient of a term on both sides of the equation, that is, 32, and you get
x2+6x+9=5+9，
∴(x+3)2= 14.
So choose a.
Comments: This question examines the matching method for solving quadratic equations in one variable. Pay attention to the accurate application of problem-solving steps when solving problems. When choosing the matching method to solve a quadratic equation with one variable, it is best to make the coefficient of the quadratic term of the equation 1 and the coefficient of the primary term a multiple of 2.
7. Every inner corner of a polygon is 108, so this polygon is ().
A. pentagon hexagon heptagon octagon
Test center: polygon inner corner and outer corner.
Analysis: By using the sum of the internal angles of the polygon =180 (n ~ 2), we can get
Solution: Solution: 108 = 180 (n-2) ÷ n
The solution is n=5.
So choose a.
Comments: This topic mainly examines the interior angles and theorems of polygons.
8. The solution of the fractional equation is ()
A.x=﹣5 B. x=5 C. x=﹣3 D. x=3
Test center: Solve fractional equation.
Special topic: calculation problems.
Analysis: The simplest common denominator is (x+ 1) (x- 1). Multiplying both sides of the equation by the simplest common denominator, the fractional equation can be transformed into an integral equation and then solved.
Solution: Solution: Multiply both sides of the equation by (x+ 1) (x- 1).
Get 3 (x+ 1) = 2 (x- 1),
The solution is x =-5.
It is proved that x=﹣5 is the solution of the original equation.
So choose a.
Comments: (1) The basic idea of solving fractional equation is "transformation idea", which transforms fractional equation into integral equation.
(2) When solving the fractional equation, we must pay attention to the root.
9. As shown in the figure, in the diamond-shaped ABCD, it is known that ∠ D = 1 10, then ∠BAC has a degree of ().
A.30 B. 35 C. 40 D. 45
Testing center: the nature of diamonds.
Special topic: calculation problems.
Analysis: First, according to the nature that the opposite sides of the diamond are parallel and the straight lines are parallel, we get ∠ bad = 70, and then divide a group of diagonals according to each diagonal of the diamond to solve it.
Solution: Solution: ∵ The quadrilateral ABCD is a diamond.
∴AD∥AB，
∴∠bad= 180 ﹣∠d= 180 ﹣ 1 10 = 70，
∫ The quadrilateral ABCD is a diamond,
∴AC split ∠ Not good,
∴∠BAC=∠BAD=35。
So choose B.
Comments: this question examines the nature of diamonds: diamonds have all the properties of parallelogram; All four sides of the diamond are equal; The two diagonals of the diamond are perpendicular to each other, and each diagonal bisects a set of diagonals; A diamond is an axisymmetric figure, which has two symmetrical axes, which are straight lines where two diagonal lines are located.
10. If the unary quadratic equation kx2-6x+9 = 0 about x has two unequal real roots, then the value range of k is ().
A.k< 1 and k≠0 B. k≠0 C. k 1
Test site: the discriminant of root; Definition of quadratic equation with one variable.
Special topic: calculation problems.
Analysis: according to the discriminant of root and the definition of quadratic equation in one variable, let △ >; 0 and the quadratic coefficient is not 0.
Solution: Solution: ∵ The quadratic equation KX2 with one variable about X6x+9 = 0 has two unequal real roots.
∴△>； 0,
That is, (-6) 2-4× 9k >; 0,
Solution, k
∫ is an unary quadratic equation,
∴k≠0，
∴k<； 1 and k≠0.
So choose a.
Comments: This topic examines the discriminant of roots and the definition of a quadratic equation. You should know: (1) △ > 0 equation has two unequal real roots;
(2) The △ = 0 equation has two equal real roots; (3)△& lt； Equation 0 has no real root.
1 1. The following figures are all composed of squares with an area of 1 according to certain rules, in which the (1) th figure has 9 squares with an area of 1, and the (2 nd) figure has 14 squares with an area of/
A.72 B. 64 C. 54 D. 50
Test center: general type: diversity of graphics.
Analysis: the 1 th graph has 9 small squares with side length of 1, the second graph has 9+5= 14 small squares with side length of 1, and the third graph has 9+5×2= 19 small squares with side length of/
Solution: Solution: 1 There are 9 small squares with the side length of 1 in the graph.
There are 9+5 = 14 small squares with a side length of 1 in the second graph.
There are 9+5× 2 = 19 small squares with the side length of 1 in the third figure.
…
There are 9+5× (n- 1) = 5n+4 small squares with sides of 1 in the nth graph.
Therefore, the number of small squares with a side length of 1 in the 10th graph is 5× 10+4=54.
So choose: C.
Comments: This topic examines the changing law of graphics, finds out the operation law between graphics and numbers, and uses the law to solve problems.
12. It is known that the quadrilateral OABC is a rectangle with the side OA on the X axis and the side OC on the Y axis. The hyperbola intersects the side BC at point D and the diagonal OB at midpoint E. If the area of △OBD is 10, the value of k is ().
A. 10 5th century BC.
Test site: the geometric meaning of the inverse proportional function coefficient K.
Analysis: Let the analytic formula of hyperbola be: y=, and the coordinate of point E be (x, y). According to the fact that E is the midpoint of OB, the coordinates of point B, point E, and k are calculated according to the area formula of triangle.
Solution: Let the analytic formula of hyperbola be: y=, and the coordinate of point E be (x, y).
E is the midpoint of OB,
∴ The coordinate of point B is (2x, 2y),
Then the coordinates of point D are (,2y),
The area of △OBD is 10,
∴×(2x﹣)×2y= 10，
Solution, k=,
Therefore, choose: d.
Comments: This question examines the geometric meaning of the inverse proportional coefficient K. If any point on the hyperbola is perpendicular to the two coordinates, the rectangular area surrounded by the coordinate axes is equal to |k|.
2. Fill in patiently (this big question is 6 small questions, with 4 points for each small question and 24 points for each small question). Please fill in the correct answer to each question in the table below.
13. Decomposition factor: 2m2-2 = 2 (m+ 1) (m- 1).
Test center: the comprehensive application of common factor method and formula method.
Topic: the finale.
Analysis: first extract the common factor 2, and then decompose the remaining polynomials with the square difference formula.
Answer: solution: 2m2 ~ 2,
=2(m2﹣ 1)，
=2(m+ 1)(m﹣ 1).
So the answer is: 2 (m+ 1) (m- 1).
Comments: This question examines the method of putting forward common factor and decomposition factor by formula. The key is to extract the common factor and continue to use the square difference formula for quadratic factorization.
14. If the score is zero, then x =-3.
Test center: the condition that the value of the score is zero.
Special topic: calculation problems.
Analysis: the value of the fraction is zero, the numerator is equal to 0, and the denominator is not 0.
Answer: Answer: According to the meaning of the question, you must
| x |-3 = 0 and x |-3 ≠ 0,
Solution, x =-3.
So the answer is: 3.
Comments: This question examines the condition that the value of the score is 0. If the score is 0, two conditions must be met at the same time: (1) molecule is 0; (2) The denominator is not 0. These two conditions are indispensable.
15. As shown in the figure, in the rectangular ABCD, diagonal AC and BD intersect at point O, AB=4, ∠ AOD = 120, and the length of diagonal AC is 8.
Test center: the nature of rectangle; A right triangle with an angle of 30 degrees.
Analysis: We can get OA=OB from the properties of rectangle, then prove that △AOB is an equilateral triangle, get OA=OB=AB=4, and get AC=2OA.
Solution: Solution: ∵ Quadrilateral ABCD is a rectangle,
∴OA=AC,OB=BD,AC=BD，
∴OA=OB，
∫∠AOD = 120，
∴∠AOB=60，
△ AOB is an equilateral triangle,
∴OA=OB=AB=4，
∴ac=2oa=8；
So the answer is: 8.
Comments: This question examines the nature of rectangle and the judgment and nature of equilateral triangle; It is the key to solve the problem to master the properties of rectangle and prove that triangle is equilateral triangle.
16. It is known that x=2 is the root of the equation x2+mx+2=0, so the value of m is ﹣3.
Test site: the solution of a quadratic equation
Analysis: Substituting x=2 into the equation can get an equation about m, and solving the equation can get the value of m. 。
Solution: Solution: Substitute x=2 into the equation to get: 4+2m+2=0,
The solution is m =-3.
So the answer is -3.
Comments: This topic mainly examines the definition of the solution of the equation, and transforms the problem of finding unknown coefficients into the problem of solving the equation.
17. Due to the hot weather, a school fumigated the classroom to prevent mosquito bites according to the school hygiene regulations. It is known that the relationship between the drug content y (mg) per cubic meter of indoor air and the combustion time x (min) is as shown in the figure (that is, the part of the line segment OA and the hyperbola at point A and its right side in the figure). When the drug is released from the burner,
Test center: the application of inverse proportional function.
Analysis: First of all, according to the meaning of the question, in the process of drug release, the drug content y (mg) per cubic meter of indoor air is directly proportional to the time x (min); After drug release, y is inversely proportional to x, and the data can be substituted into the inverse proportional function by undetermined coefficient method. Further solve the available answers.
Solution: Let the inverse proportional resolution function be y=(k≠0).
Substitute (25,6) into the analytical formula, k=25×6= 150,
The resolution function is y=(x≥ 15),
When y=2, =2,
The solution is x=75.
A: From the beginning of disinfection, teachers and students can't enter the classroom for at least 75 minutes.
Comments: This question examines the application of inverse proportional function. In real life, a large number of two variables form an inverse proportional function. The key to solve this kind of problem is to determine the functional relationship between two variables, and then use the undetermined coefficient method to find their relationship.
18. As shown in the figure, in the square ABCD, AB=2, rotate clockwise around point A ∠BADα(0
Test center: the nature of rotation; The nature of a square.
Analysis: firstly, we get ∠EAB=∠FAD=α according to the nature of rotation, then we get AB=AD, ∠ ADB = ∠ Abd = 45 according to the nature of square, and then we get ∠ EBA = ∠ FDA = 45 by using BE ∠ BD. Using the triangle area formula, we can calculate DM=2, and extend AB to m' so that BM'=DM=2, as shown in the figure. Then CM=2 can be calculated according to Pythagorean theorem, then CM'=CM=2 can be obtained by proving △ BCM △ DCM, and then it can be proved that ∠BCM'=∠DCM.
Solution: Solution: ∫∠BAD rotates clockwise around point A by α (0
∴∠EAB=∠FAD=α，
∵ quadrilateral ABCD is a square,
∴AB=AD,∠ADB=∠ABD=45 ,∵BE⊥BD，
∴∠EBD=90，
∴∠EBA=45，
∴∠EBA=∠FDA，
In △ABE and △ADF,
,
∴△ABE≌△ADF(ASA)，
∴S△ABE=S△ADF，
∴S quadrilateral aebf = s △ Abe+s △ ABF = s △ ADF+s △ ABF = s △ Abd =× 2× 2 = 4,
∫S quadrilateral AEBF=S△CDM,
∴S△CDM==2，
∴DM2=2，DM=2，
Extend AB to M' so that BM'= DM = 2, as shown in the figure,
In Rt△CDM, CM==2,
In △BCM' and △DCM.
,
∴△BCM≌△DCM(SAS)，
∴cm′=cm=2,∠bcm′=∠dcm，
∫AB∨CD，
∴∠m′nc=∠dcn=∠dcm+∠ncm=∠bcm′+∠ncm，
And NC shares ∠BCM,
∴∠NCM=∠BCN，
∴∠m′nc=∠bcm′+∠bcn=∠m′cn，
∴m′n=m′c=2，
∴bn=m′c﹣bm′=2﹣2.
So the answer is: 2-2.
Comments: This question examines the essence of rotation: the distance from the corresponding point to the center of rotation is equal; The included angle between the corresponding point and the connecting line of the rotation center is equal to the rotation angle; The figures before and after rotation are congruent. The properties of the square and the judgment and properties of congruent triangles are also studied.
3. When answering questions (this big question ***4 small questions, 19 questions 10 points, 20 questions 8 points, 2/kloc-0 questions 8 points, 22 questions 8 points, ***34 points), each small question must give the necessary calculus process or reasoning steps.
19. Solve the equation:
( 1)x2﹣6x﹣2=0
(2)=+ 1.
Test center: Solve quadratic equation with unary matching method; Solve fractional equations.
Analysis: (1) move the term, the formula, and then make a square, and then you can get two linear equations and find the solutions of the equations;
(2) First, the fractional equation is transformed into an integral equation, and the solution of the equation is obtained, and then it is tested.
Solution: Solution: (1) x2-6x-2 = 0,
x2﹣6x=2，
x2﹣6x+9=2+9，
(x﹣3)2= 1 1，
x﹣3=，
x 1=3+,x2=3﹣；