1. Multiple-choice questions: (There are 6 questions in this big question, with 2 points for each question, with a full score of 12).
1. The ordinate of the intersection of the straight line and the Y axis is ......................................... ().
(A)2； (B)？ 2; (C)3; (D)？ 3.
2. When solving the equation by substitution method, you can set it so that the original equation can be transformed into … ().
(1); (b) and:
(c) and: (4).
3. Among the following equations, the equation with real roots is.
(1); (b) and: (c) and: (4).
4. It is known that the two diagonals AC and BD of the parallelogram ABCD intersect at point O, and their lengths are equal to 8cm and 12cm respectively. If the length of BC side is equal to 6cm, the circumference of △BOC is equal to
14； (B) 15； (C) 16； 17。
5. Among the following propositions, the false proposition is .............................................. ().
(a) The two diagonal lines of the trapezoid are equal; (b) The two diagonal lines of the rectangle are equally divided;
(c) The two diagonal lines of the diamond are perpendicular to each other; (d) Each diagonal of a square bisects a set of diagonal lines.
6. In the following events, it is determined that the event is ............................................. ().
(a) The equation about x has a real number solution; (b) The equation about x has a real number solution;
(c) The equation about x has a real number solution; (d) The equation about x has a real number solution.
Two. Fill in the blanks: (This topic is entitled *** 12, with 3 points for each question, out of 36 points).
7. The solution of the equation is.
8. If the function is linear, then.
9. if point a (2, m) and point b (4, n) are on the image of the function, then the relationship between m and n is: m n. (use ">", "=" or "
10. If the equation about x has an increasing root and x=2, then the value of k is.
1 1. Please write a binary quadratic equation with a solution of 0. This equation can.
12. The sum of the internal angles of a heptagon is equal to degrees.
13. It is known that the side length of the square ABCD is equal to 8cm, so the distance from the midpoint m of the AB side to the diagonal BD is equal to
Centimeter.
14. If the height on the hypotenuse of an isosceles right-angled triangle is more than 5cm, then the length of the line segment connecting the midpoints of two right-angled sides of this triangle is equal to cm.
15. The two elements of a vector are: size and.
16. It is known that in parallelogram ABCD, if, then the vector is represented by vector sum.
= .
17. There are three red balls, five yellow balls and six black balls in the schoolbag. These balls are all the same except the color, so the probability of drawing a black ball from this bag is.
18. Choose two numbers from 2, 4 and 6 to form a two-digit number, and choose a number from all two digits. The probability that this number is divisible by 3 is.
3. Answer: (There are 7 questions in this big question, out of 52 points)
19. (The full mark of this question is 6)
solve an equation
20. (The full mark of this question is 6 points)
As shown in the figure, the known vector. Calculated as: vector (1); (2) .
2 1. (The full mark of this question is 7)
As shown in the figure, in the parallelogram ABCD, the difference between side BC and CD is 2cm, AP bisects ∠BAD, and the intersecting side BC is at point P.
Q: The length of the PC.
22. (The full mark of this question is 7 points)
Party A and Party B go to Party B, which is 35 kilometers away from Party A, to do business. Party A goes first, and Party B rides later. The relationship between travel time and distance of two people is shown in the picture. According to the information provided in the figure, the answer is as follows:
(1) B leaves a few hours later than a;
(2) B catches up with A one hour after departure;
(3) How many hours before A does B arrive at B?
23. (The full mark of this question is 8)
After learning about the Sichuan earthquake, students from a school took out their pocket money and took part in fund-raising activities for disaster relief. Students from Class A * * * donated 840 yuan, while students from Class B * * * donated 1, 000 yuan. The per capita donation of students in Class B is 5 yuan/person more than that of students in Class A, and the number of students in Class A and Class B is 2 less.
24. (The full mark of this question is 8)
It is known: As shown in the figure, AM is the center line of △ABC, D is the midpoint of the line segment AM, AM=AC, AE ‖ BC.
It is proved that the quadrangle EBCA is an isosceles trapezoid.
25. (The full mark of this question is 10)
As shown in the figure, in diamond ABCD, AB=4, ∠ B = 60, point P is the moving point on ray BC, ∠ PAQ = 60, intersecting ray CD is at point Q, and the distance from point P to point B is X, PQ = Y. 。
(1) proves that △APQ is an equilateral triangle;
(2) Find the resolution function of Y about X and write its definition domain;
(3) If PD⊥AQ, find the value of BP.
The second semester of the 2007 school year in Pudong New Area ended the second grade mathematics examination paper.
Reference answers and scoring instructions
First, multiple-choice questions:
1.d； 2.d； 3.c； 4.c； 5.a； 6.B。
Second, fill in the blanks:
7.8; 8.≠ 1; 9.& gt； 10.4; 1 1. and so on. ; 12.900; 13.； 14.5; 15. direction; 16.； 17.； 18.。
Third, answer questions:
19. solution: y = 2x .............................................................................................. (1min)
If you replace ①, you get 5x2 = 20 ............................................................................................ (1).
∴ x = 2 .................................................................. (1min)
When x=2, y = 4；; ............................... (1) when x=-2 and y =-4.
The solution of the equation is ................................. (2 points).
20. Solution: 2 points for drawing, and 1 point for conclusion.
2 1. solution: in parallelogram ABCD,
BC, ∴∠ DAP = ∠ APB .......................................... (2 points).
∠DAP=∠BAP, ∴∠ APB =∠ BAP ...................................................... (1min)
∴ AB = BP .................................................. (2 points)
Ab = cd, ∴ PC = BC-BP = 2 ........................................... (2 points)
22. Solution: (1) 2; ...................................................( 1)
(2)2; ..................................................( 1)
(3) The resolution function of Party A's distance and time is s = 5t ........................................... (1).
When S=35, t = 7 ............................................. (1min).
Let the resolution function of b's distance and time be s = kt+B.
According to the meaning of the problem, you can get a solution
∴ The resolution function of Party B's distance and time is S =10t-20 .......................................... (1).
When S=35, t = 5.5 ............................................. (1).
∴7-5.5= 1.5.
Answer: Party B arrived in ............................., B, 1.5 hours earlier than Party A (1).
23. Solution: If the number of students in Class B is X, the number of students in Class A is (x+2). .........................................................................................................................................
According to the meaning of the question, get ..................................................................... (3 points)
Finishing, get ..................................... (1 min)
Solution, .............................................. (1 min)
After testing, they are all the roots of the original equation, but they do not meet the meaning of the question, so they are discarded.
...................................................( 1)
A: There are 42 students in Class A and 40 students in Class B ................ (1).
24. Proof: ∫AE‖BC, ∴∠AED=∠MCD, ∠ EAD = ∠ cmd ......................................................................... (1minute).
Ad = md, ∴△ AED △ MCD ..................................................... (1min)
∴ AE = cm ........................................... (1min)
∵BM=CM,∴AE=BM.
∴ Quadrilateral AEBM is a parallelogram ............................................................... (1 min).
∴ EB = am ............................................ (1min)
And AM=AC, ∴ EB = AC ........................................... (1min).
∫AE‖BC, EB and AC are not parallel, ∴ quadrilateral EBCA is a trapezoidal ............................................... (1 min).
∴ trapezoid EBCA is an isosceles trapezoid ............................................................................................................. (1 min).
25. Solution: (1) Join AC. In the diamond ABCD,
Ab = bc, ∠ B = 60, ∴△ABC is an equilateral triangle ................... (1 min).
∴AC=AB,∠BAC=∠BCA=60。
∠∠paq = 60, ∴∠ BAP = ∠ CAQ ............................... (1min).
ab CD，∠B=60 ,∴∠BCD= 120。
∴∠ACQ=∠B=60。
∴△ Headquarters Base△ ACQ ............................................ (1min)
∴ AP = AQ ................................................ (1min)
∴△APQ is an equilateral triangle ........................................ (1).
(2) If △APQ is an equilateral triangle, AP = PQ = Y. 。
Let AH⊥BC be at point H, and from AB=4, BH=2, ∠ B = 60, ah = ... (1).
That is ∴ ................................. (1min).
The domain is x ≥ 0 ............................................................................... (1).
(3)(i) When the point P is on the BC side,
∵PD⊥AQ, AP=PQ, ∴PD divides AQ vertically.
∴AD=DQ.
∴ CQ = 0 .......................................... (1min)
bp = cq，∴ BP = 0。
(ii) When point P is located on the extension line of BC,
Similarly, BP = 8 .................................................. (1).
To sum up, BP=0 or BP = 8.