Examination time: 8: 30-9: 30 am on March 7th, 2065438+00.
The examination time of this volume is 60 minutes, and there are ***30 small questions, with 4 points for each small question, with a full score of 120. ※.
1. The median and mode of a set of data 4, 5, 6, 7, 7 and 8 are (A) 7, 7 (B) 7, 6.5 (C) 5.5 and 7 (D) 6.5 respectively.
7。
2. if the polynomial P=a2 4a 20 14, then the minimum value of p is (a) 2010 (b) 201(c) 2012 (d) 20/kloc-.
3. If 100 x2 x49y 2 is completely flat, then the value of k is (a) 4900 (b) 9800 (c)140 (d) 70.
4. put a three-digit m in front of a two-digit n to form a five-digit number, which can be expressed as (A) mn (B) m n.
(c) North latitude 10 meter.
5. If x, then = (A) 4x (B) 2 (C) 2 (D) 2 4x.
6. The largest of the following real numbers is (A) 5 (B) (C) (D).
7. Define the meaning of the operator "*" as: a * b = (where a and b are not 0). There are two conclusions:
(1) operation "*" satisfies the commutative law; (2) The operation "*" satisfies the associative law. Where (a) only (1) is correct and (b) only.
There are (2) correct (C) (1) and (2) all correct (D) (1) and (2) all incorrect.
8. The positive integer solution of equation x2 y2= 105 has (A) 1 group (B) 2 groups (C) 3 groups (D) 4 groups.
9. It is known that x and y satisfy 3x 4y=2, x y；; (B)y & gt； (C) x= (D) y= .
10. if abc 0 is known and = = =p, the image of the original function y=px p must pass (a) the first,
Two quadrants (b) second, third, third and fourth quadrants (c) first and fourth quadrants.
1 1. As shown in the figure, AB//CD, AC//BD, AD and BC intersect at O, AE BC intersects at E, and DF BC intersects at F, then the graph is congruent.
There are (A) 5 pairs (B) 6 pairs (C) 7 pairs (D) 8 pairs in the triangle.
12. As shown in the figure, in the square ABCD, H is a point on the extension line of BC, so that CE=CH connects DH and BE extends to DH.
In g, the following conclusions are wrong: (a) be = DH (b) hbec = 90 (c) bgdh (d) hdcable.
=90 。
13. As shown in the figure, rABC is divided into three parts with equal area by DE and FG (that is, S 1=S2=S3), while DE//FG//BC, BC=,
FG DE= (A) 1 (B) (C) (D) 2 .
14. As shown in the figure, AB is the diameter of circle O, and the following relationship among angles P, Q, R and S is correct.
( 1)p = 2q； (2)q = r； (3)p s = 180；
(a) only (1) and (2) (B) only (1) and (3) (C) only (2) and (3) (D) (1), (2) and (3).
15. As shown in the figure, after the rABC is folded along its centerline DE, the point A falls to the point A'. If C= 120 and A=26, then
The degree of A'DB is (a)120 (b)112 (c)10 (d)100.
16. As shown in the figure, the side length of equilateral rABC is 3, p is a point above BC, BP= 1, and d is a point above AC. If APD=60,
Then the length of the CD is (A) (B) (C) (D).
17. As shown in the figure, in RtrABC, ABC=90, AB=8cm, BC=6cm, with A and C as the center respectively, and the length is.
Make a circle with this radius and cut RtrABC into two sectors, then the area of the remaining (shaded) part is cm2.
(a) Article 24 (b) (c).
18. As shown in the figure, if the circumference of rhombic ABCD is 20cm, DE AB, vertical foot is E, and cosA=, the following conclusion is correct.
The number is (A) 3 (B) 2 (C) 1 (D) 0. j DE = 3cmk EB = 1cm； L S diamond ABCD =15cm2;
19. Line up the five numbers 1, 2, 3, 4, 5, and the last number is odd, so there are three consecutive numbers in it.
The sum of these three numbers can be divisible by the first number, so there are (A) 2 kinds (B) 3 kinds (C) 4 kinds of arrangement methods that meet the requirements.
Five kinds.
20. If the two sides of the rABC are A and B respectively, then the area of the rABC cannot be equal to (A) (a2 b2) (B) (a2 b2).
(c) Paragraph 2 (d).
2 1. Let x 1 and x2 be two quadratic equations x2 x 3=0, then x 13 4x22 19 is equal to (A) 4 (B) 8 (C) 6 (D) 0.
22. both a and b of rabc are acute angles and the CD is high. Whereas =( )22. RABC is a right triangle (a).
(b) isosceles triangle (c) isosceles right triangle (d) isosceles triangle or right triangle.
23. Turn the dice along one side of the dice on the chessboard with two rows and three columns (the reverse side is marked with 1 point and 6 points, 2 points and 5 points respectively.
3 o'clock, 3 o'clock and 4 o'clock), the dice can't go backwards in each flip mode. At the beginning, the dice are placed upward as shown in Figure J.
The number of points is 2; Finally, flip to the position shown in Figure K. At this time, the number of points of the dice facing upwards cannot be one of the following numbers.
5 (B) 4 (C) 3 (D) 1 .
24. As shown in the figure, in rABC, m is the midpoint of AB side and n is the point on AC side.
=2, CM and BN intersect at point K. If the area of rBCK is equal to 1,
The area of rABC is equal to (A) 3 (B) (C) 4 (D).
25. as shown in the figure, let a, b and c be the lengths of three sides of rABC respectively, and =, BD=c,
Then the relationship between CAB and CBA is (a) CBA >; 2 cabs (B) CBA=2 cabs
(C) CBA<。 2 CAB (D) is uncertain.
26. It is known that the three sides of rABC are A, B and C, and the three sides with an area of S are rA 1B 1C 1.
They are a 1, b 1, c 1, and the area is S 1, a >；; A 1, b>b 1, c>C 1, then s and S 1 are larger.
The secondary relationship must be (a) s > S 1(B)S & lt； S 1 (C) S=S 1 (D) uncertain.
27. Positive real numbers X and Y satisfy xy= 1, so the minimum value is (A) (B) (C) 1 (D).
28. Let A and B be real numbers and =, then = (A) (B) (C).
(D) 10 .
29. Let A, B and C be real numbers, a 0, parabola y=ax2 bx c intersect with X axis at points A and B, and intersect with Y axis at point C,
And the vertex of the parabola is on the straight line y= 1 If rABC is a right triangle, the maximum value of RtrABC area is
1 (B) (C) 2 (D) 3 .
30. The closest values are equal to (a), (b), (c) and (d).
answer
1.DACDB 6。 CADBB 1 1。 CBDAB 16。 BAADB
2 1.DDDCB 26。 DCDAB