1. Multiple choice questions (***5 questions, 7 points for each question, ***35 points. Only one option is correct. Please put the code of the correct option in brackets after the question. If you don't fill in, fill in too much or fill in incorrectly, you will get 0 point).

1. If is, the value of is ().

(A) (B) (C) (D)

Solution: Set by the problem.

Algebraic deformation, divided by B.

2. If real numbers A and B are satisfied, the range of A is ().

(A)a (B)a4 (C)a≤ or a≥4 (D)≤a≤4.

solution

Because b is a real number, the quadratic equation of one variable about b

The discriminant of is ≥0, and the solution is a≤ or a ≥ 4.

Equation thought, discriminant theorem; Solving quadratic inequality in one variable

3. As shown in the figure, in quadrilateral ABCD, ∠ B = 135, ∠ C = 120, AB=, BC=, CD =, then the length of the AD side is ().

(A) (B)

(C) (D)

Solution: d

As shown in the figure, the intersection points A and D are AE, DF is perpendicular to the straight line BC, and the vertical feet are E and F respectively.

Known to be available

BE=AE=，CF=，DF=2，

So ef = 4+.

The intersection point a is AG⊥DF, and the vertical foot is g. In Rt△ADG, it is obtained according to Pythagorean theorem.

AD=。

Pythagorean Theorem, Simplification and Complete Graphic Method Involving Biquadratic Roots

4. In a column number, it is known that when k≥2,

(The integer symbol indicates the largest integer that does not exceed the real number, for example,) equals ().

1 (B) 2 (C) 3 (D) 4

Solution: b

Provide by sum

,,,,

,,,,

……

Because 20 10=4×502+2, it = 2.

Gaussian function; Found a pattern.

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5. As shown in the figure, in the plane rectangular coordinate system xOy, the vertex coordinates of the isosceles trapezoid ABCD are A( 1, 1), B(2,-1), C (-2,-1) and D (-/kloc-0) respectively. Point P 1 rotates around point b 180, point P2 rotates around point c 180, point P3 rotates around point d 180, …, and repeat the operation to get point P 1, P2, …, and then point P20655.

(A)(20 10，2) (B)(20 10，)

(C)(20 12)，(D)(0，2)

Solution: The coordinates of b and point sum are (2,0) and (2,0) respectively.

Remember, among them.

According to the symmetry relation, we can get:

,,,.

Order, you can also get that the coordinates of this point are (), that is, (),

Since 20 10=4502+2, the coordinate of this point is (20 10,).

Second, fill in the blanks

6. Given a =- 1, the value of 2a3+7a2-2a- 12 is equal to.

Solution: 0

It is known that (A+ 1) 2 = 5, so A2+2A = 4, so

2 a3+7 a2-2a- 12 = 2 a3+4a 2+3 a2-2a- 12 = 3 a2+6a- 12 = 0。

7. A bus, a van and a car are driving in the same direction at a constant speed on a straight road. At a certain moment, the bus is in front, the car is behind, and the van is between the bus and the car. 10 minutes later, the car caught up with the van. After another 5 minutes, the car caught up with the bus; After another t minutes, the truck caught up with the bus, and then t =.

Solution: 15

At a certain moment, the distance between the truck and the bus and the car is S kilometers, and the speed of the car, the truck and the bus is (km/min) respectively. It is set that the truck catches up with the bus within X minutes.

, ①

, ② .③

From ① ②, you get, so, x = 30. So (points).

8. As shown in the figure, in the plane rectangular coordinate system xOy, the vertex coordinates of polygon OABCDE are O (0 0,0), A (0 0,6), B (4 4,6), C (4 4,4), D (6 6,4) and E (6 6,0) respectively. If the straight line l passes through the point M (2, 3)

Solution:

As shown in the figure, extend the X axis of BC intersection point to point f; Connect OB, AFCE, DF and intersect at N point.

It is known that point M (2 2,3) is OB and the midpoint of AF, that is, point M is the center of rectangular ABFO, so the straight line divides rectangular ABFO into two parts with equal areas. Because point N (5 5,2) is the center of the rectangular CDEF,

A straight line passing through point n (5 5,2) divides the rectangular CDEF into two parts with equal areas.

Then, the straight line is the straight line sought.

Let the function expression of a straight line be, then

Solution, so the functional expression of the straight line is.

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9. As shown in the figure, rays AM and BN are perpendicular to line AB, point E is a point above AM, perpendicular AC passing through point A intersects with BE respectively, BN is at points F and C, and perpendicular CD passing through point C is D. If CD = CF, then.

Solution:

See the picture, settings.

Because Rt△AFB∽Rt△ABC, so.

And because fc = DC = ab, that is,

Solve, or (give up).

Rt delta ∽ is rt delta again, so that's =

10. For i=2, 3, …, k, the remainder obtained by dividing the positive integer n by I is I- 1. If the minimum value of is satisfied, the minimum value of a positive integer is.

Solution: Because is a multiple of, the minimum value of is satisfied.

Where represents the least common multiple.

because

Therefore, the minimum value of the positive integer satisfied is.

III. Answer questions (***4 questions, 20 points for each question, 80 points for * * *)

1 1. As shown in the figure, △ABC is an isosceles triangle, AP is the height on the base BC, point D is the point on the line segment PC, and BE and CF are the circumscribed circle diameters of △ABD and △ACD respectively. Connect EF. Proof: (Question 12A)

.

(Question 12B)

(Question 12B)

Proof: Connect ED and FD as shown in the figure. Because Be and CF are both diameters, so

ED⊥BC，FD⊥BC，

Therefore, the three-point * * * line of D, E and F is .................. (5 points).

Connect Auto Exposure, Auto Focus and then

Therefore, △ ABC ∽△ AEF...........( 10)

Let AH⊥EF and vertical foot be h, then AH=PD. It can be obtained from △ABC∽△AEF.

I know that everyone will copy me, so let's see who gave the answer first, hoping to help you.