I. (Questions 1 to 12)

1. 1 2.3.4.5.- 1+i 6。 7.

8.5 9. 10.36 1 1.k=0，- 1 & lt； b & lt 1 12.a≤ 10

Two. (Questions 13 to 16)

13.C 14。 A 15。 A 16。 D

Three. (Question 17 to 22)

17. Solution: y = cos (x+) cos (x-)+sin2x.

=cos2x+ sin2x=2sin(2x+)

The range of the function y = cos (x+) cos (x-)+sin2x is, and the minimum positive period is π.

18. solution: connect BC with cosine theorem, bc2 = 202+102-2× 20×10cos120 = 700.

So BC= 10.

* ∴sin∠acb=，

∫∠ACB & lt； 90 ∴∠ACB=4 1

∴ Ship B should go straight to rescue in the northeast direction of 7 1.

19. solution: (1) in a quadrangular cone P-ABCD, obtained from the plane ABCD of PO⊥.

∠PBO is the angle between PB and plane ABCD, ∠ PBO = 60.

In Rt△AOB, bo = absin30 = 1,

So PO=BOtg60 =, and the area of the bottom diamond is 2.

∴ The volume of the pyramid P-ABCD V= ×2 × =2.

(2) Solution 1: Take O as the coordinate origin, and rays OB, OC and OP are the positive semi-axes of X-axis, Y-axis and Z-axis, respectively, to establish a spatial rectangular coordinate system.

OA= in Rt△AOB, so the coordinates of points A, B, D and P are A (0,-0) respectively.

B( 1，0，0)，D(- 1，0，0)P(0，0，)。

E is the midpoint of PB, so e (,0,) is = (,0), =(0,).

Let the included angle be θ, cosθ=, θ=arccos,

The angle between DE and PA is arccos.

Solution 2: Take the midpoint f of AB and connect EF and DF.

If e is the midpoint of PB, EF‖PA is obtained.

∴∠FED is the angle (or the remaining angle) formed by the straight line DE and PA.

In Rt△AOB, ao = abcos30 = = op,

Therefore, in the isosceles Rt△POA, PA=, then EF=.

In positive △ABD and positive △PBD, DE=DF=.

cos∠FED= =

The angle between DE and PA is arccos.

20. It is proved that: (1) Let the straight line passing through point T (3,0) and parabola y2=2x intersect at points A(x 1, y 1) and B(x 12, y2).

When the straight line L exists under the pheasant rate, the equation of the straight line L is x=3. At this time, the straight line L intersects the parabola at points A(3,) and B (3,-). ∴ = 3.

When the pheasant rate of the straight line L exists, let the equation of the straight line L be y = k (x-3), where k≠0.

When y2=2x

Ky2-2y-6k = 0, then y 1y2 =-6.

y=k(x-3)

And ∵x 1= y, x2= y,

∴ =x 1x2+y 1y2= =3。

To sum up, the proposition "If the straight line L passes through the point t (3,0), then =3" holds.

(2) The inverse proposition is: Let the straight line L and the parabola y2=2x intersect at two points A and B, if =3, then the straight line intersects with the point T (3,0). This proposition is false.

For example, take points A (2 2,2) and B (0, 1) on a parabola, and then =3.

The equation of straight line AB is Y= (X+ 1), but t (3,0) is not on straight line AB.

Note: y 1 y2 =-6 can be obtained from points A(x 1, y1) and B(x 12, y2) on the parabola y2=2x =3.

Or y 1y2=2, if y 1y2 =-6. , it can be proved that the straight line AB passes through point (3,0); If y 1y2=2, it can be proved that the straight line AB passes through the point (-1 0), but does not pass through the point (3,0).

2 1. Proof (1) When n= 1 and a2=2a, then = A；;

2 ≤ n ≤ 2k- 1，an+ 1 = (a- 1) sn+2，an = (a- 1) sn- 1+2，

An+ 1-an = (a- 1) an, ∴ = a, ∴ sequence {an} is geometric progression.

The solution (2) An = 2a comes from (1), ∴ A 1A2...An = 2a = 2a = a,

bn= (n= 1，2，…，2k)。

(3) let bn≤, the solution is n≤k+, and n is a positive integer, so when n≤k, bn

When n≥k+ 1, BN >:.

Original formula = (-b1)+(-B2)+…+(-bk)+(bk+1-)+…+(b2k-)

=(bk+ 1+…+b2k)-(b 1+…+bk)

= = .

When ≤4, k2-8k+4 ≤ 0, 4-2 ≤ k ≤ 4+2 and k≥2,

When k=2, 3, 4, 5, 6, 7, the original inequality holds.

22. Solve the (1) function y = x+(x >；; 0) is 2, then 2 =6, ∴b=log29.

(2) let 0 < x1< x2，y2-y 1=。

When < x1<; X2, y2>Y 1, and the function y= is increasing function.

Therefore, when x= or x=2, F(x) takes the maximum value () n+() n;

When x= 1, the minimum value of F(x) is 2n+ 1.

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