For proposition p:

① When the focus of hyperbola is on the X axis.

There is < 0,2-a >; a，2-a & gt； 0, the solution is a < 0.

② When the focus of hyperbola is on the Y axis.

You have 2-a

Since p is a false proposition, 0≤a≤2.

For proposition q:

Because it is a false proposition, that is, x 2+2ax+2a > 0 holds.

That is, Delta (a triangle symbol) = (2a) 2-8a ≤ 0, and the solution is 0≤a≤2.

To sum up, A is

19. 1. solution: from the meaning of the question, the straight line is tangent to the circle.

Bring y=x+2 into the cyclic equation.

2x 2+(4-2A)+A 2-4 = 0. Because tangency has only one focus, Delta = (4-2A) 2-8 (A 2-4) = 0, and A =-6/2.

And a>0, so the circular equation is (x-2) 2+y 2 = 8.

2. Know from the meaning of the question

The cyclic equation obtained by bringing y=x+b into 1

Equation 1: 2x 2+(2b-4) x+b 2-4 = 0.

Let the chord length of two intersections and intersections in the topic be twice the square root of 6, and let these two intersections be C and D (D >); C) and that slope of the straight line l is 1.

There is a square root of ①(d-c) radical number 2 = 2 6 ② d+c = 4-2b/2dc = b 2-4/2 (Vieta's theorem).

③(d+c)2-2dc = d-C4 delta =(2 b-4)2-8(B2-4)> 0

Solve the above one, two, three, four: 2>b & gt-6 b=(4- root number 3)/2, and b satisfies [-6,2].

20. 1. The solution is that AD=2, PA=2, PD=2 times the root number 2, and AD is perpendicular to PA.

And ABCD is rectangular, so AD is perpendicular to AB.

AB∩PA= plane PAB, so AD is perpendicular to the plane PAB.

2. According to the method of 1 and ∠PAB=60 degrees.

BC is perpendicular to the plane PAB, and PC=2 times the root number 2, so the sine value of the angle between PC and the plane PAB =BC/PC= root number 2/2.

3. Connect two diagonals of rectangular ABCD, intersect with point O, and O bisects the two diagonals.

The dihedral angle in the problem is the size of the angle POA, because AO=AC/2= AB 2+BC 2/2 = radical sign 13/2.

Then connect point f and point o in AB. Because it is parallel to AD, OF=AD/2= 1, while AD is perpendicular to the plane PAB, and OF is also perpendicular to the plane PAB.

And PF=FA/tan30 degrees =3 times the root number 3/2.

Therefore, under PO = radical number PF 2+of 2 = radical number 3 1/2.

Then it is obtained from the length of three sides, cosine theorem and sine theorem of triangle POA

Pa 2 = po 2+OA 2-2 popacos ∠ poasin ∠ poa 2+cos ∠ poa 2 =1Xin ∠ poa > 0

Solution: tan∠POA=