Morning = n b c d.

2. Let the volume of a tetrahedron be V 1, and form a convex polyhedron with the midpoint of each side as the vertex, and its volume is V2, which is ().

The CBI is not sure

3. Among the permutations A 1, a2, a3, a4 and a5 of1,2, 3, 4 and 5, the number of permutations satisfying a 1 < A2, A2 > A3, A3 < A4, A4 > A5 is ().

A.24 B. 16 C. 10 D.8

4. Given positive numbers P, Q, A, B, C, where P ≠ Q. If P, A, Q are geometric series and P, B, C, Q are arithmetic series, the unary quadratic equation bx2-2ax+c=0 ().

A. There are two equal real roots B. There are two real roots with the same sign but different from each other.

C. there are two real roots with different signs. D. there is no real root cause

5. If odd function is a decreasing function in [0,], then one value of θ is ().

a .πb .πc .-d-

6. It is known that x and y satisfy: if the maximum value of z=ax+y is 3a+9 and the minimum value is 3a-3, the value range of a is ().

A.0 ≤ A ≤ 1B。 - 1 ≤ A ≤ 0C。 - 1 ≤ A ≤ 1d。 A ≤- 1 or a≥ 1.

2. Fill in the blanks (the full score of this question is 30 points, and each small question is 5 points)

7. Let a circle Ck={(x, y)|(x-mk)2+(y-mk)2≤2k2} and k∈N+, where mk is defined as follows: m 1=0, MK+1= MK+2k+650.

8. Given the complex set d, the complex number z∈D is _ _ _ _ if and only if there is a complex number z 1 with the modulus of1,so that the real part and imaginary part in d are integers.

9. Among the cosine values of dihedral angle A-BC-D composed of any four vertices A, B, C and D that are not on the same plane of the cube, the number of values less than is _ _ _ _.

10. let a = {1, 2,3, ..., n} (n > 1, n∈N), and the mapping f: a→ a → a. satisfies f (1) ≤ F. ...

1 1. Let f be the focus of the parabola y2=2x- 1 and Q(a, 2) be the point on the straight line y=2. If only one point p on the parabola satisfies |PF|=|PQ|, then the value of a is _ _ _ _.

12. let x, y > 0, S(x, y)=min{x, y,,}, then the maximum value of S(x, y) is _ _.

Third, answer the question (this question is ***4 small questions, out of 90 points)

13. (The full mark of this small question is 20) Let the focus of 0 < α, β, γ 0 be f, and the straight line passing through (-0) intersects with the parabola at points C and D in the first quadrant. If there is a point E on the X axis, let CE⊥DE find the range of the slope k of the straight line CD.

15. (Full score for this small question is 25) AN is the bisector of △ABC, and the extension line of AN intersects the circumscribed circle of △ABC in D; M is a point on AN, and the straight lines BM and CM intersect △ABC, which circumscribe E and F respectively; DF crosses AB in P, and DE crosses AC in Q. Verification: P, M and Q three-point * * * lines.

16. (The full score of this small question is 25) Set one set.

M={n|n！ It can be expressed as the product of (n-3) consecutive positive integers, and n > 4}. It is proved that m is a finite set, and all the elements of m are obtained.

Reference answers and grading standards

I. Multiple-choice questions (the full score of this question is 30 points, and each small question is 5 points)

1.A 2。 A 3。 B 4。 D 5。 B 6。 C

2. Fill in the blanks (the full score of this question is 30 points, and each small question is 5 points)

7.8.49 9.410.1.0 or1/2.

Third, answer the question (this question is ***4 small questions, out of 90 points)

13. (Full score for this small question)

Solution: obtained from the known cos2α+cos2β+cos2γ+2cosα cosβ cosγ =1

(cosγ+cosαcosβ)2 =( 1-cos 2α)( 1-cos 2β)= sin 2αsin 2β。

Cosγ+cosαcosβ=sinαsinβ is obtained from α, β, γ∈(0,).

cosγ=-(cosαcosβ-sinαsinβ)

=-cos (α+β) = cos (π-α-β)...(5 points)

And π-α-β, γ∈(0,), so α+β+γ = π.

So α, β and γ are the three internal angles of an acute triangle.

Let x = sin α; y = sinβz=sinγ，

X, y, z, y and z can form the length of three sides of a triangle ... (10 minute)

..... (15 points)

By | x-y | < z，| y-z| < x，| z-x| < y，

So ≤...(20 points)

14. (Full score for this small question)

Solution: The linear equation is that if you substitute it into the parabolic equation, you get

...... (5 points)

Let the intersection of a straight line and a parabola be C(x 1, y 1) and D(x2, y2), then there are

,;

,,

..... (10)

The coordinate of point E is (x0,0), which is defined by CE⊥DE.

(x1-x0) (x2-x0)+y1y2 = 0, that is.

Substitute and sort it out

..... (15 points)

Because of the existence of point e,

Solution. Because c and d are in the first quadrant, k > 0,

So, ... (20 points)

15. (The full score for this short question is 25)

Proof: connect PM, QM and BD.

∠∠PAD =∠MAC，∠ADP=∠ACM，

∴△ADP∽△ACM, Associated Press: AM = PD: MC...(5 points)

∠∠BPD =∠PAD+∠ADP =∠MAC+∠ACM =∠NMC，

And < fdb = < fcn,

∴△BDP∽△NCM，

Pb: Mn = PD: MC...( 15 points)

∴AP:AM=PB:MN.

∴ PM ‖ BC...(20 points)

Similarly, it can be proved that QM‖BC,

So, the three-point * * * line of P, M and Q ... (25 points)

16. (The full score for this short question is 25)

Solution: let n∈M, then n! = 1×2×…×n = m(m+ 1)……(m+n-4)，

If m≤4, then m (m+1) ... (m+n-4) ≤ 4× 5× ... × n < n!

Therefore, there must be m ≥ 5...(5 points).

Because m (m+1) ... (m+n-5) ≥ 5× 6× n,

So m+n-4≤4! =24，n≤28-m≤23，

So m is a finite set ... (10 points)

When m=5 and n=23, 23! =5×6×…×24.

When m≥6, due to 5! = 120 cannot be expressed as the product of two consecutive positive integers.

So m (m+1) ... (m+n-4) has at least three factors.

Because m (m+1) ... (m+n-6) ≥ 6× 7××× n,

So (m+n-5)(m+n-4)≤5! = 120< 1 1× 12.

Therefore, m+n ≤ 15...( 15 points).

When m≥n, n ≤ 7; When m < n, m ≤ 7;

Therefore, 5≤m≤7 or n ≤ 7. ..............................................................................................................................................................

After verification, 23! =5×6×…×24,6! =8×9× 10,7! =7×8×9× 10,

So, m = {6 6,7,23} ... (25 points)