I. Multiple-choice questions: (7 points for each question, ***42 points)

1, simplified: the result is _ _.

A, irrational number b, true fraction c, odd number d, even number

Solution:

So choose d

2. The lengths of the four sides inscribed in the circle are 5, 10, 1 1,14; Then the area of this quadrilateral is _ _.

a、78.5 B、97.5 C、90 D、 102

Solution: from the meaning of the question:

52+ 142-2×5× 14×cosα= 102+ 12-2× 10× 1 1×cos( 180-α)

∴ 221-140cosα = 221+220cosα

∴cosα=0

∴α=90

The area of this quadrilateral is: 5× 7+5× 1 1 = 90.

Choose C.

3, let r≥4, a =, b =, c =, then the following must be true.

a、a & gtb & gtc B、b & gtc & gta C、c & gta & gtb D、c & gtb & gta

Solution 1: Using the special value method, let r=4, there is.

a=，b=，

c=

∴c>； B>a, select D.

Solution 2: a =,

b=

c=

Solution 3: r ≥ 4 ∴ < 1

∴

c=

∴a<； B<c, choose D.

4. The three shadows in the figure are divided by two circles with radius of 1 and their tangents. If the area of the middle shadow is equal to the sum of the upper and lower areas, the common chord length of the two circles is _ _.

A, B, C, D,

Solution: We know that the area of the circle is equal to the area of the rectangular ABCD by cutting and complementing the figure.

∴

According to the vertical diameter theorem, the common chord is

Option d

5. The image with known quadratic function f (x) = AX2+BX+C is shown in the figure.

P = | A-B+C | +| 2A+B |, Q = | A+B+C |+2A-B |, then _ _.

The relationship among a, p>q B, p=q C, p<q D, p and q cannot be determined.

Solution: from the meaning of the question: a

∴p=|a-b|+|2a+b|,q=|a+b|+|2a-b|

and

∴p=|a-b|+|2a+b|=b-a+2a+b=a+2b=2b+a，

q=|a+b|+|2a-b|= a+b+b-2a=2b-a

∴p<； Q, choose C.

6. If x 1, x2, x3, x4 and x5 are mutually unequal positive odd numbers, (2005-x1) (2005-x2) (2005-x4) = 242.

a、 1 B、3 C、5 D、7

Solution: (2005-x 1), (2005-x2), (2005-x3), (2005-x4) and (2005-x5) are different positive odd numbers.

The product of 242 decomposed into five mutually unequal even numbers has only one form: 242 = 2 (-2) 4.6 (-6).

Therefore, (2005-x 1), (2005-x2), (2005-x3), (2005-x4) and (2005-X5) are equal to 2, (-2), 4, 6 and (-6) respectively.

so(2005-x 1)2+(2005-X2)2+(2005-X3)2+(2005-X5)2 = 22+(-2)2+42+62+(。

Expand:

Two. Fill in the blanks (***28 points)

In 1 and not more than 100, the sum of all numbers that are multiples of 3 or 5 is _ _.

Solution: (3×1+3× 2+) ... 3× 33)+(5×1+5× 2+... 5× 20)-(15×1+/kloc.

2、x=___ .

Answer: Molecules are organized:

∵x≠0，

∴

For simplicity, both sides are squares:

Reorganize and simplify:

3. If the real numbers x and y are satisfied, then x+y = _ _.

Solution 1: suppose X+Y = A, then y = a-X.

Solution 2: Easy to know

Simplify:

4. It is known that the three internal angles A, B and C of acute triangle A>B>C satisfy the following conditions: A > B > C. If A is the minimum among A-B, B-C and 90° A, the maximum value of A is _ _.

Solution:

Three. Problem solving (1 20 points for the question, 25 points for the second and third questions)

1, a, b and c are real numbers, and AC < 0, which proves that the unary quadratic equation AX2+BX+C = 0 has roots greater than and less than 1.

Solution: Suppose.

∴

∴ The unary quadratic equation AX2+BX+C = 0 has roots greater than and less than 1

2. In acute angles Δ AB > AC and AB > AC, CD and BE are the heights of AB and AC, respectively. The extension lines of DE and BC intersect at T, the perpendicular lines of D and BC intersect at F, and the perpendicular lines of E and CD intersect at G. It is proved that F, G and T are three-point lines.

Prove 1: Let the perpendicular lines passing through D and E intersect BC at M and N, respectively. In Rt△BEC and Rt△BDC, we can get from the projective theorem:

CE2=CN CB，BD2=BM BC

∴

Rt△CNG ∽Rt△DCB，Rt△BMF ∽Rt△BEC，

∴

∴

In Rt△BEC and Rt△BDC, the relationship of area is: Be Ce = en BC, BD CD = DM BC.

∴

From (1)(2):

Prove 2: Let CD and BE intersect at point H, then H is the vertical center of △ABC, and the intersections of DF, EG, AH and BC are M, N and R respectively.

∫DM∨AR∨EN

∴

According to the ratio theorem:

Proof 3: In △ABC, the straight line DET intersects BC, CA and AB at t, e and d respectively, which is obtained by Menelaus theorem:

Let CD and BE intersect at point H, then H is the center of △ABC, and AH⊥BC.

∵DF⊥BC、EG⊥BC

∴ah∨df∨eg

∴

From the inverse theorem of Menelaus's theorem, we can get: F, G and T three-point * * * lines.

Proof 4: Connecting FT and EN in G' is simple and clear.

In order to prove the three-point line of f, g and t, just prove it.

∵

and

∴

∵CD⊥AB, BE⊥CA, ∴B, D, E and C * * * cycles.

∴∠ABE=∠ACD (2)

And (3)

Substitute (2) and (3) into (1) to get:, so f, g and t are three lines.

3. Let A, B and C be positive integers, A2+B3 = C4, and find the minimum value of C. ..

Solution: obviously, C > 1. Set by the topic: (c2-a)(c2+a)=b3.

Ruoqu

When we examine B from the largest to the smallest, let it be a complete square number. It is easy to know that when B = 8, C2 = 36, then c=6, so a=28. The following shows that C has no positive integer solution less than 6, and the list is as follows:

cc4x 3(x3 & lt； C4)C4-x3

2 16 1,8 17,8

3 8 1 1,8,27,64 80,73,54, 17

4 256 1,8,27,64, 125,2 16 255,248,229, 192, 13 1,40

5 625 1,8,27,64, 125,2 16,343,5 12 624,6 17,598,56 1,500,409,282, 1 13

Obviously, the values of c4-x3 in the table are not completely square. So the minimum value of c is 6.

Reference answer: 1. 1, d Original formula =

2. C ∶ 52+142 = 221=102+12 ∠ A and ∠C are right angles.

3、D

4、D 5、C 6、A

2. 1, 24182,3, X+Y = 33+43+53+63 = 4324, 15.

3. 1, omit 2, omit 3, and the minimum value of c is 6.