First attempt
1. Multiple-choice question: (The full score of this question is 42 points, and each small question is 7 points)
1. If they are all integers and satisfy, then (b)
A. 1。 b . 2 c . 3d . 4。
2. If the real number satisfies the equation, then the possible maximum value is (c).
A.0. B. 1。 C.2. D.3
3. If it is two positive numbers, and (c)
A.。 B. c。 d。
4. If the two roots of the equation are also the roots of the equation, the value of is (a).
A.- 13.b-9。 C.6. D. 0。
5. In △, it is known that D and E are points on AB side and AC side respectively, then (b)
A. 15。 b20。 C.25D.30
6. For natural numbers, the sum of their numbers is, for example, then (d).
A.28062. B.28065. C.28067. D.28068
Fill in the blanks: (The full score of this question is 28 points, and each small question is 7 points)
1. Known real numbers satisfy equations, 13.
2. The image of quadratic function intersects with the positive axis direction at points A and B, and intersects with the positive axis direction at point C. If,, is known, then.
3. In the isosceles right angle △ABC, AB = BC = 5, P is a point in △ABC, PA =, PC = 5, then Pb = _ _ _ _ _
4. Arrange several red and black balls in a row, requiring both balls to appear, and any two balls with five or 10 balls in the middle must be of the same color. According to this requirement, at most _ _ _15 _ _ balls can be placed.
The second test (1)
1. (The full mark of this question is 20) Let the integer () be the length of three sides of a triangle, and satisfy it, and find the number of triangles with a circumference not exceeding 30.
This solution can be obtained from the known equation.
①
Order, then, is all natural numbers.
Then, equation (1) becomes, that is
②
Because they are all natural numbers, the judgment is easy to know, so there are only two groups in Equation ②: and.
(1) when,, is the length of three sides of a triangle, so, that is, solving. And because the perimeter of the triangle does not exceed 30, it is solved. So you can take the values of 4, 5, 6, 7 and 8, and you can get five qualified triangles accordingly.
(2) When, is the length of three sides of a triangle, then, it is the solution. And because the perimeter of the triangle does not exceed 30, that is, the solution. So you can take the values of 2, 3, 4, 5, 6 and 7, and you can get six qualified triangles accordingly.
Generally speaking, the number of triangles with perimeter less than 30 is 5+6 = 1 1.
2. (The full mark of this question is 25) It is known that the bisector of ∠C in the isosceles triangle △ABC intersects with the AB side at point P, m is the tangent point of the inscribed circle ⊙I of △ABC, and MD//AC intersects with ⊙I at point D, which proves that PD⊙.
It is proved that the tangent PQ (tangent point Q) of point P is ⊙ I, and the intersection BC is at point N. 。
Because CP is the bisector of ∠ACB, ACP = BCP.
And because PA and PQ are tangents of ≥I, ∠ APC = ∠ NPC.
And CP is male, so △ ACP △ NCP, so ∠ PAC = ∠ PNC.
From nm = qn and ba = BC, so △ qnm ∽△ BA=BC, so ∠ NM=QN = ∠ ACB, so MQ//AC.
Because MD//AC, MD and MQ are a straight line.
Both points Q and D are on ⊙i, so points Q and D coincide, so PD is the tangent of ⊙ I. 。
3. (The full mark of this question is 25) The image of the known quadratic function passes through two points, p and q.
(1) If they are all integers, and.
(2) Let the intersection of the image of the quadratic function with the axis be A and B, and the intersection with the axis be C. If both roots of the equation are integers, find the area of △ABC.
The solution points p and q are on the image of quadratic function, so,
Solve,.
(1) is known by.
It's another integer, so ...
(2) Let be two integer roots of the equation, and.
From the relationship between roots and coefficients, we can get, eliminate, get,
Multiply both sides by 9 at the same time to get, decompose the factor, and get.
So either or or or or or or.
Solve or or or or or or or
It is also an integer, so the last three sets of solutions are discarded, so.
Therefore, the analytical formula of quadratic function is.
The coordinates of point A and point B are (1, 0) and (2,0) respectively, and the coordinates of point C are (0,2), so the area of △ABC is.
The second test (b)
1. (The full mark of this question is 20) Let an integer be the length of three sides of a triangle, and satisfy, and find the number of triangles with a perimeter not exceeding 30 (congruent triangles only calculates 1 time).
Solutions can be hypothetical and can be obtained from known equations.
①
Order, then, is all natural numbers.
Then, equation (1) becomes, that is
②
Because they are all natural numbers, the judgment is easy to know, so there are only two groups in Equation ②: and.
(1) when,, is the length of three sides of a triangle, so, that is, solving. And because the perimeter of the triangle does not exceed 30, it is solved. So you can take the values of 4, 5, 6, 7 and 8, and you can get five qualified triangles accordingly.
(2) When, is the length of three sides of a triangle, then, it is the solution. And because the perimeter of the triangle does not exceed 30, that is, the solution. So you can take the values of 2, 3, 4, 5, 6 and 7, and you can get six qualified triangles accordingly.
Generally speaking, the number of triangles with perimeter less than 30 is 5+6 = 1 1.
2. (The full mark of this question is 25) The question type and solution are the same as the second question in Volume (A).
3. (The full mark of this question is 25) The question type and solution are the same as the third question in Volume (A).
The second test (c)
1. (The full mark of this question is 20) Questions and solutions are the same as the first question in Volume (B).
2. (The full mark of this question is 25) The question type and solution are the same as the second question in Volume (A).
3. (The full score of this question is 25) Let it be a prime number greater than 2, and k is a positive integer. If at least one abscissa of the two intersections between the function image and the X axis is an integer, find the value of k. 。
According to the meaning, at least one of the two roots of the equation is an integer.
According to the relationship between roots and coefficients, there are
①
(1) If, the equation is and has two integer roots.
(2) If, then.
Because it is an integer, if at least one of them is an integer, then they are all integers.
Because it is a prime number, we can know from the formula (1).
If set, you can set (where m is a non-zero integer), which can be obtained from the formula (1).
Therefore, that is.
Again, so, this is
②
If m is a positive integer, then,, therefore, contradicts Equation 2.
If m is a negative integer, then,, therefore, contradicts Equation 2.
So an equation can't have integer roots.
To sum up,.