April 2008, 65438+ 0: 00 a.m.-165438+0: 30 a.m.
The second test (1)
1. (The full mark of this question is 20) It is known that A 2+B 2 =1.For all real numbers X that satisfy the condition 0 ≤ x ≤ 1, the inequality A (1–x)–b x (b–x–b x).
Solution: sort out the inequality (1) and substitute it into a 2+b 2 = 1 to get (1+a+b) x2-(2a+1) x+a ≥ 0 (2).
In (2), let x = 0 and get a ≥ 0; ; Let x = 1 and get b ≥ 0. Yizhi 1+a+b > 0，0 & lt； & lt 1,
Therefore, the image (parabola) of the quadratic function y = (1+a+b) x2-(2a+1) x+a has an upward opening, and the abscissa of the vertex is between 0 and1. According to the problem, inequality (2) holds for all real numbers X that satisfy the condition 0 ≤ x ≤ 1, so its discriminant △ = (2a+1) 2–4a (1+a+b) ≤ 0, that is, a b ≥ is determined by equation group (3).
If b is excluded,16a4–16a2+1= 0, so a 2 = or a 2 = And because a ≥ 0,
So a 1 = or a 2 =, so b 1 = or b 2 =. Therefore, the minimum value of a and b is 0, and the values of a and b are a =, b = and a =, b = respectively.
As shown in the figure, circle O and circle D intersect at points A and B, BC is the tangent of circle D, point C is on circle O, and AB = BC.
(1) Prove that point O is on the circumference of circle D;
(2) Let the area of △ABC be S, and find the minimum value of the radius r of circle D. ..
Solution: (1) even OA, OB, OC, AC, because O is the center of the circle and AB = BC, so △OBA∽△OBC, thus ∠OBA =∠OBC, because OD⊥AB, DB⊥BC, so.
(2) Let the radius of circle O be a, and the extension line of BO and AC intersect at point E, so it is easy to know that BE⊥AC. Let AC = 2 y (0
L2 = y2+(a+x)2 = y2+a2+2a x+x2 = 2a 2+2a x = 2a(a+x)= .
Because ∠ABC = 2∠OBA = 2∠OAB =∠BDO, AB = BC, DB = DO, so △BDO∽△ABC.
So =, that is =, so r =, so r 2 = =? = ? () 3 ≥, that is, r ≥, where the equal sign holds when a = y, and when AC is the diameter of the circle O. So the minimum value of the radius r of the circle D is.
3. (The full mark of this question is 25) Let A be a prime number, B be a positive integer, and 9 (2a+b) 2 = 509 (4a+511b) (1).
Find the values of a and b.
Solution: (1) formula is () 2 =, let m =, n =, then n = m 2,
B = = (2 a+b) 2 so 3n–5 11m+6a = 0, so 3m2–51m+6a = 0 (3), from the formula (1), (2
That is, the univariate quadratic equation (3) about m has integer roots, so its discriminant △ = 5112–72a is a complete square number.
Let △ = 5112–72a = t2 (natural number), then 72a = 5112–t2 = (511+t).
Because the parity of 5 1 1+t and 51–t is the same, 511+t ≥ 511,there are only the following situations.
①, ②, ③ and ④ can be obtained by adding two formulas respectively.
36 a+2 = 1022, 18 a+4 = 1022, 12 a+6 = 1022, 6 a+12 =/kloc.
⑤, ⑤, the two formulas are added to get 4 a+ 18 = 1022 respectively, and the solution is a = 251;
2 a+36 = 1022, get a = 493,493 =17× 29 is not a prime number, and discard it. A = 25 1。
At this time, the solution of equation (3) is m = 3 or m = (rounding).
Substitute a = 25 1 and m = 3 into equation (2) to get b = = 7.
The second test (b)
1. A 2+B 2 =1is known. For all real number pairs (x, y) that satisfy the condition of x+y = 1 and x y ≥ 0, the inequality ay2–xy+bx2 ≥ 0 (1) holds. When the product A+Y ≥ 0,
Solution: According to x+y = 1 and x y ≥ 0, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. In the formula (1), let x = 0, y = 1, and get a ≥ 0; ; Let x = 1 and y = 0, and b ≥ 0 is obtained. Substitute y = 1–x into the formula (1) to get a (1–x) 2–x (1–x)+bx2 ≥ 0.
That is, (1+a+b) x2-(2a+1) x+a ≥ 0 (2), Yizhi1+a+b >: 0,0 <; & lt 1,
Therefore, the image (parabola) of the quadratic function y = (1+a+b) x2-(2a+1) x+a has an upward opening, and the abscissa of the vertex is between 0 and1. According to the problem, inequality (2) holds for all real numbers x that satisfy the condition 0 ≤ x ≤ 1
So its discriminant △ = (2a+1) 2–4a (1+a+b) ≤ 0, that is, a b ≥ If B is eliminated from the equation set (3),16a4–16a2+/kloc-. And because a ≥ 0,
So a 1 = or a 2 =, so b 1 = or b 2 =. Therefore, the minimum value of a and b is 0, and the values of a and b are a =, b = and a =, b = respectively.
2. (The full score of this question is 25 points) The question type and solution are the same as the second question in Volume (A).
3. (The full score of this question is 25 points) 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 25 points) The question type and solution are the same as the first question in Volume (B).
2. (The full score of this question is 25 points) The question type and solution are the same as the second question in Volume (A).
3. (The full score of this question is 25 points) Let A be a prime number, B and C be positive integers, and satisfy, and find the value of A (b+c).
Solution: (1) formula is () 2 =, let m =, then 2b–c = (3), so 3n–511m+6a = 0, and n = m 2,
Therefore, 3m2–511m+6a = 0 (4). According to the formula (1), (2a+2b–c) 2 can be divisible by 509.
And 509 is a prime number, so 2a+2b–c can be divisible by 509, so m is an integer.
That is, the univariate quadratic equation (4) about m has integer roots, so its discriminant △ = 5112–72a is a complete square number.
Let △ = 5112–72a = t2 (natural number), then 72a = 5112–t2 = (511+t).
Because the parity of 5 1 1+t and 51–t is the same, 511+t ≥ 511,there are only the following situations.
①, ②, ③ and ④ can be obtained by adding two formulas respectively.
36 a+2 = 1022, 18 a+4 = 1022, 12 a+6 = 1022, 6 a+12 =/kloc.
⑤, ⑤, the two formulas are added to get 4 a+ 18 = 1022 respectively, and the solution is a = 251;
2 a+36 = 1022, get a = 493,493 =17× 29 is not a prime number, and discard it. A = 25 1。
At this time, the solution of equation (3) is m = 3 or m = (rounding).
Substitute a = 25 1 and m = 3 into formula (3) to get 2 b-c = = 7, that is, c = 2 b-7, and substitute into formula (2) to get b-(2 b-7) = 2, so b = 5 and c = 3, so a (b+c).