Generalized Duan Zhiyi
Like the previous three sessions, this year's national high school mathematics competition is still divided into two parts: league and extra test, but this year's test questions are obviously more difficult than last year, and the average score of Shaanxi division has dropped by nearly 60 points. In order to reflect the purpose of this column, the following only explains this year's league test questions for reference.
? First, multiple-choice questions (the full score of this question is 36 points, and each small question is 6 points)
? 1. given a is a given real number, and the number of subsets of the set m = {x | x2-3x-a2+2 = 0, x ∈ r} is ().
? A.1b.2c.4d. Not sure.
? Note: M represents the solution set of equation X2-3x-A2+2 = 0 in the real number range. Because δ = 1+4a2 > 0, m contains two elements. So the set m has 22 = 4 subsets, so choose C.
? 2. Proposition 1: In a cuboid, there must be a point whose height is equal to each vertex.
? Proposition 2: In a cuboid, there must be points with equal distances to all sides;
? Proposition 3: In a cuboid, there must be points with equal distances to all faces.
? Of the above three propositions, the correct one is ().
? A.0 B. 1 C.2 D.3
? Explanation: The proposition 1 is correct because the distance from the center of a cuboid to each vertex is equal. For Proposition 2 and Proposition 3, there are no points in a general cuboid (except a cube) with equal distances to all sides, and there are no points with equal distances to all sides. So this question only has the proposition 1, so choose B.
? 3. Among the four functions y = sin | x |, y = cos | x |, y = | ctgx |, y = LG | sinx |, the even function with π as the cycle and monotonically increasing on (0, π/2) is ().
? A.y=sin|x| B.y=cos|x|
? C.y=|ctgx| D.y=lg|sinx|
? Explanation: Exclusion method can be considered. Y = sin | x | is not a periodic function (it can be judged by drawing), and a is excluded; The minimum positive period of y = cos | x | is 2π, and it is a decreasing function at (0, π/2) without B; Y = | ctgx | is the subtraction function on (0, π/2), excluding C, so D should be selected.
? 4. If ∠ ABC = 60, AC = 12, BC = K, and there is exactly one △ABC, then the range of K is ().
? A.k=8 B.024，
①
4x+5y 24，b 1 1×24- 12×22=0.
? ∴ 2x > 3y, choose one.
Figure 1
? Solution 2: The plane area composed of inequalities ①, ②, x > 0 and y > 0 is the shaded part (excluding the boundary) in the figure 1. Let 2x-3y = 2c, then c represents the intercept of straight line L: 2x-3y = 2c is on the X axis. Obviously, when l passes through point (3,2), the minimum value of 2c is 0.
? Description: (1) This question is similar to the following 1983 national high school mathematics league question:
? Given that the function m = f (x) = ax2-c satisfies: -4 ≤ f (1) ≤- 1,-1 ≤ f (2) ≤ 5, then f(3) should satisfy ().
? A.-7≤f(3)≤26 B.-4≤f(3)≤ 15
? c .- 1≤f(3)≤20d .-28/3≤f(3)≤35/3
? (2) If the value ranges of x and y are obtained from conditions ① and ② respectively, and then the conclusion is drawn from the value range of 2x-y, it is easy to make mistakes. The above scheme 1 adopts the overall idea, and scheme 2 is intuitive and reliable. See [1] for details.
? 2. Fill in the blanks (the full score of this question is 54 points, and each small question is 9 points)
? 7. The short axis length of an ellipse ρ = 1/(2-cos θ) is equal to _ _ _ _ _ _ _ _.
? Note: if you notice that the pole is at the left focus of the ellipse, you can use the special value method; If we pay attention to the geometric significance of eccentricity e and focus parameter p (the distance from the focus to the corresponding directrix), we can also directly calculate the length of the short semi-axis in this question.
Solution 1: by
ρ(0)=a+c= 1，
get
a=2/3，
ρ(π)=a-c= 1/3，
c= 1/3。
? So b =/3, so 2b = 2/3.
? Solution 2: From E = C/A = 1/2, P = B2/C = 1 and B2 = A2-C2, B =/3. So 2b = 2/3.
? Note: this is a test that conforms to the outline and is not within the scope of the college entrance examination.
? 8. If the complex numbers z 1 and z2 satisfy | Z 1 | = 2 and | z3 | = 3,3z1-2z2 = (3/2)-i, then z1z2 = _ _ _ _ _.
? Note: The solution given by referring to the answer is skillful. According to the characteristics of the problem, the triangular form of the complex number seems to be more in line with the students' thinking characteristics and not complicated.
? Let z 1 = 2 (COS α+ISIN α) and Z2 = 3 (COS β+ISIN β), then the necessary and sufficient conditions for 3Z 1-2Z2 = (3/2)-I and complex numbers to be equal are obtained.
6(cosα-cosβ)=3/2，
6(sinα-sinβ)=- 1
that is
- 12 sin((α+β)/2)sin((α-β)/2)= 3/2，
12 cos((α+β)/2)sin((α-β)/2)=- 1。
? Divided by two formulas, TG (α+β)/2) = 3/2.
? From the general formula, we get
? sin(α+β)= 12/ 13，cos(α+β)=-5/ 13。
? So z1.z2 = 6 [cos (α+β)+isin (α+β)]
? =-(30/ 13)+(72/ 13)i。
? Note: this problem can also be solved by using the geometric meaning of complex numbers.
? 9. If the side length of the cube ABCD-A 1 B11is1,the distance between the straight line A 1C 1 and BD 1 is _ _ _ _ _ _
? Note: This is a problem of finding the distance between two straight lines on different planes. There are many solutions. Here is a basic solution.
Figure 2
? In order to ensure that the line segment representing the distance is perpendicular to A 1C 1 and BD 1, it is suggested that one of the straight lines be placed on the vertical plane of the other straight line first. Therefore, if BDD 1B 1 is the diagonal plane of a cube, BDD 1 on a1⊥. and BD 1. Let a1c1∩ b1d1= 0, so that OH⊥BD 1 is in BDD 1b 1.
? 10. The solution set of inequality | (1/log1/2x)+2 | > 3/2 is _ _ _ _ _ _ _ _.
? Note: From the appearance, this is an absolute inequality. First, get log 1/2x 0.
? So x > 4, or 1 < x < 22/7, or 0 < x < 1
? 1 1. The range of the function y = x+ is _ _ _ _ _ _ _ _.
? Note: Square the root sign first.
? If (y-x) 2 = x2-3x+2, then x = (y2-2)/(2y-3).
? If y≥x, y ≥ (y2-2)/(2y-3).
? The solution is 1 ≤ y < 3/2 or y ≥ 2.
? Since the lower limit of 0 can be reached, the value range of the function is [1, 3/2]∩[2, +∞).
? Note: (1) After the reference answer is 1 ≤ y < 3/2 or y≥2, it takes a long time to verify, which is really unnecessary.
? (2) This problem can also be solved by trigonometric substitution method and mirror image method, but it is more complicated, so readers might as well give it a try.
Figure 3
? 12. Planting ornamental plants in six areas of a regular hexagon (as shown in Figure 3) requires planting the same plant in the same block and planting different plants in two adjacent blocks. There are four different plants to choose from, so there are _ _ _ _ _ _ _ _ kinds of planting schemes.
? Note: For the convenience of description, we mark six areas with letters A, B, C, D, E and F, and the number of plants planted in blocks A, C and E is divided into the following three categories.
? (1) If A, C and E are the same strain, there are four methods. After planting A, C and E, B, D and E can choose one plant from the remaining three plants (repetition is allowed), and each plant has three methods. At this point, * * * has 4× 3× 3 = 65438+.
? (2) If A, C and E are two plants, there is P42 method. When planting A, C and E, if A and C are the same, there are three methods for B and two methods for D and F; If c, e or e and a are the same, they are the same (just in different order). At this time, * * has P43× 3 (3× 2× 2) = 432 methods.
? (3) If there are three kinds of plants, A, C and E, there is P43 method. At this time, there are two methods for B, D and F. At this time, * * has P43× 2× 2 = 192 methods.
? According to addition principle, there are n =108+432+192 = 732 planting schemes.
? Note: this topic is called circular arrangement problem.
? Third, solve the problem (the full score of this question is 60 points, and each small question is 20 points)
? 13. let {an} be arithmetic progression, {{bn}} be geometric progression, B 1 = A 12, B2 = A22, B3 = A32 (A 1 < A2). And (B 1+B2+…+.
? Explanation: This is a basic question about arithmetic and geometric series. The first three terms of series {an} and {{bn}} satisfy bi = AI2 (I = 1, 2,3), so the relationship between the first term of series {an} a 1 and the tolerance d can be determined. The values of a 1 and d can be obtained from (B 1+B2+…+BN) =+ 1.
? Let the tolerance of {an} be d, and from A 1 < A2, d > 0 is obtained.
? From B22 = B 1b3, we get A24 = A 12a32.
? ∴ A22 = a 1=a2=a3 (leave it alone, otherwise A1= A2 = A3)
? Or a22 =-a 1a3.
? ∴(a 1+d)2=-a 1(a 1+2d)，
? That is 2A 12+4A 1d+D2 = 0.
? D = (-2) A 1。
? If D = (-2-) A 1, then q = a22/a12 = (+1) 2 >1,which does not meet the requirements.
? If d = (-2+) a 1, then q = a22/a 12 = (- 1) 2.
? From (B 1+B2+…+BN) =+ 1, we get.
? b 1/( 1-q)=+ 1，
? That is, a12/(1-(-1)) 2 =+1. The solution is A 12 = 2.
? A 1 0, so yp =
? YP should be maximized and xP should be minimized. It is easy to know that when m = a, (XP) min = a-2a2. Therefore, (YP) max = 2. Therefore, (s △ OAP) max = a.
? When m = (A2+ 1)/2, XP =-A2. Therefore YP =, so s △ OAP = (1/2) a.
? Let's compare the size of A and (1/2) A. 。
? ∵()2-(( 1/2))2
? =…=-( 1/4)(3a- 1)(a- 1)。
? ∴ when 0 < a ≤ 1/3, a ≤ (1/2) a;
? When 1/3 < A < 1/2, a > (1/2) a.
Therefore, (s △ OAP) max =
( 1/2)a
(0 A3 > A4 > A5 > A6) to form a module as shown in Figure 4. How to choose the resistance in assembly to minimize the total resistance of the module? Prove your conclusion.
Note: Assume that the total resistance of a 6-resistor module (as shown in Figure 5) is RFG. When RI = AI (I = 3, 4, 5, 6) and R 1, R2 is any arrangement of a 1 and a2, and RFG is the smallest.
? Through the gradual adjustment method, it is proved as follows:
Figure 5
? (1) When R 1 and R2 are connected in parallel, the resistance value of the obtained element R satisfies1/r = (1r1)+(1/R2). If R 1, R2 and R are interchanged, R remains unchanged.
? (2) Let the total resistance of the assembly with three resistors (as shown in Figure 6) be RAB, then
? RAB =(r 1r 2/(r 1+R2))+R3 =(r 1r 2+r 1r 3+r2r 3)/(r 1+R2)。
? Obviously, the greater R 1+R2, the smaller the RAB. Therefore, in order to minimize RAB, R3 must be the smallest of the three resistances.
Figure 6
Figure 7
? (3) If the total resistance of the assembly with four resistors (as shown in Figure 7) is RCD, then
? 1/ RCD =( 1/RAB)+( 1/R4)
? =(r 1r 2+r 1r 3+r 1 R4+r2r 3+r2r 4)/(r 1r2r 4+r 1r3r 4+r2r3r 4)。
? Remember that S 1 =, S2 =, then S 1, S2 are fixed values. So rcd = (S2-r1r2r3)/(s1-r3r4).
? Obviously, when R3