The idea of ??combining numbers and shapes occupies a very important position in the college entrance examination. The combination of "number" and "shape" penetrates each other, combining the precise characterization of algebraic expressions with the intuitive description of geometric figures, making algebra Problems and geometric problems are transformed into each other, so that abstract thinking and image thinking are organically combined. Applying the idea of ??combining numbers and shapes is to fully examine the internal connection between the conditions and conclusions of mathematical problems, analyze their algebraic meanings and reveal their geometric meanings, and combine quantities Relationships and spatial forms are cleverly combined to find problem-solving ideas and solve the problem. To use this mathematical idea, one must be proficient in the geometric meaning of some concepts and operations and the algebraic characteristics of common curves. Difficult magnetic field? 1. Curve y= When 1 (–2≤x≤2) has two intersections with the straight line y=r(x–2) 4, the value range of the real number r. 2. Suppose f(x)=x2–2ax 2, when x∈[ –1, ∞), f(x)>a is always true, find the value range of a. ●Case study? [Example 1] Suppose A={x|–2≤x≤a}, B={y| y=2x 3, and x∈A}, C={z|z=xlt;supgt;2lt;/supgt;, and x∈A}, if C B, find the value range of the real number a. Propositional intention: This question uses The combination of numbers and shapes examines questions about set relation operations. It is a ★★★★ level question. Knowledge base: The key to solving this question is to rely on the method of finding the value range of a quadratic function on the interval to determine the set C. Then use the inequality of C B Transform it into a mathematical language. Error analysis: Candidates are prone to making mistakes when determining the value range of z=x2, –2 This is a special situation. Skills and methods: To solve set problems, first see clearly what the elements are, then "translate" the set language into general mathematical language, then analyze the conditions and conclusion characteristics, and then transform it into graphics Language, use the idea of ??combining numbers and shapes to solve. Solution: ∵y=2x 3 is an increasing function ∴–1≤y≤2a 3 on [–2, a], that is, B={y|–1≤y≤2a 3} Make the image of z=x2. The right end point x=a of the function domain has three different positions as follows: ① When –2≤a≤0, a2≤z≤4, that is, C={z|zlt; supgt; 2lt; / supgt; ≤ z ≤ 4} To make C B, it is necessary and only necessary that 2a 3≥4, a≥ and –2≤a<0. ② When 0≤a≤2, 0≤z≤4 That is, C = {z | 0 ≤ z ≤ 4}. To make C B, it can be seen from the figure: it must and only needs to be solved to ≤ a ≤ 2 ③ When a > 2, 0 ≤ z ≤ a2, that is, C = {z | 0 ≤z≤alt;supgt;2lt;/supgt;}, to make C B must and only need to solve 2＜a≤3④When a＜–2, A= At this time, B=C=, then C B is established. In summary As mentioned above, the value range of a is (–∞, –2)∪[,3].[Example 2] It is known that acosα bsinα=c, acosβ bsinβ=c(ab≠0, α–β≠kπ, k∈ Z) Verification: Proposition intention: This question mainly tests the ability to convert the geometric meaning of mathematical algebraic expressions. It is a ★★★★★-level question. Knowledge support: The key to solving this question is to associate the structure of the conditional expression with the straight-line equation. Then from A The coordinate characteristics of the two points B and B are known to be on the unit circle. Analysis of wrong solutions: It is difficult for candidates to associate the geometric meaning of the conditional expression, which is one of the bottlenecks. How to skillfully use its geometric meaning is the second bottleneck. Skills and methods: good at To discover the geometric meaning of the conditions, we must also analyze the geometric meaning of the conclusion clearly based on the properties of the graph. Only in this way can we skillfully use the method of combining numbers and shapes to solve the problem. Proof: In the plane rectangular coordinate system, point A (cosα, sinα) and point B (cosβ, sinβ) is the two intersection points of the straight line l: ax by=c and the unit circle x2 y2=1 as shown in the figure. Thus: |AB|2=(cosα–cosβ)2 (sinα–sinβ)2=2–2cos (α–β) and ∵The distance from the center of the unit circle to the straight line l is known from the knowledge of plane geometry｜OA｜2–(｜AB｜)

2=d2 is ∴ .● Tips? When applying the idea of ??combining numbers and shapes, you should pay attention to the following transformations of numbers and shapes: (1) Set operations and Venn diagrams (2) Functions and their images (3) Sequence terms and Functional characteristics and function images of summation formulas (4) Equations (mostly binary equations) and the curves of equations are commonly used to formulate numbers: with the help of the number axis; with the help of function images; with the help of the unit circle; with the help of the structural characteristics of mathematical expressions ; With the help of analytic geometry methods. Commonly used numbers to assist shapes include: with the help of the quantitative relationships followed by geometric trajectories; with the help of the combination of operation results and geometric theorems. ● Training to eliminate difficult points? 1. Multiple-choice questions 1. (★★★ ★) The number of real solutions to the equation sin(x– )= x is ( )A.2 B.3 C.4 D. None of the above are true 2. (★★★★★) It is known that f(x)=( x–a)(x–b)–2 (where a＜b, and α and β are the two roots of the equation f(x)=0 (α＜β), then the relationship between the real numbers a, b, α, and β is ( )A.α＜a＜b＜β B.α＜a＜β＜bC.a＜α＜b＜β D.a＜α＜β＜b 2. Fill in the blanks 3. (★★★★★) (4cosθ 3–2t)2 (3sinθ–1 2t)2, (θ, t are parameters), the maximum value is .4. (★★★★★) Known set A={x|5–x≥ }, B= {x | cosx a=0 has different solutions α and β in (0, π). (1) Find the value range of a; (2) Find the value of tan (α β). 6. (★★★★) Let A={(x, y)|y= , a>0}, B={(x, y)|(x–1)2 (y–3)2=a2, a>0}, and A∩B ≠, find the maximum and minimum values ??of a. 7. (★★★★) It is known that A (1, 1) is a point inside the ellipse = 1, F1 is the left focus of the ellipse, and P is the moving point on the ellipse. Find | PF1 The maximum and minimum values ??of || What should be the minimum side length of the window? Reference answer ● Difficult magnetic field 1. Analysis: The curve of the equation y=1 is a semicircle, and y=r(x–2) 4 is a straight line passing through (2, 4). Answer: ( ] 2. Solution 1: From f(x)>a, x2–2ax is always true on [–1, ∞) and 2–a>0 is always true on [–1, ∞). Examine the function g(x)=x2 The graph of –2ax 2–a is located above the x-axis when [–1, ∞]. As shown in the two cases: the conditions for the establishment of the inequality are: (1) Δ=4a2–4(2–a)<0 a∈ (–2, 1) (2) a∈(–3,–2), in summary, a∈(–3, 1). Solution 2: From f(x)＞a x2 2＞a(2x 1) let y1=x2 2, y2=a(2x 1), draw the graphs of two functions in the same coordinate system. As shown in the figure, the straight line l that meets the conditions is located between l1 and l2, and the a values ??corresponding to the straight lines l1 and l2 ( That is, the slope of the straight line) are 1, –3 respectively, so the straight line l corresponds to a∈(–3, 1).●Annihilation Difficulty Training 1. 1. Analysis: Make y1=sin(x–) and The image of y2= x is as shown in the figure. Answer: B2. Analysis: a and b are the two roots of the equation g(x)=(x–a)(x–b)=0. In the same coordinate system, the function f( The images of x) and g(x) are as shown in the figure: Answer: A 2. 3. Analysis: Thinking of the distance formula, the coordinates of the two points are A (4cosθ, 3sinθ), B (2t–3, 1–2t) The geometric shape of point A is an ellipse, and point B represents a straight line. Consider using the distance from the point to the straight line.

Solve from the formula. Answer: 4. Analysis: Solve to get A={x|x≥9 or x≤3}, B={x|(x–a)(x–1)≤0}, and draw the number axis to get. Answer: a＞3 3. 5. Solution: ① Make the graph of y=sin(x)(x∈(0,π)) and y=–. It is known that when |– |＜1 and – ≠, the curve is The straight line has two intersection points, so a∈(–2,–)∪(–,2).②Subtract sinα cosα=–a, sinβ cosβ=–a to get tan, so tan(α β)=3.6. Solution : ∵ The graphic formed by the elements in set A is a semicircle with the origin O as the center and a as the radius; the elements in set B are a circle with the point O′ (1, ) as the center and a as the radius. As shown in the figure ∵A∩B≠, ∴ semicircle O and circle O′ have a common point. Obviously, when semicircle O and circle O′ are circumscribed, a is minimum a a=|OO′|=2, ∴amin=2 –2 When semicircle O and circle O′ are inscribed, the radius of semicircle O is the largest, that is, a is the largest. At this time, a–a=|OO′|=2, ∴amax=2 2.7. Solution: It can be seen that a=3, b=, c=2, left focus F1(–2,0), right focus F2(2,0). Defined by ellipse, |PF1|=2a–|PF2|=6–|PF2|, ∴|PF1||PA| =6-|PF2| When P is at P2 on the extension line of AF2, take the right "=" sign; when P is at P1 on the reverse extension line of AF2, take the left "=" sign. That is, the maximum and minimum of |PA|–|PF2| The values ??are, respectively, - . So the maximum value of |PF1| |PA| is 6, and the minimum value is 6- .8. Solution: This question is actually to find the minimum side length of the square window. Since the width and height of each face of the cuboid are The side length of the face is the smallest, so this face should pass through the window symmetrically to make the side length of the square window as small as possible. As shown in the figure: assuming AE=x, BE=y, then AE=AH=CF=CG= x，BE=BF=DG=DH=y∴ ∴ .