高中竞赛教程4.4.5 典型例题分析

高中数学竞赛中的解题技巧案例分析高中数学竞赛中的解题技巧案例分析数学竞赛作为一种常见的学科竞赛，对于参加者来说是一项较具挑战性的挑战，尤其对于高中生来说更是如此。

在数学竞赛中，解题技巧是至关重要的，下面我们将分析几个典型的数学竞赛题目，分析其解题技巧。

1. 周长为50的矩形的面积最大是多少？解题技巧：由于周长为50，设矩形长为x，宽为y，则2x+2y=50，即x+y=25。

又因面积为xy，根据均值不等式，可得到最大面积为625。

2. 方程x^2-10x+m=0，若有两个实根，则m的取值范围是多少？解题技巧：由于方程的两个实根，因此判别式D>0，即10^2-4m>0，解得m<25。

又由于方程只有两个实根，因此判别式D≠0，即10^2-4m≠0，解得m≠25。

所以m的取值范围为（-∞，25）。

3. 在平面直角坐标系中，以原点O为圆心，以3为半径画一个圆，以（1，2）为圆心画一个半径为4的圆，则这两个圆相交的面积是多少？解题技巧：先求出两个圆的面积，由于圆的面积公式为S=πr^2，因此原点O为圆心、3为半径的圆的面积为9π，（1，2）为圆心、4为半径的圆的面积为16π。

然后再求出其公共部分的面积，由于这两个圆心之间的距离为√（2^2+1^2）=√5，因此所求面积为弓形的面积，由于弓形的面积公式为S=（θ/360）πr^2，其中θ为弧度，即弧度角AOB和A'OB'的角平分线所成的角度，可算出θ=2sin^-1（1/4√5），代入公式求解可得到所求面积为4π-√15。

4. 若a,b,c,d,x均为正整数且满足abcd=4x^4，则a+b+c+d的最小值是多少？解题技巧：由于abcd=4x^4，因此可以将abcd分解质因数，得到a=b=c=d=2x，因此a+b+c+d=8x。

因此所求最小值为8。

总结：以上四道数学竞赛题目涉及的解题技巧涵盖了常见的数学解题方法，如均值不等式、判别式等。

专题01 坐标系【知识网络】【考情分析】 考纲要求①理解坐标系的作用。

②了解在平面直角坐标系伸缩变换作用下平面图形的变化情况。

③能在极坐标系中用极坐标表示点的位置，理解在极坐标系和平面直角坐标系中表示点的位置的区别，能进行坐标和直角坐标的互化。

④能在极坐标系中给出简单图形（如过极点的直线、过极点或圆心在极点的圆）的方程，通过比较这些图形在极坐标系和平面直角坐标系中的方程，理解用方程表示平面图形时选择适当坐标系的意义。

⑤了解柱坐标系、球坐标系中表示空间中点的位置的方法，并与空间直角坐标系中表示点的位置的方法相比较，了解它们的区别。

考情分析高频考点 常见曲线的极坐标方程、直角坐标和极坐标的互化考查形式 通过近几年高考命题趋势看，本部分重点考查直角坐标方程和极坐标方程的互化，常见曲线的极坐标方程也是考查的重点，主要考查基础知识、基本技能， 题型一般为解答题，难度中等．命题角度 结合直线与圆、圆锥曲线、三角函数及恒等变换、向量等知识考查 常见题型 解答题备考要求对知识点进行归纳整理、掌握常见曲线的极坐标方程、直角坐标和极坐标之间的互化公式及其运用等．【知识详单】1.平面直角坐标系的作用通过平面之间坐标系，实现了平面上的点与坐标(有序实数对)，曲线与方程建立联系，从而使得数与形的结合．2. 平面直角坐标系中的伸缩变换(1)平面直角坐标系中方程表示图形，那么平面图形的伸缩变换就可归结为坐标伸缩变换，这就是用代数方法研究几何变换．(2)平面直角坐标系中的坐标伸缩变换：设点P (x ，y )是平面直角坐标系中任意一点，在变换φ：⎩⎪⎨⎪⎧x ′＝λx ，λ>0y ′＝μy ，μ>0的作用下，点P (x ，y )对应到点P ′(x ′，y ′)，称φ为平面直角坐标系中的坐标伸缩变换，简称伸缩变换．坐标系直角坐标系 柱坐标系和球坐标系极坐标系极坐标方程及其应用极坐标和极坐标系的概念直角坐标和伸缩变换极坐标与直角坐标的互化3.极坐标与极坐标系极点、极轴、长度单位、角度单位和它的方向构成极坐标系的四要素，缺一不可。

4.4免疫学的应用（教学设计） -2023-2024学年高二上学期生物人教版（2019）选择性必修1一、教学内容分析本节课的主要教学内容是免疫学的应用，这是2023-2024学年高二上学期生物人教版（2019）选择性必修1的第四章第四节。

在这一节中，我们将学习免疫学的应用，包括免疫预防、免疫诊断、免疫治疗等方面的内容。

教学内容与学生已有知识的联系主要体现在学生已经学习了免疫系统的基本结构和功能，以及免疫反应的类型和特点。

这些基础知识为学生学习免疫学的应用奠定了基础。

通过学习免疫学的应用，学生可以更好地理解和运用免疫学知识，解决实际问题。

二、教学目标本节课的教学目标针对的是高二学生，他们已经具备了一定的生物学基础知识，但可能对免疫学的应用理解不够深入。

因此，本节课的教学目标设定为：1. 让学生了解免疫学的应用，包括免疫预防、免疫诊断和免疫治疗等方面的内容。

2. 培养学生运用免疫学知识解决实际问题的能力，提高他们的实践操作能力。

3. 激发学生对免疫学研究的兴趣，培养他们的科学探究精神和创新思维能力。

4. 帮助学生建立正确的免疫学观念，提高他们的科学素养。

为了达到以上教学目标，我们将采取以下教学策略：1. 通过实例讲解，帮助学生理解免疫学的应用，提高他们的理解和应用能力。

2. 开展实践活动，让学生亲身体验免疫学的应用，培养他们的实践操作能力。

3. 引入科学探究，激发学生的兴趣，培养他们的科学探究精神和创新思维能力。

4. 加强课堂互动，鼓励学生提问和发表观点，提高他们的参与度和积极性。

三、学习者分析本节课的授课对象是高二学生，他们已经掌握了以下相关知识：1. 生物学基础知识：学生已经学习了生物学的基本概念、原理和方法，包括细胞结构与功能、生物的多样性、遗传与进化、生态学等。

2. 免疫系统的基本结构和功能：学生已经了解了免疫系统的基本组成，包括抗原、抗体、免疫细胞和免疫分子等，以及免疫系统的作用机制，包括抗原识别、免疫应答等。

演练篇核心考点AE卷高考数学2021年6月■安徽省利辛高级中学胡彬1.在平面直角坐标系scOy中，以坐标原点O为极点，工轴的非负半轴为极轴建立极坐标系，点MSo,仇)(卩0>0)在曲线C-.p=4cos(9上，直线I过点A(4,今)且与OM垂直，垂足为P。

⑴当久=专时，求在直角坐标系下点M的坐标和直线I的方程；(2)当点M在曲线C上运动且点P在线段OM上时，求点P在极坐标系下的轨迹方程。

2.在平面直角坐标系工Oy中，直线I的f4rc=1H-—J9o参数方程为彳&为参数)。

以坐3)=1+亍标原点o为极点，工轴的非负半轴为极轴建立极坐标系，曲线c的极坐标方程为°=愆COS(°o(1)求直线I的极坐标方程和曲线C的直角坐标方程；(2)已知直线Z与曲线C交于A,E两点，试求两点间的距离。

3.已知曲线C的极坐标方程是°=1,以极点为坐标原点，极轴为鼻轴的非负半轴建立平面直角坐标系，曲线C经过伸缩变换=愆工，得到曲线E。

直线Z：\y/=y,cc=l-\-t, y=/3i为参数)与曲线E交于A,B两点。

(1)设曲线C上任意一点为M(鼻，夕)，求*—3夕的最小值；(2)求曲线E的直角坐标方程，以及直线I被曲线E截得的弦AJB的长。

4.在平面直角坐标系a:Oy中，曲线C的参数方程为<W)心参数i1坐标原点O为极点，工轴的非负半轴为极轴建立极坐标系。

(1)写出曲线C的普通方程和极坐标方程；(2)M,N为曲线C上两点，若OM丄ON,求|MN|的最小值。

5.在平面直角坐标系acOy中，直线I的(工=1+icos9,参数方程为(t为参数)。

以b=tsin0坐标原点O为极点，工轴的非负半轴为极轴建立极坐标系，点P的极坐标为(1,0),曲线2一123cos20+4sin20°(1)求直线I的普通方程和曲线C的直角坐标方程；(2)若直线Z与曲线C交于A,B两点，求\PA\+\PB\的取值范围。

§3．5 典型例题分析例1、绷紧的肥皂薄膜有两个平行的边界，线AB 将薄膜分隔成两部分(如图3-5-1)。

为了演示液体的表面张力现象，刺破左边的膜，线AB 受到表面张力作用被拉紧，试求此时线的张力。

两平行边之间的距离为d ，线AB 的长度为l (l ＞πd/2)，肥皂液的表面张力系数为σ。

解：刺破左边的膜以后，线会在右边膜的作用下形状相应发生变化(两侧都有膜时，线的形状不确定)，不难推测，在l ＞πd/2的情况下，线会形成长度为)2/(21d l x π-=的两条直线段和半径为d/2的半圆，如图3-5-2所示。

线在C 、D 两处的拉力及各处都垂直于该弧线的表面张力的共同作用下处于平衡状态，显然 ∑=i f T 2式中为在弧线上任取一小段所受的表面张力，∑i f 指各小段所受表面张力的合力，如图3-5-2所示，在弧线上取对称的两小段，长度均为r △θ，与x 轴的夹角均为方θ，显然θσ∆⋅==r f f 221而这两个力的合力必定沿x 轴方向，(他们垂直x 轴方向分力的合力为零)，这样θθσ∆⋅==cos 221r f f x x所以∑∑==∆=d r r f i σσθθσ24cos 2图3-5-1图3-5-2因此 d T σ=说明 对本题要注意薄膜有上下两层表面层，都会受到表面张力的作用。

例2、在水平放置的平玻璃板上倒一些水银，由于重力和表面张力的影响，水银近似呈圆饼形状(侧面向外凸出)，过圆盘轴线的竖直截面如图3-5-3所示。

为了计算方便，水银和玻璃的接触角可按180º计算，已知水银密度33106.13m kg ⨯=ρ，水银的表面张力系数m N a 49.0=。

当圆饼的半径很大时，试估算厚度h 的数值大约是多少(取一位有效数字)？分析：取圆饼侧面处宽度为△x ，高为h 的面元△S ，图3-5-3所示。

由于重力而产生的水银对△S 侧压力F ，由F 作用使圆饼外凸。

但是在水银与空气接触的表面层中，由于表面张力的作用使水银表面有收缩到尽可能小的趋势。

典型例题分析例1．分别从方差为20和35的正态总抽取容量为8和10的两个样本，求第一个样本方差是第二个样本方差两倍的概率的范围。

解 以21S 和22S 分别表示两个（修正）样本方差。

由222212σσy x S S F =知统计量2221222175.13520S S S S F ==服从F 分布，自由度为（7，9）。

1） 事件{}22212S S =的概率 {}{}05.320352352022222122212221===⎭⎬⎫⎩⎨⎧⨯==⎭⎬⎫⎩⎨⎧===F P S S P S S P S S P因为F 是连续型随机变量，而任何连续型随机变量取任一给定值的概率都等于0。

2） 现在我们求事件{}二样本方差两倍第一样本方差不小于第=A 的概率：{}{}5.322221≥=≥=F P S S P p 。

由附表可见，自由度9,721==f f 的F 分布水平α上侧分位数),(21f f F α有如下数值：)9,7(20.45.329.3)9,7(025.005.0F F =<<=。

由此可见，事件A 的概率p 介于0.025与0.05之间；05.0025.0<<p 。

例2．设n X X X ，，， 21是取自正态总体),(2σμN 的一个样本，2s 为样本方差，求满足不等式95.05.122≥⎭⎬⎫⎩⎨⎧≤σS P 的最小n 值。

解 由随机变量2χ分布知，随机变量σ/12S n ）（-服从2χ分布，自由度1-=n v ，于是，有{}{}95.0)1(5.1)1(5.1)1(2,05.02222=≤≥-≤=⎭⎬⎫⎩⎨⎧-≤-=v v v P n P n S n P χχχσ 其中2v χ表示自由度1-=n v 的2χ分布随机变量，2,05.0v χ是自由度为1-=n v 的水平05.0=α的2χ分布上侧分位数（见附表）。

我们欲求满足2,05.015.1v n χ≥-）（的最小1+=v n 值，由附表可见226,05.0885.3839)127(5.1χ=>=-， 22505.0652.375.401265.1，）（χ=<=-。

竞赛数学高中试题及答案试题一：多项式问题题目：已知多项式 \( P(x) = x^3 - 3x^2 + 2x - 5 \)，求 \( P(2) \) 的值。

解答：将 \( x = 2 \) 代入多项式 \( P(x) \) 中，得到：\[ P(2) = 2^3 - 3 \times 2^2 + 2 \times 2 - 5 = 8 - 12 + 4 -5 = -5 \]试题二：几何问题题目：在直角三角形 ABC 中，角 C 是直角，若 \( AB = 10 \) 且\( AC = 6 \)，求斜边 BC 的长度。

解答：根据勾股定理，直角三角形的斜边 \( BC \) 可以通过以下公式计算：\[ BC = \sqrt{AB^2 - AC^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \]试题三：数列问题题目：给定数列 \( a_n = 2n - 3 \)，求数列的前 5 项。

解答：根据数列公式 \( a_n = 2n - 3 \)，我们可以计算出前 5 项：\[ a_1 = 2 \times 1 - 3 = -1 \]\[ a_2 = 2 \times 2 - 3 = 1 \]\[ a_3 = 2 \times 3 - 3 = 3 \]\[ a_4 = 2 \times 4 - 3 = 5 \]\[ a_5 = 2 \times 5 - 3 = 7 \]数列的前 5 项为：-1, 1, 3, 5, 7。

试题四：概率问题题目：一个袋子里有 5 个红球和 3 个蓝球，随机抽取 2 个球，求抽到一个红球和一个蓝球的概率。

解答：首先计算总的可能组合数，即从 8 个球中抽取 2 个球的组合数：\[ \text{总组合数} = \binom{8}{2} = \frac{8 \times 7}{2} = 28 \]然后计算抽到一个红球和一个蓝球的组合数：\[ \text{有利组合数} = \binom{5}{1} \times \binom{3}{1} = 5 \times 3 = 15 \]所以，抽到一个红球和一个蓝球的概率为：\[ P = \frac{\text{有利组合数}}{\text{总组合数}} =\frac{15}{28} \]试题五：函数问题题目：若函数 \( f(x) = x^2 - 4x + 4 \)，求 \( f(x) \) 的最小值。

代数竞赛试题及答案高中试题一：设\( a \)和\( b \)是实数，且满足\( a^2 - 4ab + 4b^2 = 0 \)。

求\( a \)和\( b \)的值。

答案：将给定的方程\( a^2 - 4ab + 4b^2 = 0 \)视为关于\( a \)的一元二次方程，可以写成\( (a - 2b)^2 = 0 \)。

因此，\( a - 2b = 0 \)，得到\( a = 2b \)。

试题二：解不等式：\( |x - 3| + |x + 1| \geq 4 \)。

答案：根据绝对值的性质，我们可以将不等式分为三个区间进行讨论：1. 当\( x < -1 \)时，不等式变为\( -(x - 3) - (x + 1) \geq 4 \)，即\( -2x + 2 \geq 4 \)，解得\( x \leq -1 \)。

2. 当\( -1 \leq x < 3 \)时，不等式变为\( (x - 3) - (x + 1)\geq 4 \)，即\( -4 \geq 4 \)，这是不成立的，所以这个区间没有解。

3. 当\( x \geq 3 \)时，不等式变为\( (x - 3) + (x + 1) \geq 4 \)，即\( 2x - 2 \geq 4 \)，解得\( x \geq 3 \)。

综合以上三个区间，不等式的解集为\( x \in (-\infty, -1] \cup [3, \infty) \)。

试题三：已知\( \frac{1}{x - 1} + \frac{1}{x} = 1 \)，求\( x \)的值。

答案：将方程\( \frac{1}{x - 1} + \frac{1}{x} = 1 \)进行合并，得到\( \frac{x + (x - 1)}{x(x - 1)} = 1 \)。

化简得\( 2x - 1 = x^2 - x \)。

整理后得到\( x^2 - 3x + 1 = 0 \)。

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If I exist, find the value of m; If not, explain why.In the middle, already, the area is equal to 6.The length of the trilateral ( I );(II) is set P (including boundary), P to the trilateral AB, BC, AB, and o value range.In the sequence, a given non~zero integer.⑴if,(2) proof: there must be an infinite number of two different constants.We know that the ellipse, at (0, 1), is a right Angle, and the sides AB, BC, and ellipse are handed in B and C. If the maximum value of the ABC area of the triangle is equal to the value of this.Figure, ellipse:, the vertex of the ellipse.(I ), if if and only if the points on the ellipse in the oval vertex, obtain the maximum and minimum values, the value range of;(II) if the point on the ellipse to the maximum distance between the focus, the minimum value for, and fellowship with straight line, two points (not elliptical or vertex), and meet. Try to research: a fixed line? If you have a fixed point, ask for a fixed point coordinate. If it is only a fixed point, please explain why.As shown in the four pyramid, the base is the square of the length of the side, the side is a positive triangle, and is perpendicular to the base.(1)the volume of the four pyramid:(2)is there a point on the side? Please explain why.(=15 out of 10)The equation:.(1)if the equation represents the circle, the range of the real number;(2)when the equation represents the circle, if the circle and theline are intersecting at two points, and the value of the number of real Numbers is true:(3)in (2), if the coordinates of the fixed point are (1, 0), the point is the moving point on the line segment, and the slope of the line is evaluated.Known elliptic C (), the centrifugal rate is the distance between the two lines.(1) the value of the request: (2) instruction in A coordinate (6, 0), for the fixed point on the ellipse C B, based on A rectangular vertices, A isosceles right-angle delta ABP (letters A, B, P) clockwise, strives for the trajectory equation of point P.Tri-prism, as shown in figure 15, is neutral. ( I ) proof: / / plane: If (II), the distance from point to plane for: (III) when why value, the sine value of dihedral Angle E - BC1 - C?(=15 out of 15)There's a bunch of points on the plane. . . For each positive integer, the point is located on the image of the function.(1)to prove: the series is an arithmetic sequence:(2)the area of the set is,Proof: for any, both.0 17.In the quadratic function, the real number is equal to 0.(1) (2) the equation has a solution in (0, 1).All the edges of the triangular prism are,Lateral surface, and.(1)the distance between the surface and the distance:(2)the degree of dihedral Angle between the sides and the bottom surface.19.Let's say that the vector is the unit vector of the X-axis in the rectangular coordinate plane, and the Y-axis is going up, and if the vector, and then.(1)the equation of the path that satisfies the above conditions;(2)let's ask if there is a constant. Prove your conclusion.We know the parabola. I'11 do it in a straight line, and I' 11 have a parabola in B and C.I want to find the equation of the trajectory of the center of the ABC of ABC.In the series of columns, and satisfying. Test certificate:The two focus of the ellipse C: = 1 (a > b > 0) is Fl (~c, 0), and M is a point on the ellipse, and satisfies = 0. ( I ) the scope of eccentricity e: (II) set up the slope to k (k indicates 0) linear 1 with elliptical C in different points A, B, Q is the midpoint of AB, can ask A, B two symmetrical about point P, Q) straight line? If you can find out the range of k, if you can't, please explain why.The function f (x), which is defined on R, is also met:(1)(R, a is constant);(2); (3) at that time, it was less than 2.0: ( I ) function analytic expression; (II) constant a scope.Put the number of positive and odd Numbers in the following trianglelist:one3 57, 9, 11Set is the number of first rows from the top to the bottom of the triangle table.(I) if, the value of the request: (II) the inverse function of a known function is, if you take the sum of the number of rows from the top to the next in the triangle table, the first n terms of the sequence.If, if, satisfying, the maximum value of the request.Set the function defined on [0, 2] to satisfy the following conditions:There is always, and there is:For, if, then.Proof: (1) () : (2),.Solve the inequality.The tolerances of the non-negative arithmetic sequence, the first nentries of the sequence, and the proof:(1)if, and:(2)if the.We know the number line satisfies.(I ) series of general term formula;(II), sequence and referred to in the preceding paragraph: (III), referred to in the preceding paragraph and to sequence.Proof: to arbitrary.The answer to the high school math contest, answer the questionIt's two different positive Numbers, and they' re satisfied, and all possible integers, c, make it.Solution: by the way, so,The resulting.Again, because of it,... Four pointsAgain, because of the order, ... 6 pointsAt the time, the t monotonically increasing, so.Therefore, you can take 1, 2, 3 .............. 10 pointsF (n) = monotonically increasing, f (1) = leastSo the >,Three solutions:We can tell from the topic.So, the tolerance is equal to 2. Thus availableProve that: (1), 4 .Similarly,The five solutions (1) group the Numbers:Because 1 + 2 + 3 +. . . + 62 = 1953; 1 + 2 + 3 +. . . + 63 = 2016,So the 2010 item in the sequence is the seventh of the 63. 一一TO points(2) the grouping by above may know, every odd number groups in a 1, so the 2010th 1 appeared in the 4019 group, is located in the group and 4019 in the group 1 4019th, so a value of 1 2010 serial number of the item for the (1 + 2 + 3 +. . . + 4018) + 2010 =809428. 17Solution: the number of balls of red, black and white in a bag is, and it is(* 1)------------------------- 5Itis. That isIt is. (* 2)And there. So you have a 5 in the middle. So let's do that. Let's put it in (* 1)-10 pointsAt this point, y is going to be 1, 2, 8, 9 (accordingly, 9, 8, 2, 1),A total of 9 kinds of method. And you could say that when y is equal to 5 or z is equal to 5, there are 9 ways to do it, but sometimes there are two ways to do it. So there's a total9 times 3 minus 2 is 25. 一 一 一 一 一 一 一 -177 solution: at the time,Namely,So, the first term is equal to the ratio of the common ratio,, hence.So when it's even, when it's odd, it's going to be odd.So if there is a positive integer that satisfies the condition, it is even.At the time,At that time, that is:At the time, it was.So there's a positive integer, so it has to be any positive integer.8. Solution: ( I ) set three angles in A triangle, corresponding trilateral respectively. A, B, C, B, C,,「，/. by sine theorem,By the cosine theorem have /., i. e.,So it's Rt, and it's.againIf you want to get a = 4k, b = 3k (k >, 0).Then, /. three sides respectively three, four, five.(II) with C as the origin of coordinates, the rays CA in order to establish the right Angle coordinate system x is half shaft, the coordinates of A and B (3, 0), (0, 4), for the line AB equationLet's say that the point P is (x, y), the distance from P to 3, AB, BC, AB is dl, d2, and d3And therefore,Let's say that the linear programming knowledge tells us that 0 isless than or equal to m is less than or equal to 8, so the range of dl plus d2 plus d3 is going to be9: (1) the solution,/. since 22 items, each item three adjacent cycle value 1, 0, so the = 1 .................. Four points(2) the first proof that the sequence must be zero after a finite item.As a result of this, all of us have. . . 6 pointsAt the time,();At the time,(),The value is either at least 1, or at least 1. . . . Eight pointsThat, then.Because of the positive integers is determined, along the way, there must be a, this and contradictions, which will be zero ............. .................... 10 pointsIf the first zero term is, remember, it starts with the first term, every three adjacent term cycles, that is,So there's going to be an infinite number of these two different constants in the array. 12 pointsSolution: the equation of the equation is the equation.Will:Will:Thus there are————~5So.To make,一一-10 pointsBecause the equals sign is true.So when it,s 一一一-14make一一-1711. ( I )•The equation of the line of symmetry, by meaning or by means.or or.(II) by the known and the ( I )The standard equation of the ellipse is 一let's say,Stand,Again,Because the right vertex of the ellipse is,The solution:, all satisfaction,At that time, the equations were linear over fixed points, with known contradictions:At that time, the equation is a straight line.(1) it's a little bit.Because the side is perpendicular to the base,So the bottom.The height of the four pyramid. 1 minute,The side is the positive triangle, and the length is,So... Two pointsAs a result,..... Four pointsSo the volume of the four pyramid is. . . 5 points(2) there is a point on the side which makes. . . 6 pointsTake the midpoint of the edge, connect, and hand in. . . 7 pointsBecause, in the middle of the square, the center of the square, therefore, and the right triangle for the whole, and... Eight pointsAnd therefore, therefore. . . 10 pointsBecause of the base, so, the plane, ... 11 pointsSo. . . 12 pointsThe solution (1) equation can be reduced to... 1 minute,I want to make this equation a circle,All that is needed is... Three pointsTherefore, when the equation represents the circle, the range of the real Numbers is... Four points(1)when the equation is round, the circle is centered in radius. 5 pointsThe vertical line with the center of the circle is perpendicular to the vertical..... 6 pointsIt is also known that. . . 7 pointsSo... Eight pointsSolve ...... 10 points(3) the equation of the circle is:.Then again, ... 12 pointsSo... 13 points are known by images, or. . . 14 pointsSo the slope of the line is... 15 pointsSolution: (1) set c as the focal radius of the ellipse. So we have a is equal to 5, and b is equal to 3.(2)solution 1: set point B coordinates, P point coordinates.So there areBecause, so there isIt is. (Al)And because ABP is an isosceles right triangle, there is AB = APIt is. (A2)It's going to be (Al), and it's going to be (A2)And so there is, that is, (unceiy) or.So we plug in the elliptic equation, and we have the equation of the trajectory of PSolution 2: set, the parameter equation of the circle of radius of Ais the circle of AIt is.Let's say that the Angle between AB and the x axis is going to be the parameter of B,The parameter of P is represented by the parameter.From the top two, you get.Because B is on the ellipse. That's the equation of the trajectory for P.15. Solution: ( I ) connection to point, the connection.In the middle, because of the midpoint.Because the plane, the plane, the plane.Method (II) a: the distance from point to plane problem know that point to the distance of the plane,It' s the positive triprism, the plane,Plane, plane plane,r m going to do the dot, the plane,The distance from the point to the plane.In delta, it's equal to, well, it's equal to the area.The distance from the point to the plane is zero.Method 2: the distance from the point to the plane is h, in the delta,—,.The distance from the point to the plane is zero.Method 3: take the midpoint, connect,To establish a space rectangular coordinate system, as shown here Then,Then.r m going to set the normal vector of the plane, That is, the order, that is.The distance from the point to the plane is, The distance from the point to the plane is zero.Method (III) one: in, consists of three vertical theorem, So let'scall it the plane Angle of the dihedral Angle. When AA1 = 2a, AB = b, In the delta.Solution b = 2 a,At that time, the sine of the dihedral Angle is zero.Method two: set, take center point, connect,To establish a space rectangular coordinate system, as shown in the right imageThen,Then.The normal vector of the plane is the normal vector of the plane, And then there is,Set, then,A = 1, solution.And that' s the sine of the dihedral Angle.A: (1) the radius of. . . 1 minute,With one another,..... Two points..... Three pointsLet's square it, let's simplify it,That is,... Four pointsThat is, . . . 6 points/. sequence is arithmetic progression ...............................7 points(2) by topic, , /., i. e. , ..................... Eight points ..... Nine points..... 10 points..... 12 points= .... 13 points..... 14 points..... 15 pointsProve (1)... Three pointsBecause it is a quadratic function, therefore, so, < 0 Four points (2) in the case of... 5 pointsAt the time, it was (1) knowledge of (1). 7 pointsIf, in the case of zero,)in solution: .............. Nine pointsSo if I say minus, minus, minus, plus,Again < 0, there is a solution in (1)... 11 pointsAt the time, it was (1) known as >. . . 13 pointsIf, then (-) + = < 0,So we have a solution in (1): ........... 15 pointsIf, then, > is 0, so in (0,) there is a solution. . . 17 points Therefore, the equation is in (0, 1). 18 points18 solutions: (1) to take the midpoint D.By the way ....... Four points// // plane.So the distance between the plane and the plane is equal to...6 points (2) as shown in figure, .................... Eight points........... 12 pointsSolution: (1) the condition is known:.By the hyperbolic definition, the trajectory equation of the point P: .............. Four points(2) in the first quadrant, at this time ............................6 pointsThe following proof is true when PF is not perpendicular to the x axis and P is in the first quadrant.It's going to be.Plug in and simplify. . . 10 pointsBy symmetry, when P is in the fourth quadrant, same thing.Therefore, there is a constant, which keeps the constant 12 .............................. pointsSolution: (1) the equation of the line that has been set. Let's saythat this is going to cancel out. To have,Let's say that the center of mass of the triangle ABC is zeroCancel out k.(2) because, soThe upper right equals sign is when and only if. Suppose,The upper right equals sign is when and only if. So you get (). Thus there areIt is.21. Solution: ( I ) are the coordinates of the instruction in M (x, y),X plus c, y is equal to x minus c, y, which is equal to 0,X2 minus c2 plus y2 is equal to 0, x2 minus c2 is equal to minus y2. 1.And then the point M is on the ellipse, so y2 is equal to b2X2 - c2 = the x2 = a2 - the 0 x2 a2, or less or less /. 0or less a2 - a2 or less.So 0 is less than or less than or equal to 1, or less than or equal to 1And 0 < < 1 e /. acuities were e < 1 (10 points) (II) linear equation for y = kx + m 1, substitution = 1,So it's going to be (1 + 2k2) x2 + 4kmx + (2x2-32) = 0The line 1 and the ellipse C cross two points: delta = (4km) 2-4 (1 + 2k2) (2m2~32) > 0,m2 < 32 k2 + 16. (2)So you have to have A, B, and B have to be symmetric between P and Q Let's say A (xl, yl), B (x2, y2), and then xQ = k,kPQ = 3.(2), (3) by 16 /. < < 32 k2 + k2 <. K indicates zero again/. <k<0or0<k< (20 points)22. The solution. ( I ) in the medium,Separate orders: Have toBy the name of the two/. (10 points)(II) at that time?.(1) 2 or less, when a < 1, 2 or less or less or less.Namely. Or less or less or less or less. (2) the 2 or less, when a 1 or more? 2 or less or less 1 or less.That is, 1 is less than or equal to a.The number of steps in the triangle count table,The final number of the first row should be the first term in the odd column.So the last number in the first row isSo, the m is the least positive integer solution for the inequality.willSo the first number of line 45 is(10 points)(n).Therefore,The last number of line n is, and there are n Numbers,If you' re going to view this as the first number of row n, then you，re going to have an arithmetic sequence of minus 2 for the n line.So (15 points)The two subtract:(20 points)If, if, satisfying, the maximum value of the request.Solution: by the mean inequalityThe equal sign is equal to if and only if,The maximum value is 100.As is known, the graph of a function is symmetrical with respect to the line. Two pointsLet's say, and then./. on [0, 1] is not decreased function.... Four points............... E ight points(2) for any one, there must be positive integers.Because in (0, 1) it's not a minus function, so,By (1).By (1), in (2), that, too, .And, /. /., (12)points,「，and /.,Therefore, when ............................... 14 points.Solve the inequality.Solution: (I) case. This is the inequality.So there are(1).So at that time, there was: At that time:At the time, there was: At the time, the empty set.(2).There was then, there was: At that time: At that time: At that time. (II). This is the inequality.So there are(3).So at that time, there was : At that time: At the time, the empty set.(4).So at that time, there was: At the time, the empty set.Synthesis (1) - (4)At that time; At that time: At that time.Solution: the first term for a non-negative sequence is the tolerance.(1)because, therefore,.Thus there is. Because, so there issoIt is.(2)And because, so there isWe know the number line satisfies.(I ) series of general term formula;(II), sequence and referred to in the preceding paragraph:(III), referred to in the preceding paragraph and to sequence. Proof: to arbitrary.28. ( I ),And, the sequence is the first term, which is the geometric sequence of the ratio.That is. (II).(Ill),At that time,For any one. one。