数学问题解答

数学问题解答１９９５年６月号问题解答（解答由问题提供人给出）９５６设实数ｘ，ｙ，ｚ满足求３ｘ＋４ｙ＋５ｚ的范围．解设ｘ＋２ｙ＋３ｚ＝ａ（１）２ｘ＋３ｙ＋４ｚ＝ｂ（２）则．解由（１），（２）组成的方程组得：ｘ＝ｚ＋２ｂ－３ａ，ｙ＝２ａ－ｂ－２ｚ．则：３...     

新题征展(1)

Ａ．题组新编１．设函数ｆ（ｘ）＝ｌｇ（ａｘ２－４ｘ＋ａ－３）．（１）若ｆ（ｘ）的定义域是Ｒ,则ａ的取值范围是　　;（２）若ｆ（ｘ）的值域是Ｒ,则ａ的取值范围是　　;（３）若ｆ（ｘ）在区间（－４,－１）上递减,则ａ的取值范围是　　．２．设０＜θ＜π．（１）若ｓｉｎθ＋ｃｏｓθ＝１５,则ｔｇθ＝　　;（２）若ｓｉｎθ＋ｃｏｓθ＝－１５,则ｔｇθ＝　　;（３）若ｓｉｎθ－ｃｏｓθ＝１５,则ｔｇθ＝　　;（４）若ｓｉｎθ－ｃｏｓθ＝－１５,则ｔｇθ＝　　．３．如图,向高为Ｈ的水瓶（Ａ）、（Ｂ）、（Ｃ）、…     

复合函数的性质及应用

复合函数这一概念,现行高中课本未直接定义．而复合函数的应用实例,在现行高中课本《代数》上册（以下简称课本）中大量出现．如　ｙ＝ｌｏｇ０．５（４ｘ－３）（课本Ｐ１００）;ｙ＝１６－５ｘ－ｘ２（课本Ｐ１０８）;ｙ＝Ａｓｉｎ（ωｘ＋φ）（课本Ｐ１７８）;ｙ＝ｓｉｎ（ａｒｃｓｉｎｘ）（课本Ｐ３００）．在全国高考数学试题中,复合函数的应用问题成为命题热点．本文试对复合命题的性质作点介绍．设ｙ＝ｆ（ｕ）,ｕ＝ｇ（ｘ）,则ｙ＝ｆ［ｇ（ｘ）］为复合函数．根据函数单调性的定义,容易得出下列结论．（１）若ｙ＝ｆ（ｕ…     

QUALITATIVE THEORY OF THE QUADRATIC DIFFERENTIAL SYSTEMS (II)-ERGODICITY OF LIMIT CYCLES

Ａｓｉｓｗｅｌ－ｋｎｏｗｎｗｈｅｎａｒｅａｌｑｕａｄｒａｔｉｃｄｉｆｅｒｅｎｔｉａｌｓｙｓｔｅｍ：ｘ＝－ｙ＋δｘ＋ｌｘ２＋ｍｘｙ＋ｎｙ２＝Ｐ（ｘ，ｙ），ｙ＝ｘ（１＋ａｘ＋ｂｙ）＝Ｑ（ｘ，ｙ）（１）ｈａｓｆｏｕｒｆｉｎｉｔｅｃｒｉｔｉｃａｌｐｏｉｎｔｓ...     

教学参考

●目标检测因式分解（Ａ）一、填空题（１）提公因式法、公式法、分组分解法、十字相乘法；（２）４；（３）－４ｘ；（４）ｍ２－２ｍ＋４；（５）ｘ＋１、ｘ－１；（６）ａｘ－６；（７）ａ－３；（８）ｙ、５ｙ；（９）２５；（１０）原式＝２５（５０２２－４９８２）＝２５（５０２＋４９８）（５０２－４９８）＝２５×１０００×４＝１０００００．二、选择题（１）Ｄ；（２）ｘ４－８１＝（ｘ２＋９）（ｘ＋３）（ｘ－３），ｘ３－２７＝（ｘ－３）（ｘ２＋３ｘ＋９），ｘ２－６ｘ＋９＝（ｘ－３）２，所以公因式是ｘ－３，选Ｂ；（…     

巧解一次方程组

解一次方程组的思想是消元，消元后转化为一元一次方程．但还要注意仔细观察，认真分析题目的特征、巧妙、灵活地运用消元法来解题．例１　解方程组（１）２ｘ＋ｙ－ｚ＝２，ｘ＋２ｙ＋３ｚ＝１３，－３ｘ＋ｙ－２ｚ＝－１１；　①②③（２）ｘ＋２ｙ－３ｚ＝－４，４ｘ＋８ｙ＋９ｚ＝５，２ｘ＋６ｙ－９ｚ＝－１５．　①②③分析　上面两题若逐步消元，都比较麻烦．仔细观察，发现方程组（１）三式相加可得ｙ；而方程组（２）呢，可先整体消元求出ｘ和ｚ，于是得妙解．（１）解　由①＋②＋③得４ｙ＝４，即ｙ＝１．把ｙ＝１代入①、②，得…     

一种新的二次曲线系方程及其应用

确定一个二次曲线：Ａｘ２＋Ｂｘｙ＋Ｃｙ２＋Ｄｘ＋Ｅｙ＋Ｆ＝０一般需五个独立条件,因此,经过四点的二次曲线一般情况下有无数条,它们组成一个二次曲线系;本文以定理形式介绍一种新的二次曲线系,并举例说明其应用,并以此引伸出一种新的解题方法;１．定理的证明定理　若直线ＡＢ的方程为Ｆ１（ｘ,ｙ）＝０;直线ＢＣ的方程为Ｆ２（ｘ,ｙ）＝０;直线ＣＤ的方程为Ｆ３（ｘ,ｙ）＝０;直线ＤＡ的方程为Ｆ４（ｘ,ｙ）＝０;则方程Ｆ１（ｘ,ｙ）·Ｆ３（ｘ,ｙ）＋λＦ２（ｘ,ｙ）·Ｆ４（ｘ,ｙ）＝０表示过Ａ、Ｂ、Ｃ、Ｄ四点的…     

十二省市(辽宁、云南、湖南、乌鲁木齐、昆明、哈尔滨、山西、广西、西宁、河南、嘉兴等、甘肃)选择题集萃

数与式１．若ａ≠０,则下列运算正确的是（　）．（Ａ）ａ４·ａ２＝ａ８　　（Ｂ）ａ２＋ａ２＝ａ４（Ｃ）（－３ａ４）２＝９ａ６（Ｄ）（－ａ）４÷（－ａ）２＝ａ２２．下列各式中计算错误的是（　）．（Ａ）ａｂ＝ａｃｂｃ（ｃ≠０）（Ｂ）ａ＋ｂａｂ＝ａ２＋ａｂａ２ｂ（Ｃ）０．５ａ＋ｂ０．２ａ－０．３ｂ＝５ａ＋１０ｂ２ａ－３ｂ（Ｄ）ｘ－ｙｘ＋ｙ＝ｙ－ｘｙ＋ｘ３．化简１２－３的结果是（　）．　　（Ａ）－２＋３　　（Ｂ）－２－３（Ｃ）２＋３（Ｄ）２－３４．２ｘ２·３ｘ３等于（　）．（Ａ）６ｘ５　（Ｂ）６ｘ６　（Ｃ…     

错题·错解两例

错题·错解两例万兴灿（湖北省宜昌市第一中学４４３０００）１错题贵刊１９９７．１．Ｐ２５例３［１］中给出：若正数ｘ，ｙ满足ｘ＋ｙ＝ｍ（定值），则函数ｆ（ｘ，ｙ）＝（ｘ＋１ｘ）（ｙ＋１ｙ）取得最小值（ｍ２＋２ｍ）２（当ｘ＝ｙ＝ｍ２时）此题最小值是不对的．...     

一元一次方程目标检测题(一)

一、填空１．方程１３ｘａ＋２＝３是一元一次方程,则ａ＝．２．３ｘ－２与２ｘ－３互为相反数,则ｘ＝．３．（２ｘ－１）２＋｜３ｙ＋２｜＝０,则ｘ＝,ｙ＝．４．当ｍ＝时,关于ｘ的方程ｍｘ－８＝１７＋ｍ的解是－５．５．若５ｘｍｙ与１２ｙｎ＋２ｘ３是同类项,则ｍ＝,ｎ＝．６．把浓度为９５％的酒精１５００克稀释为７５％的酒精,需加水克．二、单项选择题１．已知ｙ＝１是方程２－１３（ｍ－ｙ）＝２ｙ的解,那么关于ｘ的方程ｍ（ｘ－３）－２＝ｍ（２ｘ－５）的解是（　）（Ａ）ｘ＝－２　　（Ｂ）ｘ＝－１（Ｃ）ｘ＝０　　（…     

ON A THEOREM OF A.A. LYAPUNOV

Abstract

In this note we obtain some extensions and an approximation of the Lyapunov convexity theorem by means of the bilinear integration of a set-valued function. The integration is performed successively with respect to a non-atomic, a direct sum and a Darboux vector measure. The necessary counterexamples are provided.     

A Priori and A Posteriori: A Bootstrapping Relationship

The distinction between a priori and a posteriori knowledge has been the subject of an enormous amount of discussion, but the literature is biased against recognizing the intimate relationship between these forms of knowledge. For instance, it seems to be almost impossible to find a sample of pure a priori or a posteriori knowledge. In this paper, it will be suggested that distinguishing between a priori and a posteriori is more problematic than is often suggested, and that a priori and a posteriori resources are in fact used in parallel. We will define this relationship between a priori and a posteriori knowledge as the bootstrapping relationship. As we will see, this relationship gives us reasons to seek for an altogether novel definition of a priori and a posteriori knowledge. Specifically, we will have to analyse the relationship between a priori knowledge and a priori reasoning, and it will be suggested that the latter serves as a more promising starting point for the analysis of aprioricity. We will also analyse a number of examples from the natural sciences and consider the role of a priori reasoning in these examples. The focus of this paper is the analysis of the concepts of a priori and a posteriori knowledge rather than the epistemic domain of a posteriori and a priori justification.     

有限群的一个类函数的模构造

本文研究了有限群上的一个类函数.通过计算它和不可约特征标的内积,证明了它是特征标并且通过复群代数的中心的正则表示给出了它的一个模构造.     

A REMARK ON A BMO MARTINGALE

Let M = (Mt,Ft) be a uniformly integrable continuous martingale with MO = 0. For1 5 p < cot we setIIMllBMO. = '3p II[E[IMoo ~ MTIplFT]]'/Pll.,where the supremum is taken over all stopping times T.Set BMO. = {M: IIMllBMO. < co}. It is well known that BMO. = BMO, (VI S p 5 q).F'urthermore, all 11.llBMO. norms are equivalent andIIi ~~if;llMllBMO. = SUP T P(T < co)i'where the supremum is taken over all stopping times T satisfying P(T < co) > 0. In the laterwe shall simply …     

NAMING A COLUMN ON A SPREADSHEET

Spreadsheets use a meaningful algebra-like notation which, research suggests, can support pupils in developing an understanding of variables. This paper discusses the activity of Year 8 pupils who were taught to name a column on a spreadsheet, and who were asked to reflect upon their activity in a stimulated recall interview. More specifically, it considers the pupils' understanding of notation, such as 'A2' and 'm', which they used when constructing spreadsheet formulae. It is suggested that experience of naming columns may help pupils to develop a clearer sense of the notation as a variable, and to make links between their spreadsheet activity and use of standard algebraic notation [1].     

A new characterization of A 5

Let G be a group and τ _e (G) the set of numbers of elements of G of the same order. In this paper, by τ _e (G), we give a new characterization of A ₅, where A ₅ is the alternating group of degree 5. We get the theorem following: Theorem. Let G be a group, ${G\cong A_5}$ if and only if τ _e (G) = τ _e (A ₅) = {1, 15, 20, 24}.     

简化摩擦接触问题的对称弱超内罚间断Galerkin方法的先验和后验误差估计

本文讨论了简化摩擦接触问题的一类对称弱超内罚间断Galerkin方法.首先,在能量范数意义下得到最优先验误差估计.进一步,我们推导了一类残量型后验误差估计子,并证明了它的可靠性和有效性.