2006年高考数学广东卷

一、选择题:共10小题,每小题5分,共50分.1.函数f(x)=3x21-x+lg(3x+1)的定义域是A.-31,+∞B.-13,1C.-13,31D.-∞,-312.若复数z满足方程z2+2=0,则z3=A.±22B.-22C.-22iD.±22i3.下列函数中,在其定义域内既是奇函数又是减函数的是A.y=-x3,x∈RB.y=sinx,x∈RC.y=x,x∈RD.y=21x,(x∈R)4.如图1所示,D是△ABC的边AB上的中点,则向量CD=A.-BC+21BAB.-BC-21BZC.BC-21BAD.BC+21BA5.给出以下四个命题:①如果一条直线和一个平面平行,经过这条直线的平面和这个平面相交,那么这条直线和交线平行.②如果一条直线和一个平面内的两条相交直…     

点到空间直线距离的一个公式

利用求条件极值的拉格朗日乘数法给出了空间中点P(x0,y0,z0)到直线A1x+B1y+C1z+D1=0A2x+B2y+C2z+D2=0距离的一个公式d=|(A1x0+B1y0+C1z0+D1)n2-(A2x0+B2y0+C2z0+D2)n1|/|n1×n2|,其中ni={Ai,Bi,Ci},(i=1,2)     

考点5 反函数

1.(江苏卷,2)函数y=21-x+3(x∈R)的反函数的解析表达式为().(A)y=log2x-23(B)y=log2x-23(C)y=log23-2x(D)y=log23-2x2.(山东卷,2)函数y=1-x x(x≠0)的反函数的图像大致是().(A)(B)(C)(D)3.(全国卷,3)函数y=3x2-1(x≤0)的反函数是().(A)y=(x+1)3(x≥-1)(B)y=-(x+1)3(x≥-1)(C)y=(x+1)3(x≥0)(D)y=-(x+1)3(x≥0)4.(辽宁卷,5)函数y=ln(x+x2+1)的反函数是().(A)y=ex+2e-x(B)y=-ex+2e-x(C)y=ex-2e-x(D)y=-ex-2e-x5.(天津卷,9)设f-1(x)是函数f(x)=12(ax-a-x)(a>1)的反函数,则使f-1(x)>1成立的x的取值范围为().(A)(a22-a1,+∞)(B)(-∞,a22-…     

简解一道高考题

大纲卷高考题(2011.重庆.文.15)若实数a,b,c满足2a+2b=2a+b,2a+2b+2c=2a+b+c,则c的最大值是.(答案:2-log23.)简解设x=2a,y=2b,z=2c,得该题等价于"已知正实数x,y,z满足x+y=xy,x+y+z=xyz,求z的最大值".由"正实数x,y,z满足x+y=xy"及     

点到直线的距离

空间解析几何的教学中 ,空间点、直线、平面之间的关系是学习的一个重点。点和直线的位置关系包括两种 :点在直线上 ,点在直线外。当点在直线外时 ,点到直线距离的计算随之出现。笔者在教学中发现 ,这一问题的解决可以涵盖空间解析几何教学中两点间距离、向量运算、直线方程、平面方程等诸多知识点。下面以一具体例题说明。例　求点 A( 2 ,4,1 )到直线 L:x+12 =y2 =z-2-3 的距离。解法一　先求过 A点与直线 L垂直的平面方程 .用点法式 ,得2 ( x -2 ) +2 ( y -4) -3 ( z -1 ) =0即 2 x +2 y +3 z -9=0 .　　将直线方程用参数方程表示为x =2…     

考点24 双曲线

1.(全国卷,6)已知双曲线x62-y32=1的焦点为F1、F2,点M在双曲线上且MF1⊥x轴,则F1到直线F2M的距离为().(A)356(B)566(C)56(D)652.(全国卷,9)已知双曲线x2-y22=1的焦点为F1、F2,点M在双曲线上且MF1·MF2=0,则点M到x轴的距离为().(A)34(B)35(C)233(D)33.(福建卷,10)已知F1、F2是双曲线x2a2-yb22=1(a>0,b>0)的两焦点,以线段F1F2为边作正三角形MF1F2,若边MF1的中点在双曲线上,则双曲线的离心率是().(A)4+23(B)3-1(C)32+1(D)3+14.(上海卷,5)若双曲线的渐近线方程为y=±3x,它的一个焦点是(10,0),则双曲线的方程是.5.(山东卷,14)设双…     

复数的一道复习题

在《复数》这一章的复习课上 ,我给出这样一道题 :若复数z适合 |z| =1 ,求复数 2z+3 - 4i所对应的点的轨迹方程与轨迹 .同学们讨论非常热烈 .有同学当即回答 :“由于考虑的是复平面上复数所对应点的轨迹方程 ,即考虑复数实部、虚部之间所满足的代数关系 ,再通过轨迹方程判断是何种轨迹 .所以只要设所求复数2z+3- 4i的实部为x虚部为 y,找出x ,y之间的代数关系即可 .解 :设w =2z+3 - 4i=x +yi(x,y∈R)令 :z=a+bi(a,b∈R)则 :w =(2a +3) +(2b- 4 )i∴ x=2a +3y=2b- 4a=x - 32b=y +42　∵ |z|=1　∴a2 +b2 =1∴ x - 322 +y+422 =1即 :(x - 3) 2…     

2006年高考数学安徽卷

一、选择题:共12小题,每小题5分,共60分.1.复数1+3i3-i等于A.i B.-i C.3+i D.3-i2.设集合A={x||x-2|≤2,x∈R},B={y|y=-x2,-1≤x≤2},则R(A∩B)等于A.RB.{x|x∈R,x≠0}C.{0}D.3.若抛物线y2=2px的焦点与椭圆x62+y22=1的右焦点重合,则p的值为A.-2B.2C.-4D.44.设a,b∈R,已知命题p∶a=b;命题q∶(a2+b)2≤a22+b2,则p是q成立的A.必要不充分条件B.充分不必要条件C.充分必要条件D.既不充分也不必要条件5.函数y=2x,x≥0,-x2,x<0的反函数是A.y=x2,x≥0-x,x<0B.2x,x≥0-x,x<0C.y=x2,x≥0--x,x<0D.2x,x≥0--x,x<0第(6)题图6.将函数y=sinωx(…     

解析几何中的四类对称变换

[复习说明 ]由于平面解析几何中所研究的许多图形是对称图形 ,于是相关的对称变换问题经常在全国高考试卷与各地模拟试卷中出现 ,它是高考复习的一个热点专题 .本专题复习的重点是两点关于直线成轴对称问题 ;难点是两曲 (直 )线关于直线成轴对称问题 .[内容提要 ]1 .点 P(x,y)关于点 M(a,b)成中心对称的点是 P′(2 a - x,2 b - y) .2 .两点 P(x1,y1)、Q(x2 ,y2 )关于直线 Ax+By +C=0 (AB≠ 0 )成轴对称的充要条件是　 A .x1+x22 +B .y1+y22 +C =0 ,且 (- AB) .y1- y2x1- x2=- 1 .特例　点 P(x,y)依次关于直线 x =a,y =b,y =x,y =- x…     

含两参数的平均值代换法的应用

含一参数的平均值代换法是 ,若数学题中含有x + y =2a的条件 ,则令x =a +t,y =a -t(t为参数 ) .除此以外 ,还有含两参数的平均值代换法 ,即若数学题中含有x + y +z =3a的条件 ,则令x =a +t1,y =a +t2 ,z =a - (t1+t2 ) (t1,t2 为参数 ) .此法也可以推广到更多个参数的情形 .下面只举例说明含两参数的平均值代换法的应用 .1　求值例 1　已知 xa + yb + zc =1,ax + by + cz =0 ,求 x2a2 + y2b2 + z2c2 的值 .解　因 ax + by + cz =0 ,故可令 ax =t1,by =t2 ,cz =- (t1+t2 ) ,则有1t1+ 1t2- 1t1+t2=1.将上式两边平方 ,得1t21+ 1t22+ 1(t1+t2 ) …     

Universal sums of three quadratic polynomials

Let a,b,c,d,e and f be integers with a≥ c≥ e> 0,b>-a and b≡a(mod 2),d>-c and d≡c(mod 2),f>-e and f≡e(mod 2).Suppose that b≥d if a=c,and d≥f if c=e.When b(a-b),d(c-d) and f(e-f) are not all zero,we prove that if each n∈N={0,1,2,...} can be written as x(ax+b)/2+y(cy+d)/2+z(ez+f)/2 with x,y,z∈N then the tuple(a,b,c,d,e,f) must be on our list of 473 candidates,and show that 56 of them meet our purpose.When b∈[0,a),d∈[0,c) and f∈[0,e),we investigate the universal tuples(a,b,c,d,e,f) over Z for which any n∈N can be written as x(ax+b)/2+y(cy+d)/2+z(ez+f)/2 with x,y,z∈Z,and show that there are totally 12,082 such candidates some of which are proved to be universal tuples over Z.For example,we show that any n∈N can be written as x(x+1)/2+y(3y+1)/2+z(5z+1)/2 with x,y,z∈Z,and conjecture that each n∈N can be written as x(x+1)/2+y(3y+1)/2+z(5z+1)/2 with x,y,z∈N.     

探讨Pdx+Qdy+Rdz为某函数u(x,y,z)全微分的积分因子

就微分形式P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz为某函数u(x,y,z)的全微分的积分因子进行了探讨,提出了积分因子的必要条件,以及P(x,y,z),Q(x,y,z),R(x,y,z)是齐次函数时,方程Pdx+Qdy+Rdz=0具有积分因子的充分条件进行了初步探讨.     

The Semi-global Isometric Imbedding in R3 of Two Dimensional Riemannian Manifolds with Gaussian Curvature Changing Sign Cleanly

An abstract Riemannian metric ds²= Edu² + 2Fdudv + Gdv² is given in (u, v) ∈ [0, 2&Pi] × [-&delta, &delta] where E, F, G are smooth functions of (u, v) and periodic in u with period 2&Pi. Moneover K|_{v=0} = 0. K_r|_{v=0} ≠ 0. when> K is the Gaussian curvature. We imbed it semiglobally as the graph of a smooth surface x = x(u, v ), y = y(u, v), z = z(u, v) of R³ in the neighborhood of v = 0. In this paper we show that, if [K_rΓ²_{11}]_{v=0}, and three compatibility conditions are satisified, then there exists such an isometric imbedding.     

方程xy+yz+zx=n的正整数解

本文用 Ｓｉｅｇｅｌ－Ｔａｔｕｚａｗａ定理证明了：当ｎ＞１．２×１０~１１时,至多有两个正 整数ｎ。使方程ｘｕ＋ｙｚ＋ｚｘ＝ｎ无适合（ｘ,ｙ,ｚ）＝１且０＜ｘ＜ｙ＜ｚ的解（ｘ,ｙ,ｚ）, 并给出类数为２的二次域与多项式表素数的一个结果．     

三维椭球的光线障碍问题

对任给的正实数ｕ，令Ｃ＿ｕ＝｛（ｘ，ｙ，ｚ）：ｕ￣２ｘ￣２＋ｙ￣２＋ｚ￣２≤１｝．α为一正实数，使得对于任给的射线ｘ＿ｉ＝α＿ｉｔ（α＿ｉ＞０，ｉ＝１，２，３，ｔ≥０）都和相交。关于Ｃ＿ｕ的光线障碍问题就是要确定Ｋ（Ｃ＿ｕ）＝ｉｎｆα．本文对所有的实数ｕ，确定了Ｋ（Ｃ＿ｕ）的值。     

Trivial nuclear ideals of alternative algebras

Let A be a free alternative Φ-algebra, where Φ is an associative commutative ring with 1, containing 1/6, and g(y, z, t, v, x, x)=2[J_({[y, z], t, x}_, x, v)+J_({[y,x], z, x}_, t, v)], where [x, y]=xy−yx, J_(x, y, z)=[[x, y], z]+[[z, x], y]+[[y, z], x], {x, y, z}_=J_(x, y, z)+3[x, [y, z]]. We construct trivial nuclear ideals of A, that is, nonzero ideals with zero multiplication, lying in the associative center of A. In particular, it is shown that if G and B are fully invariant ideals of A on k≥7 free generators, generated by a function g and by double commutators, respectively, then GB+BG is a nuclear ideal of A. This implies that an unmized alternative algebra satisfies GB=BG=0. If an unmixed algebra is finitely generated, then G=0. In addition, we prove that if R is an unmixed solvable alternative algebra then (R^N)²=0 for some N. Translated fromAlgebra i Logika, Vol. 36, No. 1, pp. 97–115, January–February, 1997.     

关于Thue方程的解数

设ｍ是正整数，ｆ（Ｘ，Ｙ）＝ａ０Ｘｎ＋ａ１Ｘ（ｎ－１）Ｙ＋．．．＋ａｎＹｎ∈Ｚ［Ｘ，Ｙ］是Ｑ上不可约化的叫ｎ（ｎ≥３）次齐次多项式。本文证明了：当ｇｃｄ（ｍ，ａ０）＝１，ｎ≥４００且ｍ≥１０（３５）时，方程｜ｆ（ｘ，ｙ）｜＝ｍ，ｘ，ｙ∈ｚ，ｇｃｄ（ｘ，ｙ）＝１，至多有６ｎｖ（ｍ）组解（ｘ，ｙ），其中ｖ（ｍ）是同余式Ｆ（ｚ）＝ｆ（ｚ，１）≡０（ｍｏｄｍ）的解数。特别是当ｇｃｄ（ｍ，ＤＦ）＝１时，该方程至多有６ｎ（ω（ｍ）＋１）组解（ｘ，ｙ），其中ＤＦ是多项式Ｆ的判别式，ω（ｍ）是ｍ的不同素因数的个数．     

一类平面方程的几何意义

根据点P（x0,y0,z0）与椭球面x^2/a^2＋y^2/b^2＋z^2/c^2=1的三种位置关系,给出平面方程x0x/a^2＋y0y/b^2＋y0y/b^2＋z0z/c^2=1的三种几何意义．     

Liouvillian and analytic integrability of the quadratic vector fields having an invariant ellipse

We characterize the Liouvillian and analytic integrability of the quadratic polynomial vector fields in R2 having an invariant ellipse.More precisely,a quadratic system having an invariant ellipse can be written into the form x=x2＋y2-1＋y（ax＋by＋c）,y=x（ax＋by＋c）,and the ellipse becomes x2＋y2=1.We prove that（i） this quadratic system is analytic integrable if and only if a=0;（ii） if x2＋y2=1 is a periodic orbit,then this quadratic system is Liouvillian integrable if and only if x2＋y2=1 is not a limit cycle;and（iii） if x2＋y2=1 is not a periodic orbit,then this quadratic system is Liouvilian integrable if and only if a=0.     

ON THE RELATIVE POSITION OF LIMIT CYCLES FOR THE EQUATION OF TYPE $\[{(II)_{l = 0}}\]$

In this paper, we consider the relative position of limit cycles for the system $$\[\begin{array}{*{20}{c}} {\frac{{dx}}{{dt}} = \delta x - y + mxy - {y^2}}\{\frac{{dy}}{{dt}} = x + a{x^2}} \end{array}\]$$ under the condition $$\[a < 0,0 < \delta \le m,m \le \frac{1}{a} - a\]$$ The main result is as follows: (i)Under Condition (2), if $\[\delta = \frac{m}{2} + \frac{{{m^2}}}{{4a}} \equiv {\delta _0}\]$, then system $\[{(1)_{{\delta _0}}}\] $ has no limit cycles and on singular closed trajectory through a saddle point in the whole plane, (ii)Under condition (2), the foci 0 and R'' cannot be surrounded by the limit cycles of system (1) simultaneously.