“f(x)g(x)=of(x)=0或g(x)=0”对吗?——从一道错题谈起

首先看一道选择题:设全集为实数集R,M={x|f(x)=0},N={x|g(x)=0},那么集合P={x|f(x)g(x)=0}可表示为(A)M∩N;(B)M∪N;(C)M∪N;(D)M∪N.这是一道广为流传的题目.如1998年福州市高中毕业班质量检查卷(理科)第一题.参考答案都选(D).其实这是一道错题.例如,设f(x)=x2-1,g(x)=lg(x-1).则M={x|f(x)=0}={-1,1},N={x|g(x)=0}={2},M∪N={-1,1,2},但P={x|f(x)g(x)=0}={x|(x2-1)lg(x-1)=0}={2}≠M∪N.又如设f(x)=sinx,g(x)=cosx,M={x|f(x)=0}={x|x=kπ,k∈Z},N={x|g(x)=0}={x|cosx=0}={x|x=kπ π2,k∈Z}.M∪N={x|x=kπ或kπ π2,k∈Z}…     

二、幂函数、指数函数和对数函数

1.已知全集I={实数对(x,y)},集合A={(x,y)|(y-4)/(x-2)=3},B={(x,y)|y==3x-2},求A∩B。 2.设全集I={2,4,a~2-a+1}及集合A={a+1,2},A={7},求实数a。 3.设集合A={(x,y)|x∈Z,y∈N,x+y,<3},集合B={0,1,2},从A到B的对应法则f:(x,y)→x+y,试画出对应图,判断这个对应是不是映射? 4.已知集合A={x|x∈R},B={y|y∈R},从A到B的对应法则f:x→y=tg2x,(1)求A的元素arctg2的象;(2)求B里元素5的原象;(3)上述对应f是否一一映射?为什么? 5.已知函数y=2/3(9-x~2)~(1/2)(-3≤x≤0),求它     

Inverse of a Function f（x）=y

For a function y= f（x） to have an inverse function, f must be one-to-one. Then for each x in its domian there is exactly one y in its range; furthermore, to each y in the range, there corresponds exactly one x in the domain. The correspondence from the range of f onto the domian of f is, therefore, also a function. It is this function that is the inverse of f.     

综合题新编选登

题188已知函数y=(2 x)(3-x)的定义域为集合A,函数y=lg(kx2 4x k 3)的定义域为集合B,分别在下列情况下求k的取值范围.1)B A;2)A B;3)A∩B=.解A={x|(2 x)(3-x)≥0}={x|-2≤x≤3},B={x|kx2 4x k 3>0}.1)当k≥0时,显然B A,∴k<0,设f(x)=kx2 4x k 3=0的两根为x1,x2(x1相似文献   

2006年高考数学全国卷Ⅰ

一、选择题:共12小题,每小题5分,共60分.1.设集合M={x|x2-x<0},N={x||x|<2},则A.M∩N=B.M∩N=MC.M∪N=MD.M∪N=R2.已知函数y=ex的图像与函数y=f(x)的图像关于直线y=x对称,则A.f(2x)=e2x(x∈R)B.f(2x)=ln2·lnx(x>0)C.f(2x)=2ex(x∈R)D.f(2x)=lnx+ln2(x>0)3.双曲线mx2+y2=1的虚轴长是实轴长的2倍,则m=A.-41B.-4C.4D.414.如果复数(m2+i)(1+mi)是实数,则实数=A.1B.-1C.2D.-25.函数f(x)=tanx+4π的单调增区间为A.kπ-2π,kπ+2π,k∈ZB.(kπ,(k+1)π),k∈ZC.kπ-34π,kπ+4π,k∈ZD.kπ-4π,kπ+34π,k∈Z6.△ABC的内角A、B、…     

CHINESE ANNALS OF MATHEMATICS Vol.6 Ser.A No.4 CONTENTS AND ABSTRACTS

Approximation of Nonlinear Dirichlet Problem by Finite Element MethodsLi Likang(李立康)The author obtains approximate solution of the nonlinear Dirichlet problem-▽·(a(x,u)▽u)=f(x),x∈Ω,u(x)=g(x),x∈Γby finite element method.In this paper finite element spaces V_h is defined by the followingV_h={v_h|v_h∈C~0(Ω),v_h is a linear function on each K_i},     

高中代数总复习(一)

1.选择题: (1)集合P={s|s=x~2 3x 1, x∈R}与集合Q={t|t=y~2-3y 1,y∈R},则P,Q的关系是( ) (A)P\Q (B)P=Q (C)PQ (D)P≠Q,且pQ,PQ (2)已知f(x)=8 2x-x~2,如果g(x)=f(2-x~2),那么g(x)( ) (A)在区间(-2,0)上是增函数     

Frequency Assignment through Combinatorial Optimization Approach

An L(2,1)-labeling of a graph G is a function f from the vertex set V(G) to the set of all nonnegative integers such that |f(x)-f(y)|(?)2 if d(x, y)=1 and |f(x)-f(y)|(?)1 if d(x,y)=2. The L(2,1)-labeling numberλ(G) of G is the smallest number k such that G has an L(2,1)-labeling with max{f(v) : v∈V(G)}=k. We study the L(3,2,1)-labeling which is a generalization of the L(2,1)-labeling on the graph formed by the (Cartesian) product and composition of 3 graphs and derive the upper bounds ofλs(G) of the graph.     

高中数学综合练习题(五)

一、选择题 1.集合A={x|x≠1,x∈R}U{y|y≠1,y∈R},集合B={x|x<-1,或-1},则A,B之间的关系是( )。 (A)A=B (B)AB (C)AB (D)无法判定 2.若函数y=x-n(x∈Z)的图象过原点,并是增函数,则n为( )。     

一个结论的是与非

文[1]给出了集合中几个似是而非的结论,并举例说明了它们的错误之处.仔细研读了文章之后,笔者认为该文对其中一个结论的判断值得商榷.结论若A={x|f(x)=0},B={x|g(x)=0},C={x|f~2(x)+g~2(x)=0},则A n B=C.文[1]认为该结论是错误的(只有在f(x),g(x)的定义域为R时成立),并给出了反例如下:     

BMO ESTIMATES FOR MULTILINEAR FRACTIONAL INTEGRALS

In this paper,the authors prove that the multilinear fractional integral operator T A 1,A 2 ,α and the relevant maximal operator M A 1,A 2 ,α with rough kernel are both bounded from L p (1 p ∞) to L q and from L p to L n/(n α),∞ with power weight,respectively,where T A 1,A 2 ,α (f)(x)=R n R m 1 (A 1 ;x,y)R m 2 (A 2 ;x,y) | x y | n α +m 1 +m 2 2 (x y) f (y)dy and M A 1,A 2 ,α (f)(x)=sup r0 1 r n α +m 1 +m 2 2 | x y | r 2 ∏ i=1 R m i (A i ;x,y)(x y) f (y) | dy,and 0 α n, ∈ L s (S n 1) (s ≥ 1) is a homogeneous function of degree zero in R n,A i is a function defined on R n and R m i (A i ;x,y) denotes the m i t h remainder of Taylor series of A i at x about y.More precisely,R m i (A i ;x,y)=A i (x) ∑ | γ | m i 1 γ ! D γ A i (y)(x y) r,where D γ (A i) ∈ BMO(R n) for | γ |=m i 1(m i 1),i=1,2.     

解集合题常见的错误剖析

1　忽视特殊的集合空集致误例1　已知A={x|x2-3x 2=0},B={x|x2-bx 2=0},若BA,求实数b的范围.错解　A={1,2}把x=1和x=2分别代入方程x2-bx 2=0均有b=3,这时B={1,2}满足BA∴b=3.剖析　因为空集是任何集合的子集,所以上面的解答忽视了空集的特殊情形,而当B=时,Δ=b2-8<0,即-220,所以x≠0,y≠0,故由A=B知lg(xy)=0x=yxy=|x|　或　lg(xy)=0x=|x|xy=y解得x=y=1或x=y=-1.剖析　当x=y=1时,A…     

BOUNDEDNESS AND CONVERGENCE FOR THE NON-LINARD TYPE DIFFERENTIAL EQUATION

In this article, the author studies the boundedness and convergence for the non-Lienard type differential equation (x|·)=a(y)-f(x) (y|·)=b(y)β(x)-g(x) e(t) where a(y),b(y),f(x),g(x),β(x) are real continuous functions in y∈R or x∈R,β(x)≥0 for all x and e(t) is a real continuous function on R = {t: t≥0} such that the equation has a unique solution for the initial value problem. The necessary and sufficient conditions are obtained and some of the results in the literatures are improved and extended.     

一些集的基数(Ⅷ)

Let X be an infinite set, C={B:B is a Boolean algebra defined on the X},Define B={D:D is isomorphie to B}, C={C & B is atomic},C_2= {C & B is not atomic},then |C_1|=|C_2|=2~(|x|). The power set of X is denoted by P(X), P(X) is a field of sets (Under Union and Intersection of sets, and Complement of set), Suppose K={F:F js a field of sets & FP(X)}, K_1={K},K_2{K & F is not atomie} then |K_1|=|K_2|=2~(2~(|x|)).     

Positive Solutions of Hammerstein Nonlinear Integral Equations

In this paper we consider the Hammerstein nonlinear integral equation (x)=integral from G (K(x,y) f(y,(y))dy=A(x)) (1) where G is a bounded closed domain in R"; the function f(x,u) is non-negative,continuous on G×[0,+∞) and f(x,0)≡0; the kernel k(x,y) is non-negative continuous on G×G. Obviously, A acts in the space C(G) and is completely continuous. Theorem 1 Suppose that (i) lim u~(-1)f(x,u)=0 and lim u~(-1)f(x, u)=+∞     

广义Lienard方程边值问题

By coincidence degree,the existence of solution to the boundary value problem of a generalized Liénard equation a(t)x＂+F(x,x′)x′+g(x)=e(t),x(0)=x(2π),x′(0)=x′(2π)is proved,where a∈C1[0,2π],a(t)>0(0≤t≤2π),a(0)=a(2π),F(x,y)=f(x)+α| y|β,α>0,β>0 are all constants,f∈C(R,R),e∈C[0,2π]. An example is given as an application.     

Almansi decomposition for Dunkl operators

Let Ωbe a G-invariant convex domain in RN including 0, where G is a Coxeter group associated with reduced root system R. We consider functions f defined in Ωwhich are Dunkl polyharmonic, i.e. (△h)nf =0 for some integer n. Here △h=∑j=1N Dj2 is the Dunkl Laplacian, and Dj is the Dunkl operator attached to the Coxeter group G, where kv is a multiplicity function on R and σv is the reflection with respect to the root v. We prove that any Dunkl polyharmonic function f has a decomposition of the form f(x)=f0(x) |x|2f1(x) … |x|2(n-1)fn-1(x),(?)x∈Ω, where fj are Dunkl harmonic functions, i.e. △hfj = 0. This generalizes the classical Almansi theorem for polyharmonic functions as well as the Fischer decomposition.     

世界各地数学奥林匹克试题选登

1找到所有映射f:R→R,满足f(f(x) y)=f(x2-y) 4f(x)y,其中x,y∈R.解映射f(x)=0和f(x)=x2显然符合条件.下面证明不存在其它的映射符合要求.设映射f:R→R满足f(f(x) y)=f(x2-y) 4f(x)y(1)其中x,y∈R.令a=f(0).在(1)中取x=0则对任意y∈R,f(a y)=f(-y) 4ay(2)在(2)式中先取y=0,则有f(a)=a.取y=-a,则有a=a-4a2,即a=0.因此由(2)式知f是一个偶函数.在(1)式中令y=-f(x)及y=x2.比较其结果有4(f(x))2=4x2f(x).因而f(x)=0或f(x)=x2.现假设存在x0使得f(x0)≠0,则x0≠0及f(x0)=x02.因为f是偶函数.我们假设x0>0.令x为任意非零实数,在(1)式中令y=-x0,则…     

关于Hammerstein型非线性积分方程固有值的一点注记

Let Aφ(x)=∫_GK(x,y)f(y,φ(y))dy, where G is a bounded closed domain in Euclidean space, K(x,y) is continuous on G×G, f(x,u) is continuous on G×R, and f(x,0)≡0. Set G_x={x|x∈G,K(x,y)≠0},G_y={y|y∈G,K(x,y)≠0},G₁=G_x∩G_y≠φ.Let K₁(x,y) be the restriction of K(x,y) on G₁×G₁,f₁(x,u)be the restriction of f(x, u) on G₁× R, and A₁φ=∫_G1K₁(x,y)f₁(y,φ₁(y))dy, The main result of this paper is Theorem λ≠0 is an eigenvalue of A, if and only if λ is an eigenvalue of A₁.     

两参数分数Wiener过程增量的一般形式

§1Introductionandresults A2-parameterGaussianprocess{Z(t,s);t,s≥0}iacalleda2-parameterfractional Wienerprocesswithorderα(0<α<1),ifZ(0,0)=0a.s.EZ(s,t)=0anditscovariance EZ(t1,s1)Z(t2,s2)={|t1|2α+|t2|2α-|t2-t1|2α}{|s1|2α+|s2|2α-|s2-s1|2α}/4.LetR=[x1,x2]×[y1,y2],DT={(x,y)∶0≤x,y≤bT,xy≤T}.Let0相似文献