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题目(2010年全国高中数学联赛一试(A卷)第9题)已知函数f(x)=ax3+bx2+cx+d(a≠0),当0≤x≤1时,|f’(x)|≤1,试求a的最大值. 相似文献
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本文将对2013年全国高中数学联赛江苏赛区初赛第13题的解法及本质探究,与读者交流.试题设实数a,b满足0≤a≤1/2≤b≤1,证明:2(b-a)≤cosπa-cosπb.1.解答方法证法1要证2(b-a)≤cosπa-cosπb,只要证2b+cosπb≤2a+cosπa.即设f(x)=2x+cosπx,下证f(b)≤f(a); 相似文献
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众所周知,不等式a≤c≤a中蕴涵着等量关系c=a,不等式g(x)≤f(x+k)-f(x)≤g(x)(x∈R)中蕴涵着等量关系f(x+k)-f(x)-g(x).若函数g(x)已知,再给出f(x0)的值以及n(n∈R且n≥2),就可以求出f(x0+nk)=f(x0)+∑i=0^n-1g(x0+ik)这一函数值. 相似文献
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2011年高考湖北理科压轴题(第21题):
(Ⅰ)已知函数f(x)=lnx—x+1,x∈(0,+∞),求函数f(x)的最大值;
(Ⅱ)设ak,bk(k=1,2,…,n)均为正数,证明:
(1)若a1b1+a2b2+…+anbn≤b1+b2+…+bn,则a1^b1a^b2^2≤1; 相似文献
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性质1y=f(x)关于x=a轴对称<=>f(a+x)=f(a—x)(或f(x)=f(2a-x),f(-x)=f2+x)等)性质2y=f(x)关于(a,b)中心对称<=>f(a+x) f(a-x)=2b(或f(x)+f(2e-x)=2b,f(-x) f(2a+x)=2b等)特别地有:(1)y=f(x)关于(a,0)对称b八a+x)—一人a-x)(或人x)—一人如一动,人一X)—一人加十X)等)(2)y一人工)关于(0,b)对称白人工)+*(一X)一Zb证明1.y一人工)关于x=a轮对称hoJ一人。+*关于x—0对称edy一人x+a)为偶函数今户八一x+a)一人x+。),通过提元面得人)一人加一),人一)一八b+*等.2.… 相似文献
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This article is concerned with the global existence and large time behavior of solutions to the Cauchy problem for a parabolic-elliptic system related to the Camassa-Holm shallow water equation {ut+(u^2/2)x+px=εuxx, t〉0,x∈R, -αPxx+P=f(u)+α/2ux^2-1/2u^2, t〉0,x∈R, (E) with the initial data u(0,x)=u0(x)→u±, as x→±∞ (I) Here, u_ 〈 u+ are two constants and f(u) is a sufficiently smooth function satisfying f" (u) 〉 0 for all u under consideration. Main aim of this article is to study the relation between solutions to the above Cauchy problem and those to the Riemann problem of the following nonlinear conservation law It is well known that if u_ 〈 u+, the above Riemann problem admits a unique global entropy solution u^R(x/t) u^R(x/t)={u_,(f′)^-1(x/t),u+, x≤f′(u_)t, f′(u_)t≤x≤f′(u+)t, x≥f′(u+)t. Let U(t, x) be the smooth approximation of the rarefaction wave profile constructed similar to that of [21, 22, 23], we show that if u0(x) - U(0,x) ∈ H^1(R) and u_ 〈 u+, the above Cauchy problem (E) and (I) admits a unique global classical solution u(t, x) which tends to the rarefaction wave u^R(x/t) as → +∞ in the maximum norm. The proof is given by an elementary energy method. 相似文献
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进入初三年级,我们学习了二次方程ax^2+bx+c=0根的判别式△=b^2-4ac,学习了二次函数f(x)=ax^2+bx+c与x轴有无交点的判别方法,将二次函数f(x)=ax^2+bx+c化简变形得到f(x)=a[(x+b/2a)^2-△/4a^2],当a〉0,△=b^2-4ac≤0时,有f(x)≥0. 相似文献
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Let Ω IR^N, (N ≥ 2) be a bounded smooth domain, p is Holder continuous on Ω^-,
1 〈 p^- := inf pΩ(x) ≤ p+ = supp(x) Ω〈∞,
and f:Ω^-× IR be a C^1 function with f(x,s) ≥ 0, V (x,s) ∈Ω × R^+ and sup ∈Ωf(x,s) ≤ C(1+s)^q(x), Vs∈IR^+,Vx∈Ω for some 0〈q(x) ∈C(Ω^-) satisfying 1 〈p(x) 〈q(x) ≤p^* (x) -1, Vx ∈Ω ^- and 1 〈 p^- ≤ p^+ ≤ q- ≤ q+. As usual, p* (x) = Np(x)/N-p(x) if p(x) 〈 N and p^* (x) = ∞- if p(x) if p(x) 〉 N. Consider the functional I: W0^1,p(x) (Ω) →IR defined as
I(u) def= ∫Ω1/p(x)|△|^p(x)dx-∫ΩF(x,u^+)dx,Vu∈W0^1,p(x)(Ω),
where F (x, u) = ∫0^s f (x,s) ds. Theorem 1.1 proves that if u0 ∈ C^1 (Ω^-) is a local minimum of I in the C1 (Ω^-) ∩C0 (Ω^-)) topology, then it is also a local minimum in W0^1,p(x) (Ω)) topology. This result is useful for proving multiple solutions to the associated Euler-lagrange equation (P) defined below. 相似文献
1 〈 p^- := inf pΩ(x) ≤ p+ = supp(x) Ω〈∞,
and f:Ω^-× IR be a C^1 function with f(x,s) ≥ 0, V (x,s) ∈Ω × R^+ and sup ∈Ωf(x,s) ≤ C(1+s)^q(x), Vs∈IR^+,Vx∈Ω for some 0〈q(x) ∈C(Ω^-) satisfying 1 〈p(x) 〈q(x) ≤p^* (x) -1, Vx ∈Ω ^- and 1 〈 p^- ≤ p^+ ≤ q- ≤ q+. As usual, p* (x) = Np(x)/N-p(x) if p(x) 〈 N and p^* (x) = ∞- if p(x) if p(x) 〉 N. Consider the functional I: W0^1,p(x) (Ω) →IR defined as
I(u) def= ∫Ω1/p(x)|△|^p(x)dx-∫ΩF(x,u^+)dx,Vu∈W0^1,p(x)(Ω),
where F (x, u) = ∫0^s f (x,s) ds. Theorem 1.1 proves that if u0 ∈ C^1 (Ω^-) is a local minimum of I in the C1 (Ω^-) ∩C0 (Ω^-)) topology, then it is also a local minimum in W0^1,p(x) (Ω)) topology. This result is useful for proving multiple solutions to the associated Euler-lagrange equation (P) defined below. 相似文献
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构造二次函数巧用判别式解一类题 总被引:1,自引:1,他引:0
判别式△=b^2-4ac是二次函数f(x)=ax^2+bx+c(a≠0)的一个重要的特征数字,其一条性质:若f(x)=ax^2+bx+c且a〉0,则f(x)≥0对x∈R恒成立 △≤0,为我们利用二次函数解决一些数学问题提供了突破IZl.本文将利用这一性质,构造适当二次函数,灵活解决一类问题. 相似文献
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2010年全国数学联赛第9题:
已知函数f(x)=ax3+bx2+cx+d(a≠0),当0≤x≤1时,f'(x)≤1,试求a的最大值. 相似文献
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问题1设1≤a≤e,函数f(x)=x+x^-a^2,x∈[1,e]有f(x)〉2e-1成立,求实数a的取值范围. 相似文献
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Bao-huai Sheng 《应用数学学报(英文版)》2005,21(4):529-536
Let S^1-1,q≥2,be the surface of the unit sphere in the Euclidean space R^1,f(x)∈L^p(S^q-1),f(x)≥0,f absohutely unegual to 0,1≤p≤+∞,Then,it is proved in the present paper that there is a spherical harmonics PN(x) of order≤N and a constant C〉0 such that where ω(f,δ)L^p=sup 0〈t≤δ‖St(f)-f‖L^p is a kind of moduli of continuity and ^‖f-1/PN‖L^p≤Cω(f,N^-1)L^p,St(f,μ)=1/|S^q-2|Sin^2λt ∫-μμ’=t f(μ')dμ' is a translation operator. 相似文献
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2006年全国高中数学联合竞赛一试最难的题是第15题:
设f(x)=x^2+a,记f^1(x)=f(x),f^n(x)=f(f^n-1(x)), 相似文献
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问题:已知f(x)=ax~2+bx+c(a≠0),且方程f(x)=x无实数解,下列命题:①方程f[f(x)]=x也一定没有实数解;②若a〉0,则不等式f[f(x)]〉x对一切实数x都成立;③若a〈0,则必存在实数x_0,使f[f(x_0)]④若a+b+c=0, 相似文献