“两边夹”方法解题例说

两边夹方法:若A≤a≤A,则a=A,它实现了由不等到等的转化.下面结合例题谈谈它在数学解题中的妙用.     

例谈导数法可能引起的错解

导数作为高中数学的新增内容,为解决函数单调性、最值(极值)、零点及交点问题提供了有力的工具.但借助导数工具解决某些特殊函数时还有一些注意的地方,否则会导致一些不易察觉的错误,下面举例说明.     

Effect of Ni and vacancy concentration on initial formation of Cu precipitate in Fe–Cu–Ni ternary alloys by molecular dynamics simulation

In the present work, the effects of Ni atoms and vacancy concentrations(0.1%, 0.5%, 1.0%) on the formation process of Cu solute clusters are investigated for Fe–1.24%Cu–0.62%Ni alloys by molecular dynamics(MD) simulations. The presence of Ni is beneficial to the nucleation of Cu precipitates and has little effect on coarsening rate in the later stage of aging. This result is caused by reducing the diffusion coefficient of Cu clusters and the dynamic migration of Ni atoms. Additionally, there are little effects of Ni on Cu precipitates as the vacancy concentration reaches up to 1.0%,thereby explaining the embrittlement for reactor pressure vessel(RPV) steel. As a result, the findings can hopefully provide the important information about the essential mechanism of Cu cluster formation and a better understanding of ageing phenomenon of RPV steel. Furthermore, these original results are analyzed with a simple model of Cu diffusion, which suggests that the same behavior could be observed in Cu-containing alloys.     

一道映射个数题的隔板解法

在高三数学复习中,有一道求映射个数题,题目如下: 已知A={1,2,3,4,5},B={6,7,8},从集合A到集合B的映射中,满足f(1)≤f(2)≤f(3)≤f(4)≤f(5)的映射有多少个? 通常的解法是:根据A中的元素在B中对应的元素个数分类.第1类,A中的元素对应B中的同一个元素,有C31=3个映射;第2类,A中的元素对应B中的两个元素,这里需要先确定是哪两个元素,所以要选出这两个元