金科大联考·河南省2023-2024学年高一年级第二学期4月联考数学.考卷答案试卷答案,我们目前收集并整理关于金科大联考·河南省2023-2024学年高一年级第二学期4月联考数学.考卷答案得系列试题及其答案，更多试题答案请关注微信公众号：考不凡/直接访问www.kaobufan.com(考不凡)
金科大联考·河南省2023-2024学年高一年级第二学期4月联考数学.考卷答案试卷答案
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分析根据题中的新定义，结合函数值域的概念，从而得到本题的结论
解答解：①“f（x）∈A”即函数f（x）值域为R，
“?b∈R，?a∈D，f（a）=b”表示的是函数可以在R中任意取值，
故有：设函数f（x）的定义域为D，则“f（x）∈A”的充要条件是“?b∈R，?a∈D，f（a）=b”
∴命题①是真命题；
②若函数f（x）∈B，即存在一个正数M，使得函数f（x）的值域包含于区间[-M，M]．
∴-M≤f（x）≤M．例如：函数f（x）满足-2＜f（x）＜5，则有-5≤f（x）≤5，此时，f（x）无最大值，无最小值．
∴命题②“若函数f（x）∈B，则f（x）有最大值和最小值．”是假命题；
③若函数f（x），g（x）的定义域相同，且f（x）∈A，g（x）∈B，
则f（x）值域为R，f（x）∈（-∞，+∞），
并且存在一个正数M，使得-M≤g（x）≤M．
∴f（x）+g（x）∈R．
则f（x）+g（x）∉B．
∴命题③是真命题．
④∵函数f（x）=aln（x+2）+$\frac{x}{{x}^{2}+1}$（x＞-2，a∈R）有最大值，
∴假设a＞0，当x→+∞时，$\frac{x}{{x}^{2}+1}$→0，ln（x+2）→+∞，
∴aln（x+2）→+∞，则f（x）→+∞．与题意不符；
假设a＜0，当x→-2时，$\frac{x}{{x}^{2}+1}$→$-\frac{2}{5}$，ln（x+2）→-∞，
∴aln（x+2）→+∞，则f（x）→+∞．与题意不符．
∴a=0．
即函数f（x）=$\frac{x}{{x}^{2}+1}$（x＞-2）
当x＞0时，x+$\frac{1}{x}$≥2，∴0$＜\frac{1}{x+\frac{1}{x}}$$≤\frac{1}{2}$，即 0＜f（x）≤$\frac{1}{2}$；
当x=0时，f（x）=0；
当x＜0时，x+$\frac{1}{x}$≤-2，∴-$\frac{1}{2}$≤$\frac{1}{x+\frac{1}{x}}$＜0，即-$\frac{1}{2}$≤f（x）＜0．
∴-$\frac{1}{2}$≤f（x）＜$\frac{1}{2}$..即f（x）∈B．
故命题④是真命题．
故选：D
点评本题考查了函数值域的概念、基本不等式、充要条件，还考查了新定义概念的应用和极限思想．本题计算量较大，也有一定的思维难度，属于难题．
