2018年高考题和高考模拟题数学(文)分项版汇编：专题04数列与不等式文(含解析).pdf

2018 年高考题和高考模拟题数学（文）分项版汇编4数列与不等式1【2018 年浙江卷】已知成等比数列，且若，则A.B.C.D.【答案】B 点睛：构造函数对不等式进行放缩，进而限制参数取值范围，是一个有效方法.如2【2018 年文北京卷】“十二平均律”是通用的音律体系，明代朱载堉最早用数学方法计算出半音比例，为这个理论的发展做出了重要贡献.十二平均律将一个纯八度音程分成十二份，依次得到十三个单音，从第二个单音起，每一个单音的频率与它的前一个单音的频率的比都等于.若第一个单音的频率f，则第八个单音频率为A.B.C.D.【答案】D【解析】分析：根据等比数列的定义可知每一个单音的频率成等比数列，利用等比数列的相关性质可解.详解：因为每一个单音与前一个单音频率比为，所以，又，则，故选 D.点睛：此题考查等比数列的实际应用，解决本题的关键是能够判断单音成等比数列.等比数列的判断方法主要有如第 1 页，共 20 页下两种：（1）定义法，若（）或（），数列是等比数列；（2）等比中项公式法，若数列中，且（），则数列是等比数列.3【2018 年浙江卷】已知集合，将的所有元素从小到大依次排列构成一个数列记为数列的前n项和，则使得成立的n的最小值为 _【答案】27，所以只需研究是否有满足条件的解，此时，为等差数列项数，且.由得满足条件的最小值为.点睛：本题采用分组转化法求和，将原数列转化为一个等差数列与一个等比数列的和.分组转化法求和的常见类型主要有分段型（如），符号型（如），周期型（如）.4【2018 年浙江卷】已知等比数列an 的公比q1，且a3+a4+a5=28，a4+2 是a3，a5的等差中项数列bn满足b1=1，数列 （bn+1-bn）an的前n项和为 2n2+n（）求q的值；（）求数列bn 的通项公式【答案】（）（）（）设第 2 页，共 20 页文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2文档编码:CJ10B3R3T3W7 HX9W5A5Y5O8 ZI10R1S2I4Q2，数列前n项和为.由解得.由（）可知，所以，故，.设，所以，因此，又，所以.点睛：用错位相减法求和应注意的问题：(1)要善于识别题目类型，特别是等比数列公比为负数的情形；(2)在写出“”与“”的表达式时应特别注意将两式“错项对齐”以便下一步准确写出“”的表达式；(3)在应用错位相减法求和时，若等比数列的公比为参数，应分公比等于1 和不等于1 两种情况求解.5【2018 年天津卷文】设an 是等差数列，其前n项和为Sn（nN*）；bn是等比数列，公比大于0，其前n项和为Tn（nN*）已知b1=1，b3=b2+2，b4=a3+a5，b5=a4+2a6（）求Sn和Tn；（）若Sn+（T1+T2+Tn）=an+4bn，求正整数n的值【答案】()，；()4.详解：（I）设等比数列的公比为q，由b1=1，b3=b2+2，可得因为，可得，故所以，设等差数列的公差为 由，可得 由，可得从而，故，所以，（II）由（I），有由可得，整理得解得（舍），第 3 页，共 20 页文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3或所以n的值为 4点睛：本小题主要考查等差数列、等比数列的通项公式及前n项和公式等基础知识.考查数列求和的基本方法和运算求解能力.6【2018 年文北京卷】设是等差数列，且.（）求的通项公式；（）求.【答案】（I）（II）【解析】分析：（1）设公差为，根据题意可列关于的方程组，求解，代入通项公式可得；（2）由（1）可得，进而可利用等比数列求和公式进行求解.详解：（I）设等差数列的公差为，又，.（II）由（I）知，是以 2 为首项，2 为公比的等比数列.点睛：等差数列的通项公式及前项和共涉及五个基本量，知道其中三个可求另外两个，体现了用方程组解决问题的思想.7【2018 年江苏卷】设，对 1，2，n的一个排列，如果当s0时，所以单调递减，从而f（0）=1当时，因此，当时，数列单调递减，故数列的最小值为因此，d的取值范围为点睛：对于求不等式成立时的参数范围问题，一般有三个方法，一是分离参数法,使不等式一端是含有参数的式子，另一端是一个区间上具体的函数，通过对具体函数的研究确定含参式子满足的条件.二是讨论分析法，根据参数取值情况分类讨论，三是数形结合法，将不等式转化为两个函数，通过两个函数图像确定条件.9【2018 年新课标I 卷文】已知数列满足，设（1）求；第 6 页，共 20 页文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3（2）判断数列是否为等比数列，并说明理由；（3）求的通项公式【答案】(1)b1=1，b2=2，b3=4(2)bn 是首项为1，公比为2 的等比数列理由见解析.(3)an=n2n-1【解析】分析：(1)根据题中条件所给的数列的递推公式，将其化为an+1=，分别令n=1 和n=2，代入上式求得a2=4 和a3=12，再利用，从而求得b1=1，b2=2，b3=4（2）bn是首项为1，公比为 2 的等比数列由条件可得，即bn+1=2bn，又b1=1，所以 bn是首项为1，公比为2 的等比数列（3）由（2）可得，所以an=n2n-1点睛：该题考查的是有关数列的问题，涉及到的知识点有根据数列的递推公式确定数列的项，根据不同数列的项之间的关系，确定新数列的项，利用递推关系整理得到相邻两项之间的关系确定数列是等比数列，根据等比数列通项公式求得数列的通项公式，借助于的通项公式求得数列的通项公式，从而求得最后的结果.10【2018 年全国卷文】等比数列中，（1）求的通项公式；（2）记为的前项和若，求【答案】（1）或（2）【解析】分析：（1）列出方程，解出q 可得；（2）求出前n 项和，解方程可得m。详解：（1）设的公比为，由题设得由已知得，解得（舍去），或第 7 页，共 20 页文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3故或（2）若，则由得，此方程没有正整数解若，则由得，解得综上，点睛：本题主要考查等比数列的通项公式和前n 项和公式，属于基础题。11【2018 年天津卷文】设变量x，y满足约束条件则目标函数的最大值为A.6 B.19 C.21 D.45【答案】C.本题选择C选项.点睛：求线性目标函数zaxby(ab0)的最值，当b0 时，直线过可行域且在y轴上截距最大时，z值最大，在y轴截距最小时，z值最小；当b0 时，直线过可行域且在y轴上截距最大时，z值最小，在y轴上截距最小时，z值最大.12【2018 年文北京卷】设集合则A.对任意实数a，B.对任意实数a，（2,1）第 8 页，共 20 页文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3C.当且仅当a0 时,（2,1）D.当且仅当时,（2,1）【答案】D 点睛：此题主要结合充分与必要条件考查线性规划的应用，集合法是判断充分条件与必要条件的一种非常有效的方法，根据成立时对应的集合之间的包含关系进行判断.设，若，则；若，则，当一个问题从正面思考很难入手时，可以考虑其逆否命题形式.13【2018 年浙江卷】若满足约束条件则的最小值是 _，最大值是 _【答案】-2 8【解析】分析:先作可行域，再平移目标函数对应的直线，从而确定最值.详解：作可行域，如图中阴影部分所示，则直线过点A(2,2)时 取最大值8，过点B(4,-2)时 取最小值-2.点睛：线性规划的实质是把代数问题几何化，即用数形结合的思想解题.需要注意的是：一，准确无误地作出可行域；二，画目标函数所对应的直线时，要注意与约束条件中的直线的斜率进行比较，避免出错；三，一般情况下，目标函数的最大或最小值会在可行域的端点或边界处取得.14【2018 年天津卷文】已知，且，则的最小值为 _.【答案】第 9 页，共 20 页文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3文档编码:CH6C8C7M8A7 HK3G8B5G9B10 ZQ7K3T2A7R3点睛：在应用基本不等式求最值时，要把握不等式成立的三个条件，就是“一正各项均为正；二定积或和为定值；三相等等号能否取得”，若忽略了某个条件，就会出现错误15【2018 年文北京卷】若,y满足，则 2y-?最