新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第4講 函數(shù)及其性質(zhì)（解析版）

第4講函數(shù)及其性質(zhì)學(xué)校____________姓名____________班級(jí)____________一、知識(shí)梳理基本概念1.函數(shù)的概念概念一般地，給定兩個(gè)非空實(shí)數(shù)集A與B，以及對(duì)應(yīng)關(guān)系f，如果對(duì)于集合A中的每一個(gè)實(shí)數(shù)x，在集合B中都有唯一確定的實(shí)數(shù)y與x對(duì)應(yīng)，則稱f為定義在集合A上的一個(gè)函數(shù)，記作y＝f(x)，x∈A三要素對(duì)應(yīng)關(guān)系y＝f(x)，x∈A定義域自變量取值的范圍值域所有函數(shù)值組成的集合{y∈B|y＝f(x)，x∈A}2.同一個(gè)函數(shù)(1)前提條件：①定義域相同；②對(duì)應(yīng)關(guān)系相同.(2)結(jié)論：這兩個(gè)函數(shù)為同一個(gè)函數(shù).3.函數(shù)的表示法表示函數(shù)的常用方法有解析法、圖像法和列表法.4.分段函數(shù)(1)若函數(shù)在其定義域的不同子集上，因?qū)?yīng)關(guān)系不同而分別用幾個(gè)不同的式子來表示，這種函數(shù)稱為分段函數(shù).分段函數(shù)表示的是一個(gè)函數(shù).(2)分段函數(shù)的定義域等于各段函數(shù)的定義域的并集，其值域等于各段函數(shù)的值域的并集.單調(diào)性與最值1.函數(shù)的單調(diào)性(1)單調(diào)函數(shù)的定義增函數(shù)減函數(shù)定義一般地，設(shè)函數(shù)y＝f(x)的定義域?yàn)镈，且I?D如果對(duì)任意x1，x2∈I，當(dāng)x1<x2時(shí)，都有f(x1)<f(x2)，則稱y＝f(x)在I上是增函數(shù)如果對(duì)任意x1，x2∈I，當(dāng)x1<x2時(shí)，都有f(x1)>f(x2)，則稱y＝f(x)在I上是減函數(shù)圖像描述自左向右看圖像是上升的自左向右看圖像是下降的(2)單調(diào)區(qū)間的定義如果函數(shù)y＝f(x)在區(qū)間I上是增函數(shù)或減函數(shù)，那么就說函數(shù)y＝f(x)在區(qū)間I具有單調(diào)性，區(qū)間I稱為函數(shù)y＝f(x)的單調(diào)區(qū)間.2.函數(shù)的最值一般地，設(shè)函數(shù)f(x)的定義域?yàn)镈，且x0∈D：如果對(duì)任意x∈D，都有f(x)≤f(x0)，則稱f(x)的最大值為f(x0)，而x0稱為f(x)的最大值點(diǎn)；如果對(duì)任意x∈D，都有f(x)≥f(x0)，則稱f(x)的最小值為f(x0)，而x0稱為f(x)的最小值點(diǎn).最大值和最小值統(tǒng)稱為最值，最大值點(diǎn)和最小值點(diǎn)統(tǒng)稱為最值點(diǎn).奇偶性、周期性1.函數(shù)的奇偶性奇偶性定義圖像特點(diǎn)偶函數(shù)一般地，設(shè)函數(shù)y＝f(x)的定義域?yàn)镈，如果對(duì)D內(nèi)的任意一個(gè)x，都有－x∈D，且f(－x)＝f(x)，則稱y＝f(x)為偶函數(shù)關(guān)于y軸對(duì)稱奇函數(shù)一般地，設(shè)函數(shù)y＝f(x)的定義域?yàn)镈，如果對(duì)D內(nèi)的任意一個(gè)x，都有－x∈D，且f(－x)＝－f(x)，則稱y＝f(x)為奇函數(shù)關(guān)于原點(diǎn)對(duì)稱2.函數(shù)的周期性(1)周期函數(shù)：對(duì)于函數(shù)y＝f(x)，如果存在一個(gè)非零常數(shù)T，使得當(dāng)x取定義域內(nèi)的任何值時(shí)，都有f(x＋T)＝f(x)，那么就稱函數(shù)y＝f(x)為周期函數(shù)，稱T為這個(gè)函數(shù)的周期.(2)最小正周期：如果在周期函數(shù)f(x)的所有周期中存在一個(gè)最小的正數(shù)，那么這個(gè)最小正數(shù)就叫做f(x)的最小正周期.考點(diǎn)和典型例題1、函數(shù)的概念【典例1-1】（2022·河南·長(zhǎng)葛市第一高級(jí)中學(xué)模擬預(yù)測(cè)（理））已知SKIPIF1<0是定義在R上的奇函數(shù)，且SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．27 B．－27 C．54 D．－54【答案】A【詳解】由已知可得SKIPIF1<0，SKIPIF1<0，因此，SKIPIF1<0．故選:A．【典例1-2】（2022·河南·長(zhǎng)葛市第一高級(jí)中學(xué)模擬預(yù)測(cè)（文））若函數(shù)SKIPIF1<0則SKIPIF1<0（

）A．10 B．9 C．12 D．11．【答案】A【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0．故選：A.【典例1-3】（2022·北京·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的定義域是_______．【答案】SKIPIF1<0【詳解】由題意可得，SKIPIF1<0，解之得SKIPIF1<0則函數(shù)SKIPIF1<0的定義域是SKIPIF1<0故答案為：SKIPIF1<0【典例1-4】（2022·浙江·模擬預(yù)測(cè)）已知SKIPIF1<0，則SKIPIF1<0___________.【答案】11【詳解】由于SKIPIF1<0，從而SKIPIF1<0.故答案為：11.【典例1-5】（2022·浙江溫州·三模）已知函數(shù)SKIPIF1<0若SKIPIF1<0，則實(shí)數(shù)a的值等于___________.【答案】SKIPIF1<0【詳解】①當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（舍）②當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0Ⅰ：當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),有SKIPIF1<0Ⅱ:當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，有SKIPIF1<0SKIPIF1<0無解綜上，SKIPIF1<0.故答案為：SKIPIF1<02、單調(diào)性及其應(yīng)用【典例2-1】（2022·北京·二模）下列函數(shù)中，與函數(shù)SKIPIF1<0的奇偶性相同，且在SKIPIF1<0上有相同單調(diào)性的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0為奇函數(shù)且在SKIPIF1<0上遞增，A、B：SKIPIF1<0、SKIPIF1<0非奇非偶函數(shù)，排除；C：SKIPIF1<0為奇函數(shù)，但在SKIPIF1<0上不單調(diào)，排除；D：SKIPIF1<0，顯然SKIPIF1<0且定義域關(guān)于原點(diǎn)對(duì)稱，在SKIPIF1<0上遞增，滿足.故選：D【典例2-2】（2022·貴州遵義·三模（文））若奇函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，則滿足SKIPIF1<0的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0是奇函數(shù)在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，又SKIPIF1<0或SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0滿足SKIPIF1<0的x的取值范圍是SKIPIF1<0或SKIPIF1<0故選：D【典例2-3】（2022·河北唐山·二模）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：SKIPIF1<0定義域?yàn)镽，又SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0在定義域R上遞增，又SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故選：C【典例2-4】（2022·山西太原·二模（文））已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．SKIPIF1<0的圖象關(guān)于直線x=1對(duì)稱 D．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】C【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以A不正確；因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B不正確；因?yàn)镾KIPIF1<0SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于直線x=1對(duì)稱，故C正確；在SKIPIF1<0的圖象上取一點(diǎn)SKIPIF1<0，則其關(guān)于點(diǎn)SKIPIF1<0的點(diǎn)為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以點(diǎn)SKIPIF1<0不在函數(shù)SKIPIF1<0的圖象上，故SKIPIF1<0的圖象不關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，故D不正確.故選：C【典例2-5】（2022·貴州遵義·三模（文））已知函數(shù)SKIPIF1<0滿足：①SKIPIF1<0；②SKIPIF1<0；③在SKIPIF1<0上單調(diào)遞減，寫出一個(gè)同時(shí)滿足條件①②③的函數(shù)SKIPIF1<0_________．【答案】SKIPIF1<0（答案不唯一）【詳解】由題意可知，SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，且在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，可取SKIPIF1<0滿足條件.故答案為：SKIPIF1<0（答案不唯一）.【典例2-6】（2022·全國(guó)·三模（文））函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為__________．【答案】SKIPIF1<0【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則其在SKIPIF1<0上遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，綜上，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故答案為：SKIPIF1<03、奇偶性及其應(yīng)用【典例3-1】（2022·內(nèi)蒙古呼和浩特·二模（文））函數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，則SKIPIF1<0（

）A．-8 B．0 C．-4 D．-2【答案】B【詳解】∵SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，∴SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，即SKIPIF1<0是奇函數(shù)，令SKIPIF1<0得，SKIPIF1<0，即SKIPIF1<0SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0．∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，即函數(shù)的周期是4．∴SKIPIF1<0．故選：B.【典例3-2】（2022·內(nèi)蒙古呼和浩特·二模（文））已知函數(shù)SKIPIF1<0，則圖象為下圖的函數(shù)可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由題意，函數(shù)SKIPIF1<0，根據(jù)函數(shù)圖象可得函數(shù)圖象關(guān)于原點(diǎn)對(duì)稱，所以函數(shù)為奇函數(shù)，對(duì)于A中，函數(shù)SKIPIF1<0不是奇函數(shù)，所以A不符合題意；對(duì)于B中，函數(shù)SKIPIF1<0不是奇函數(shù)，所以B不符合題意；對(duì)于C中，函數(shù)SKIPIF1<0此時(shí)函數(shù)為奇函數(shù)，又由SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)在區(qū)間SKIPIF1<0單調(diào)遞增，而圖象中先增后減，所以C不符合題意.故選：D.【典例3-3】（2022·上海市市西中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0是定義在R上的單調(diào)增函數(shù)且為奇函數(shù)，數(shù)列SKIPIF1<0是等差數(shù)列，若前2022項(xiàng)和小于零，則SKIPIF1<0的值(

)A．恒為正數(shù) B．恒為負(fù)數(shù) C．恒為0 D．可正可負(fù)【答案】B【詳解】SKIPIF1<0函數(shù)SKIPIF1<0是R上的奇函數(shù)且是增函數(shù)，SKIPIF1<0，且當(dāng)SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0．設(shè)等差數(shù)列前n項(xiàng)和為SKIPIF1<0，由題可知SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0(1≤n≤2011，SKIPIF1<0)．所以SKIPIF1<0，結(jié)合函數(shù)SKIPIF1<0在R上的單調(diào)增和奇函數(shù)性質(zhì)，可得SKIPIF1<0，所以SKIPIF1<0∴SKIPIF1<0SKIPIF1<0<0；綜上，SKIPIF1<0的值恒為負(fù)數(shù)．故選：B．【典例3-4】（2022·河南開封·三模（理））函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B．C． D．【答案】A【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0為奇函數(shù)，圖象關(guān)于原點(diǎn)對(duì)稱，故CD不正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故B不正確.故選：A【典例3-5】（2022·安徽省蕪湖市教育局模擬預(yù)測(cè)（文））下列函數(shù)中是奇函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】對(duì)于A，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0為非奇非偶函數(shù)，對(duì)于B，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0為偶函數(shù)，對(duì)于C，SKIPIF1<0，SKIPIF1<0為偶函數(shù)，對(duì)于D，易知定義域?yàn)镽,SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為奇函數(shù)．故選：D4、函數(shù)性質(zhì)的綜合應(yīng)用【典例4-1】（2022·福建福州·三模）已知函數(shù)SKIPIF1<0，以下結(jié)論中錯(cuò)誤的是（

）A．SKIPIF1<0是偶函數(shù) B．SKIPIF1<0有無數(shù)個(gè)零點(diǎn)C．SKIPIF1<0的最小值為SKIPIF1<0 D．SKIPIF1<0的最大值為SKIPIF1<0【答案】C【詳解】對(duì)于A，SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0為偶函數(shù)，A正確；對(duì)于B，令SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0有無數(shù)個(gè)零點(diǎn)，B正確；對(duì)于C，SKIPIF1<0，SKIPIF1<0若SKIPIF1<0的最小值為SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的一個(gè)極小值點(diǎn)，則SKIPIF1<0；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0不是SKIPIF1<0的極小值點(diǎn)，C錯(cuò)誤；對(duì)于D，SKIPIF1<0，SKIPIF1<0；則當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0取得最大值SKIPIF1<0，D正確.故選：C.【典例4-2】（2022·吉林白山·三模（理））已知函數(shù)SKIPIF1<0，若對(duì)任意SKIPIF1<0，SKIPIF1<0，SKIPIF1<0恒成立，則m的最大值為（

）A．－1 B．0 C．1 D．e【答案】C【詳解】由題知SKIPIF1<0對(duì)任意SKIPIF1<0，SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0對(duì)任意SKIPIF1<0，SKIPIF1<0恒成立，不妨設(shè)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，則原式等價(jià)于SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0恒成立，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，即m的最大值為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取得最大值，故選：C.【典例4-3】（2022·江蘇南京·三模）已知SKIPIF1<0，若?x≥1，f（x＋2m）＋mf（x）＞0，則實(shí)數(shù)m的取值范圍是（

）A．（－1，＋∞） B．SKIPIF1<0C．（0，＋∞） D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0時(shí)，SKIPIF1<0，符合題意；SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0顯然SKIPIF1<0在R上遞增，則SKIPIF1<0對(duì)SKIPIF1<0恒成立SKIPIF1<0對(duì)SKIPIF1<0恒成立則：SKIPIF1<0；綜上，SKIPIF1<0，故選：B．【典例4-4】（2022·全國(guó)·高三專題練習(xí)）（多選題）已知函數(shù)SKIPIF1<0，下列說法正確的是（

）A．若SKIPIF1<0是偶函數(shù)，則SKIPIF1<0 B．若SKIPIF1<0，則函數(shù)SKIPIF1<0是奇函數(shù)C．若SKIPIF1<0，則函數(shù)SKIPIF1<0存在最小值SKIPIF1<0 D．若函數(shù)SKIPIF1<0存在極值，則實(shí)數(shù)a的取值范圍是SKIPIF1<0【答案】ACD【詳解】對(duì)于A，函數(shù)的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0恒成立，故SKIPIF1<0，故A正確；對(duì)于B，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0不成立，故B不正確；對(duì)于C，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，所以C正確；對(duì)于D，SKIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0存在極值，則SKIPIF1<0有解，令SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以D正確.故選：ACD.【典例4-5】（2022·河南·模擬預(yù)測(cè)（理））已知SKIPIF1<0的定義域?yàn)镽，若函數(shù)滿足SKIPIF1<0，則稱SKIPIF1<0為SKIPIF1<0的一個(gè)不動(dòng)點(diǎn)，有下列結(jié)論：①SKIPIF1<0的不動(dòng)點(diǎn)是3；②SKIPIF1<0存在不動(dòng)點(diǎn)；③若函數(shù)SKIPIF1<0為奇函數(shù)，則其存在奇數(shù)個(gè)不動(dòng)點(diǎn)；若SKIPIF1<0為偶函數(shù)，則其存在偶數(shù)個(gè)不動(dòng)點(diǎn)；④若SKIPIF1<0為周期函數(shù)，則其存在無數(shù)個(gè)不動(dòng)點(diǎn)；⑤若SKIPIF1<0存在不