新高考數(shù)學(xué)二輪復(fù)習(xí)專(zhuān)題講測(cè)練專(zhuān)題15 周期性、單調(diào)性、奇偶性、對(duì)稱(chēng)性的靈活運(yùn)用（精講精練）（解析版）

專(zhuān)題15周期性、單調(diào)性、奇偶性、對(duì)稱(chēng)性的靈活運(yùn)用【命題規(guī)律】從近五年的高考情況來(lái)看，本節(jié)是高考的一個(gè)重點(diǎn)，函數(shù)的單調(diào)性、奇偶性、周期性是高考的必考內(nèi)容，重點(diǎn)關(guān)注單調(diào)性、奇偶性結(jié)合在一起，與函數(shù)圖像、函數(shù)零點(diǎn)和不等式相結(jié)合進(jìn)行考查，解題時(shí)要充分運(yùn)用轉(zhuǎn)化思想和數(shù)形結(jié)合思想．【核心考點(diǎn)目錄】核心考點(diǎn)一：函數(shù)單調(diào)性的綜合應(yīng)用核心考點(diǎn)二：函數(shù)的奇偶性的綜合應(yīng)用核心考點(diǎn)三：已知SKIPIF1<0奇函數(shù)SKIPIF1<0核心考點(diǎn)四：利用軸對(duì)稱(chēng)解決函數(shù)問(wèn)題核心考點(diǎn)五：利用中心對(duì)稱(chēng)解決函數(shù)問(wèn)題核心考點(diǎn)六：利用周期性和對(duì)稱(chēng)性解決函數(shù)問(wèn)題核心考點(diǎn)七：類(lèi)周期函數(shù)核心考點(diǎn)八：抽象函數(shù)的單調(diào)性、奇偶性、周期性、對(duì)稱(chēng)性核心考點(diǎn)九：函數(shù)性質(zhì)的綜合【真題回歸】1．（2022·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0的定義域?yàn)镽，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．1【答案】A【解析】[方法一]：賦值加性質(zhì)因?yàn)镾KIPIF1<0，令SKIPIF1<0可得，SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0可得，SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，令SKIPIF1<0得，SKIPIF1<0，即有SKIPIF1<0，從而可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0的一個(gè)周期為SKIPIF1<0．因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以一個(gè)周期內(nèi)的SKIPIF1<0．由于22除以6余4，所以SKIPIF1<0．故選：A．[方法二]：【最優(yōu)解】構(gòu)造特殊函數(shù)由SKIPIF1<0，聯(lián)想到余弦函數(shù)和差化積公式SKIPIF1<0，可設(shè)SKIPIF1<0，則由方法一中SKIPIF1<0知SKIPIF1<0，解得SKIPIF1<0，取SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0符合條件，因此SKIPIF1<0的周期SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，由于22除以6余4，所以SKIPIF1<0．故選：A．【整體點(diǎn)評(píng)】法一：利用賦值法求出函數(shù)的周期，即可解出，是該題的通性通法；2．（2022·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0的定義域均為R，且SKIPIF1<0．若SKIPIF1<0的圖像關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0的圖像關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，代入得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，聯(lián)立得，SKIPIF1<0，所以SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镽，所以SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0．所以SKIPIF1<0．故選：D3．（多選題）（2022·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0及其導(dǎo)函數(shù)SKIPIF1<0的定義域均為SKIPIF1<0，記SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0均為偶函數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】[方法一]：對(duì)稱(chēng)性和周期性的關(guān)系研究對(duì)于SKIPIF1<0，因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0即SKIPIF1<0①，所以SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0，故C正確；對(duì)于SKIPIF1<0，因?yàn)镾KIPIF1<0為偶函數(shù)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，由①求導(dǎo)，和SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，因?yàn)槠涠x域?yàn)镽，所以SKIPIF1<0，結(jié)合SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，從而周期SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故B正確，D錯(cuò)誤；若函數(shù)SKIPIF1<0滿(mǎn)足題設(shè)條件，則函數(shù)SKIPIF1<0（C為常數(shù)）也滿(mǎn)足題設(shè)條件，所以無(wú)法確定SKIPIF1<0的函數(shù)值，故A錯(cuò)誤．故選：BC．[方法二]：【最優(yōu)解】特殊值，構(gòu)造函數(shù)法．由方法一知SKIPIF1<0周期為2，關(guān)于SKIPIF1<0對(duì)稱(chēng)，故可設(shè)SKIPIF1<0，則SKIPIF1<0，顯然A，D錯(cuò)誤，選BC．故選：BC．[方法三]：因?yàn)镾KIPIF1<0，SKIPIF1<0均為偶函數(shù)，所以SKIPIF1<0即SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故C正確；函數(shù)SKIPIF1<0，SKIPIF1<0的圖象分別關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，又SKIPIF1<0，且函數(shù)SKIPIF1<0可導(dǎo)，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故B正確，D錯(cuò)誤；若函數(shù)SKIPIF1<0滿(mǎn)足題設(shè)條件，則函數(shù)SKIPIF1<0（C為常數(shù)）也滿(mǎn)足題設(shè)條件，所以無(wú)法確定SKIPIF1<0的函數(shù)值，故A錯(cuò)誤．故選：BC．【整體點(diǎn)評(píng)】方法一：根據(jù)題意賦值變換得到函數(shù)的性質(zhì)，即可判斷各選項(xiàng)的真假，轉(zhuǎn)化難度較高，是該題的通性通法；方法二：根據(jù)題意得出的性質(zhì)構(gòu)造特殊函數(shù)，再驗(yàn)證選項(xiàng)，簡(jiǎn)單明了，是該題的最優(yōu)解．4．（2022·全國(guó)·統(tǒng)考高考真題）若SKIPIF1<0是奇函數(shù)，則SKIPIF1<0_____，SKIPIF1<0______．【答案】

SKIPIF1<0；

SKIPIF1<0．【解析】[方法一]：奇函數(shù)定義域的對(duì)稱(chēng)性若SKIPIF1<0，則SKIPIF1<0的定義域?yàn)镾KIPIF1<0，不關(guān)于原點(diǎn)對(duì)稱(chēng)SKIPIF1<0若奇函數(shù)的SKIPIF1<0有意義，則SKIPIF1<0且SKIPIF1<0SKIPIF1<0且SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0為奇函數(shù)，定義域關(guān)于原點(diǎn)對(duì)稱(chēng)，SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0得，SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0；SKIPIF1<0．[方法二]：函數(shù)的奇偶性求參SKIPIF1<0SKIPIF1<0SKIPIF1<0函數(shù)SKIPIF1<0為奇函數(shù)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0[方法三]：因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，所以其定義域關(guān)于原點(diǎn)對(duì)稱(chēng)．由SKIPIF1<0可得，SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0，即函數(shù)的定義域?yàn)镾KIPIF1<0，再由SKIPIF1<0可得，SKIPIF1<0．即SKIPIF1<0，在定義域內(nèi)滿(mǎn)足SKIPIF1<0，符合題意．故答案為：SKIPIF1<0；SKIPIF1<0．【方法技巧與總結(jié)】1、單調(diào)性技巧（1）證明函數(shù)單調(diào)性的步驟①取值：設(shè)SKIPIF1<0，SKIPIF1<0是SKIPIF1<0定義域內(nèi)一個(gè)區(qū)間上的任意兩個(gè)量，且SKIPIF1<0；②變形：作差變形（變形方法：因式分解、配方、有理化等）或作商變形；③定號(hào)：判斷差的正負(fù)或商與SKIPIF1<0的大小關(guān)系；④得出結(jié)論．（2）函數(shù)單調(diào)性的判斷方法①定義法：根據(jù)增函數(shù)、減函數(shù)的定義，按照“取值—變形—判斷符號(hào)—下結(jié)論”進(jìn)行判斷．②圖象法：就是畫(huà)出函數(shù)的圖象，根據(jù)圖象的上升或下降趨勢(shì)，判斷函數(shù)的單調(diào)性．③直接法：就是對(duì)我們所熟悉的函數(shù)，如一次函數(shù)、二次函數(shù)、反比例函數(shù)等，直接寫(xiě)出它們的單調(diào)區(qū)間．（3）記住幾條常用的結(jié)論：①若SKIPIF1<0是增函數(shù)，則SKIPIF1<0為減函數(shù)；若SKIPIF1<0是減函數(shù)，則SKIPIF1<0為增函數(shù)；②若SKIPIF1<0和SKIPIF1<0均為增（或減）函數(shù)，則在SKIPIF1<0和SKIPIF1<0的公共定義域上SKIPIF1<0為增（或減）函數(shù)；③若SKIPIF1<0且SKIPIF1<0為增函數(shù)，則函數(shù)SKIPIF1<0為增函數(shù)，SKIPIF1<0為減函數(shù)；④若SKIPIF1<0且SKIPIF1<0為減函數(shù)，則函數(shù)SKIPIF1<0為減函數(shù)，SKIPIF1<0為增函數(shù)．2、奇偶性技巧（1）函數(shù)具有奇偶性的必要條件是其定義域關(guān)于原點(diǎn)對(duì)稱(chēng)．（2）奇偶函數(shù)的圖象特征．函數(shù)SKIPIF1<0是偶函數(shù)SKIPIF1<0函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)；函數(shù)SKIPIF1<0是奇函數(shù)SKIPIF1<0函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)中心對(duì)稱(chēng)．（3）若奇函數(shù)SKIPIF1<0在SKIPIF1<0處有意義，則有SKIPIF1<0；偶函數(shù)SKIPIF1<0必滿(mǎn)足SKIPIF1<0．（4）偶函數(shù)在其定義域內(nèi)關(guān)于原點(diǎn)對(duì)稱(chēng)的兩個(gè)區(qū)間上單調(diào)性相反；奇函數(shù)在其定義域內(nèi)關(guān)于原點(diǎn)對(duì)稱(chēng)的兩個(gè)區(qū)間上單調(diào)性相同．（5）若函數(shù)SKIPIF1<0的定義域關(guān)于原點(diǎn)對(duì)稱(chēng)，則函數(shù)SKIPIF1<0能表示成一個(gè)偶函數(shù)與一個(gè)奇函數(shù)的和的形式．記SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．（6）運(yùn)算函數(shù)的奇偶性規(guī)律：運(yùn)算函數(shù)是指兩個(gè)（或多個(gè)）函數(shù)式通過(guò)加、減、乘、除四則運(yùn)算所得的函數(shù)，如SKIPIF1<0．對(duì)于運(yùn)算函數(shù)有如下結(jié)論：奇SKIPIF1<0奇=奇；偶SKIPIF1<0偶=偶；奇SKIPIF1<0偶=非奇非偶；奇SKIPIF1<0奇=偶；奇SKIPIF1<0偶=奇；偶SKIPIF1<0偶=偶．（7）復(fù)合函數(shù)SKIPIF1<0的奇偶性原來(lái)：內(nèi)偶則偶，兩奇為奇．（8）常見(jiàn)奇偶性函數(shù)模型奇函數(shù)：=1\*GB3①函數(shù)SKIPIF1<0或函數(shù)SKIPIF1<0．=2\*GB3②函數(shù)SKIPIF1<0．=3\*GB3③函數(shù)SKIPIF1<0或函數(shù)SKIPIF1<0=4\*GB3④函數(shù)SKIPIF1<0或函數(shù)SKIPIF1<0．注意：關(guān)于=1\*GB3①式，可以寫(xiě)成函數(shù)SKIPIF1<0或函數(shù)SKIPIF1<0．偶函數(shù)：=1\*GB3①函數(shù)SKIPIF1<0．=2\*GB3②函數(shù)SKIPIF1<0．=3\*GB3③函數(shù)SKIPIF1<0類(lèi)型的一切函數(shù)．④常數(shù)函數(shù)3、周期性技巧SKIPIF1<04、函數(shù)的的對(duì)稱(chēng)性與周期性的關(guān)系（1）若函數(shù)SKIPIF1<0有兩條對(duì)稱(chēng)軸SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0是周期函數(shù)，且SKIPIF1<0；（2）若函數(shù)SKIPIF1<0的圖象有兩個(gè)對(duì)稱(chēng)中心SKIPIF1<0，則函數(shù)SKIPIF1<0是周期函數(shù)，且SKIPIF1<0；（3）若函數(shù)SKIPIF1<0有一條對(duì)稱(chēng)軸SKIPIF1<0和一個(gè)對(duì)稱(chēng)中心SKIPIF1<0，則函數(shù)SKIPIF1<0是周期函數(shù)，且SKIPIF1<0．5、對(duì)稱(chēng)性技巧（1）若函數(shù)SKIPIF1<0關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0．（2）若函數(shù)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0．（3）函數(shù)SKIPIF1<0與SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，函數(shù)SKIPIF1<0與SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱(chēng)．【核心考點(diǎn)】核心考點(diǎn)一：函數(shù)單調(diào)性的綜合應(yīng)用【典型例題】例1．（2023春·江西鷹潭·高三貴溪市實(shí)驗(yàn)中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0是SKIPIF1<0上的減函數(shù)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】顯然當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為單調(diào)減函數(shù)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則對(duì)稱(chēng)軸為SKIPIF1<0，SKIPIF1<0若SKIPIF1<0是SKIPIF1<0上減函數(shù)，則SKIPIF1<0解得SKIPIF1<0，故選：A．例2．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)函數(shù)SKIPIF1<0，則滿(mǎn)足SKIPIF1<0的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】假設(shè)SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，而SKIPIF1<0是SKIPIF1<0向右平移1個(gè)單位長(zhǎng)度，向上平移3個(gè)單位長(zhǎng)度，所以SKIPIF1<0的對(duì)稱(chēng)中心為SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0求導(dǎo)得SKIPIF1<0因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0，取等號(hào)，所以SKIPIF1<0所以SKIPIF1<0在R上單調(diào)遞增，因?yàn)镾KIPIF1<0得SKIPIF1<0所以SKIPIF1<0，解得SKIPIF1<0，故選：B例3．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，且滿(mǎn)足SKIPIF1<0，則下列正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，或SKIPIF1<0，∴SKIPIF1<0（舍去），或SKIPIF1<0，即SKIPIF1<0，故A錯(cuò)誤；又SKIPIF1<0，故SKIPIF1<0，∴SKIPIF1<0，對(duì)于函數(shù)SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，故D錯(cuò)誤；∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴函數(shù)SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，故B正確；∵SKIPIF1<0，∴函數(shù)SKIPIF1<0單調(diào)遞增，故函數(shù)SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，即SKIPIF1<0，故C錯(cuò)誤．故選：B．核心考點(diǎn)二：函數(shù)的奇偶性的綜合應(yīng)用【典型例題】例4．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0為偶函數(shù)，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵SKIPIF1<0為偶函數(shù)，∴SKIPIF1<0，即函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，由SKIPIF1<0，可得SKIPIF1<0，整理得，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．故選：B．例5．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)SKIPIF1<0是定義在R上的奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】根據(jù)題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0在R上為增函數(shù)，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以不等式SKIPIF1<0可化為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0，故選：C例6．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知偶函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則使不等式SKIPIF1<0成立的實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，不等式SKIPIF1<0即為SKIPIF1<0．又因?yàn)镾KIPIF1<0是偶函數(shù)，所以不等式SKIPIF1<0等價(jià)于SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0．綜上可知，實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0，故選：A．例7．（2023·全國(guó)·高三專(zhuān)題練習(xí)）定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以不等式SKIPIF1<0，可化為SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在R上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0．故選：D．例8．（2023春·廣西·高三期末）SKIPIF1<0是定義在R上的函數(shù)，SKIPIF1<0為奇函數(shù)，則SKIPIF1<0（

）A．－1 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】A【解析】SKIPIF1<0是定義在R上的函數(shù)，SKIPIF1<0為奇函數(shù)，則SKIPIF1<0．∴SKIPIF1<0．故選：A例9．（2023春·甘肅蘭州·高三蘭化一中?？茧A段練習(xí)）若函數(shù)f（x）=SKIPIF1<0，則滿(mǎn)足SKIPIF1<0恒成立的實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，由SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是SKIPIF1<0上的增函數(shù)，所以SKIPIF1<0等價(jià)于：SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則問(wèn)題轉(zhuǎn)化為：SKIPIF1<0，因?yàn)镾KIPIF1<0且定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0是SKIPIF1<0上的偶函數(shù)，所以只需求SKIPIF1<0在SKIPIF1<0上的最大值即可．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，可得：SKIPIF1<0，即SKIPIF1<0，故選：A．核心考點(diǎn)三：已知SKIPIF1<0奇函數(shù)+M【典型例題】例10．（2022·重慶一中高三階段練習(xí)）已知SKIPIF1<0（a，b為實(shí)數(shù)），SKIPIF1<0，則SKIPIF1<0______．【答案】-2014【解析】SKIPIF1<0，因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，其中SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0故答案為：-2014例11．（2022·河南·西平縣高級(jí)中學(xué)模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．2 B．3 C．－2 D．－3【答案】D【解析】設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0為奇函數(shù)，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0．故選：D．例12．（2022·福建省福州第一中學(xué)高二期末）若對(duì)SKIPIF1<0，有SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在最大值和最小值，則其最大值與最小值的和為（）A．4 B．8 C．12 D．16【答案】B【解析】由題設(shè)，SKIPIF1<0且SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0為奇函數(shù)，令SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0是奇函數(shù)，∴SKIPIF1<0在SKIPIF1<0上的最小、最大值的和為0，即SKIPIF1<0，∴SKIPIF1<0．故選：B核心考點(diǎn)四：利用軸對(duì)稱(chēng)解決函數(shù)問(wèn)題【典型例題】例13．（2022·全國(guó)·高三專(zhuān)題練習(xí)）若SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0等于（

）A．2 B．3 C．4 D．5【答案】D【解析】由題意SKIPIF1<0，故有SKIPIF1<0故SKIPIF1<0和SKIPIF1<0是直線(xiàn)SKIPIF1<0和曲線(xiàn)SKIPIF1<0、曲線(xiàn)SKIPIF1<0交點(diǎn)的橫坐標(biāo)．根據(jù)函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0互為反函數(shù)，它們的圖象關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，故曲線(xiàn)SKIPIF1<0和曲線(xiàn)SKIPIF1<0的圖象交點(diǎn)關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)．即點(diǎn)（x1，5﹣x1）和點(diǎn)（x2，5﹣x2）構(gòu)成的線(xiàn)段的中點(diǎn)在直線(xiàn)y=x上，即SKIPIF1<0，求得x1+x2=5，故選：D．例14．（2021春·高一單元測(cè)試）設(shè)函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．（0，2] B．SKIPIF1<0C．[2，＋∞） D．SKIPIF1<0∪[2，＋∞）【答案】B【解析】由題意，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，所以函數(shù)SKIPIF1<0為SKIPIF1<0的偶函數(shù)，且在SKIPIF1<0上為單調(diào)遞減函數(shù)，令SKIPIF1<0，可得SKIPIF1<0，則不等式SKIPIF1<0可化為SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以不等式的解集為SKIPIF1<0．故選：B．例15．（2021春·西藏拉薩·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的大小關(guān)系（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【解析】令SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)；SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上是增函數(shù)，將SKIPIF1<0圖像向右平移一個(gè)單位得到SKIPIF1<0圖像，所以SKIPIF1<0關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，且在SKIPIF1<0單調(diào)遞增．∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，又∵SKIPIF1<0關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，∴SKIPIF1<0，∴SKIPIF1<0．故選：A核心考點(diǎn)五：利用中心對(duì)稱(chēng)解決函數(shù)問(wèn)題【典型例題】例16．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0是SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0，又SKIPIF1<0為SKIPIF1<0上的偶函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0．故選：C．例17．（2021春·安徽六安·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0為奇函數(shù)，若函數(shù)SKIPIF1<0與SKIPIF1<0圖象共有SKIPIF1<0個(gè)交點(diǎn)為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，故函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，即SKIPIF1<0，故函數(shù)SKIPIF1<0的圖象也關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，函數(shù)SKIPIF1<0與SKIPIF1<0圖象共有SKIPIF1<0個(gè)交點(diǎn)為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，且這六個(gè)點(diǎn)也關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以，SKIPIF1<0．故選：B．例18．（2021春·貴州黔東南·高一凱里一中?？计谥校┮阎瘮?shù)SKIPIF1<0是奇函數(shù)，若函數(shù)SKIPIF1<0與SKIPIF1<0圖象的交點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0，則交點(diǎn)的所有橫坐標(biāo)和縱坐標(biāo)之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題可得SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0的圖象也關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，即若點(diǎn)SKIPIF1<0為交點(diǎn)，則點(diǎn)SKIPIF1<0也為交點(diǎn)，同理若SKIPIF1<0為交點(diǎn)，則點(diǎn)SKIPIF1<0也為交點(diǎn)，……則交點(diǎn)的所有橫坐標(biāo)和縱坐標(biāo)之和為SKIPIF1<0SKIPIF1<0SKIPIF1<0，故選：D．例19．（2022春·湖北恩施·高一恩施市第一中學(xué)?？茧A段練習(xí)）已知定義在R上的奇函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)楹瘮?shù)SKIPIF1<0是定義在R上的奇函數(shù)，則SKIPIF1<0，且函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸交點(diǎn)關(guān)于原點(diǎn)對(duì)稱(chēng)，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，則不等式SKIPIF1<0，即為SKIPIF1<0，解得SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0．故選：A．例20．（2021春·四川綿陽(yáng)·高一四川省綿陽(yáng)南山中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，若函數(shù)SKIPIF1<0恰有SKIPIF1<0個(gè)零點(diǎn)，則所有這些零點(diǎn)之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，且SKIPIF1<0（1）SKIPIF1<0，函數(shù)SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0為奇函數(shù)，其圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，又函數(shù)SKIPIF1<0是由函數(shù)SKIPIF1<0向右平移一個(gè)單位得到的函數(shù)，故函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，令SKIPIF1<0，則SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0與SKIPIF1<0的圖象都關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以?xún)蓚€(gè)函數(shù)圖象的交點(diǎn)也關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，因?yàn)楹瘮?shù)SKIPIF1<0恰有2021個(gè)零點(diǎn)，所以2021個(gè)零點(diǎn)除SKIPIF1<0之外的2020個(gè)零點(diǎn)關(guān)于SKIPIF1<0對(duì)稱(chēng)，則所有這些零點(diǎn)之和為SKIPIF1<0．故選：D．核心考點(diǎn)六：利用周期性和對(duì)稱(chēng)性解決函數(shù)問(wèn)題【典型例題】例21．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0，用SKIPIF1<0代替SKIPIF1<0得：SKIPIF1<0，因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，故SKIPIF1<0①，用SKIPIF1<0代替SKIPIF1<0得：SKIPIF1<0②，由①②得：SKIPIF1<0，所以函數(shù)SKIPIF1<0的周期SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，其中SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0．故選：C例22．（2023·四川資陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0為偶函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），且SKIPIF1<0．則SKIPIF1<0（

）A．16 B．20 C．24 D．28【答案】C【解析】因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，由SKIPIF1<0及SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0所以函數(shù)SKIPIF1<0的周期為SKIPIF1<0，因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），且SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，因?yàn)镾KIPIF1<0且SKIPIF1<0，所以SKIPIF1<0．所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選：SKIPIF1<0．例23．（2023·山東濟(jì)寧·高三嘉祥縣第一中學(xué)?？茧A段練習(xí)）已知定義在R上的偶函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．若直線(xiàn)SKIPIF1<0與曲線(xiàn)SKIPIF1<0恰有三個(gè)公共點(diǎn)，那么實(shí)數(shù)a的取值的集合為（

）A．SKIPIF1<0（SKIPIF1<0） B．SKIPIF1<0（SKIPIF1<0）C．SKIPIF1<0（SKIPIF1<0） D．SKIPIF1<0（SKIPIF1<0）【答案】B【解析】定義在R上的偶函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，所以SKIPIF1<0的圖像關(guān)于SKIPIF1<0對(duì)稱(chēng)，且SKIPIF1<0為周期是2的偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以畫(huà)出函數(shù)圖像如下圖所示：①當(dāng)SKIPIF1<0時(shí)，結(jié)合圖像可知SKIPIF1<0與SKIPIF1<0（SKIPIF1<0）有兩個(gè)公共點(diǎn)；②當(dāng)SKIPIF1<0與SKIPIF1<0（SKIPIF1<0）相切時(shí)，滿(mǎn)足SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，結(jié)合圖像可知SKIPIF1<0與SKIPIF1<0（SKIPIF1<0）有兩個(gè)公共點(diǎn)；由圖像可知，SKIPIF1<0時(shí)，直線(xiàn)SKIPIF1<0與SKIPIF1<0（SKIPIF1<0）有三個(gè)公共點(diǎn)；又因?yàn)镾KIPIF1<0周期SKIPIF1<0，可知SKIPIF1<0（SKIPIF1<0）．故選：B．例24．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若函數(shù)SKIPIF1<0圖象與SKIPIF1<0的圖象恰有10個(gè)不同的公共點(diǎn)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)楹瘮?shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，所以函數(shù)SKIPIF1<0是周期為2的周期函數(shù)，又函數(shù)SKIPIF1<0的圖象可由函數(shù)SKIPIF1<0的圖象向左平移一個(gè)單位可得，所以函數(shù)SKIPIF1<0的圖象的對(duì)稱(chēng)軸為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象也關(guān)于SKIPIF1<0對(duì)稱(chēng)，在平面直角坐標(biāo)系中作出函數(shù)SKIPIF1<0與SKIPIF1<0在SKIPIF1<0右側(cè)的圖象，數(shù)形結(jié)合可得，若函數(shù)SKIPIF1<0圖象與SKIPIF1<0的圖象恰有10個(gè)不同的公共點(diǎn)，則由函數(shù)圖象的對(duì)稱(chēng)性可得兩圖象在SKIPIF1<0右側(cè)有5個(gè)交點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0．故選：D．例25．（2023春·江西鷹潭·高三貴溪市實(shí)驗(yàn)中學(xué)?？茧A段練習(xí)）已知SKIPIF1<0是定義在R上的奇函數(shù)，SKIPIF1<0，恒有SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<01，則SKIPIF1<0（

）A．1 B．-1 C．0 D．2【答案】B【解析】因?yàn)?/p>