新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第08講 函數(shù)的基本性質(zhì)Ⅱ-奇偶性、周期性和對(duì)稱(chēng)性（含解析）

第08講函數(shù)的基本性質(zhì)Ⅱ-奇偶性、周期性和對(duì)稱(chēng)性（精講）題型目錄一覽①函數(shù)的奇偶性②函數(shù)奇偶性的應(yīng)用③函數(shù)的周期性④函數(shù)的對(duì)稱(chēng)性⑤函數(shù)性質(zhì)的綜合應(yīng)用一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理1.函數(shù)的奇偶性奇偶性定義圖象特點(diǎn)偶函數(shù)如果對(duì)于函數(shù)SKIPIF1<0的定義域內(nèi)任意一個(gè)SKIPIF1<0，都有SKIPIF1<0，那么函數(shù)SKIPIF1<0就叫做偶函數(shù)關(guān)于SKIPIF1<0軸對(duì)稱(chēng)奇函數(shù)如果對(duì)于函數(shù)SKIPIF1<0的定義域內(nèi)任意一個(gè)SKIPIF1<0，都有SKIPIF1<0，那么函數(shù)SKIPIF1<0就叫做奇函數(shù)關(guān)于原點(diǎn)對(duì)稱(chēng)注意：由函數(shù)奇偶性的定義可知，函數(shù)具有奇偶性的一個(gè)前提條件是：對(duì)于定義域內(nèi)的任意一個(gè)SKIPIF1<0，SKIPIF1<0也在定義域內(nèi)(即定義域關(guān)于原點(diǎn)對(duì)稱(chēng))．2.函數(shù)的對(duì)稱(chēng)性(1)若函數(shù)SKIPIF1<0為偶函數(shù)，則函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)．(2)若函數(shù)SKIPIF1<0為奇函數(shù)，則函數(shù)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)．(3)若SKIPIF1<0，則函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)．(4)若SKIPIF1<0，則函數(shù)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)．3.函數(shù)的周期性（1）周期函數(shù)：對(duì)于函數(shù)SKIPIF1<0，如果存在一個(gè)非零常數(shù)SKIPIF1<0，使得當(dāng)SKIPIF1<0取定義域內(nèi)的任何值時(shí)，都有SKIPIF1<0，那么就稱(chēng)函數(shù)SKIPIF1<0為周期函數(shù)，稱(chēng)SKIPIF1<0為這個(gè)函數(shù)的周期.（2）最小正周期：如果在周期函數(shù)SKIPIF1<0的所有周期中存在一個(gè)最小的正數(shù)，那么稱(chēng)這個(gè)最小整數(shù)叫做SKIPIF1<0的最小正周期.【常用結(jié)論】1.奇偶性技巧（1）若奇函數(shù)SKIPIF1<0在SKIPIF1<0處有意義，則有SKIPIF1<0；（2）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.(3)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)．2.周期性技巧SKIPIF1<03.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<0.4.對(duì)稱(chēng)性技巧（1）若函數(shù)SKIPIF1<0關(guān)于直線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)．二、題型分類(lèi)精講二、題型分類(lèi)精講真題刷刷刷真題刷刷刷一、單選題1．（2021·全國(guó)·高考真題）下列函數(shù)中是增函數(shù)的為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)基本初等函數(shù)的性質(zhì)逐項(xiàng)判斷后可得正確的選項(xiàng).【詳解】對(duì)于A，SKIPIF1<0為SKIPIF1<0上的減函數(shù)，不合題意，舍.對(duì)于B，SKIPIF1<0為SKIPIF1<0上的減函數(shù)，不合題意，舍.對(duì)于C，SKIPIF1<0在SKIPIF1<0為減函數(shù)，不合題意，舍.對(duì)于D，SKIPIF1<0為SKIPIF1<0上的增函數(shù)，符合題意，故選：D.2．（2021·全國(guó)·統(tǒng)考高考真題）設(shè)函數(shù)SKIPIF1<0，則下列函數(shù)中為奇函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】分別求出選項(xiàng)的函數(shù)解析式，再利用奇函數(shù)的定義即可.【詳解】由題意可得SKIPIF1<0，對(duì)于A，SKIPIF1<0不是奇函數(shù)；對(duì)于B，SKIPIF1<0是奇函數(shù)；對(duì)于C，SKIPIF1<0，定義域不關(guān)于原點(diǎn)對(duì)稱(chēng)，不是奇函數(shù)；對(duì)于D，SKIPIF1<0，定義域不關(guān)于原點(diǎn)對(duì)稱(chēng)，不是奇函數(shù).故選：B【點(diǎn)睛】本題主要考查奇函數(shù)定義，考查學(xué)生對(duì)概念的理解，是一道容易題.3．（2021·全國(guó)·高考真題）設(shè)SKIPIF1<0是定義域?yàn)镽的奇函數(shù)，且SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題意利用函數(shù)的奇偶性和函數(shù)的遞推關(guān)系即可求得SKIPIF1<0的值.【詳解】由題意可得：SKIPIF1<0，而SKIPIF1<0，故SKIPIF1<0.故選：C.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題主要考查了函數(shù)的奇偶性和函數(shù)的遞推關(guān)系式，靈活利用所給的條件進(jìn)行轉(zhuǎn)化是解決本題的關(guān)鍵.4．（2021·浙江·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0，則圖象為如圖的函數(shù)可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由函數(shù)的奇偶性可排除A、B，結(jié)合導(dǎo)數(shù)判斷函數(shù)的單調(diào)性可判斷C，即可得解.【詳解】對(duì)于A，SKIPIF1<0，該函數(shù)為非奇非偶函數(shù)，與函數(shù)圖象不符，排除A；對(duì)于B，SKIPIF1<0，該函數(shù)為非奇非偶函數(shù)，與函數(shù)圖象不符，排除B；對(duì)于C，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，與圖象不符，排除C.故選：D.5．（2022·全國(guó)·統(tǒng)考高考真題）如圖是下列四個(gè)函數(shù)中的某個(gè)函數(shù)在區(qū)間SKIPIF1<0的大致圖像，則該函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由函數(shù)圖像的特征結(jié)合函數(shù)的性質(zhì)逐項(xiàng)排除即可得解.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，故排除B;設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，故排除C;設(shè)SKIPIF1<0，則SKIPIF1<0，故排除D.故選：A.6．（2021·全國(guó)·統(tǒng)考高考真題）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)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】推導(dǎo)出函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)，由已知條件得出SKIPIF1<0，結(jié)合已知條件可得出結(jié)論.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0，可得SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，所以，SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，故函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)，因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，故SKIPIF1<0，其它三個(gè)選項(xiàng)未知.故選：B.7．（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【分析】法一：根據(jù)題意賦值即可知函數(shù)SKIPIF1<0的一個(gè)周期為SKIPIF1<0，求出函數(shù)一個(gè)周期中的SKIPIF1<0的值，即可解出．【詳解】[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；法二：作為選擇題，利用熟悉的函數(shù)使抽象問(wèn)題具體化，簡(jiǎn)化推理過(guò)程，直接使用具體函數(shù)的性質(zhì)解題，簡(jiǎn)單明了，是該題的最優(yōu)解.8．（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【分析】根據(jù)對(duì)稱(chēng)性和已知條件得到SKIPIF1<0，從而得到SKIPIF1<0，SKIPIF1<0，然后根據(jù)條件得到SKIPIF1<0的值，再由題意得到SKIPIF1<0從而得到SKIPIF1<0的值即可求解.【詳解】因?yàn)镾KIPIF1<0的圖像關(guān)于直線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：D【點(diǎn)睛】含有對(duì)稱(chēng)軸或?qū)ΨQ(chēng)中心的問(wèn)題往往條件比較隱蔽，考生需要根據(jù)已知條件進(jìn)行恰當(dāng)?shù)霓D(zhuǎn)化，然后得到所需的一些數(shù)值或關(guān)系式從而解題.9．（2021·全國(guó)·統(tǒng)考高考真題）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0為奇函數(shù)，SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】通過(guò)SKIPIF1<0是奇函數(shù)和SKIPIF1<0是偶函數(shù)條件，可以確定出函數(shù)解析式SKIPIF1<0，進(jìn)而利用定義或周期性結(jié)論，即可得到答案．【詳解】[方法一]：因?yàn)镾KIPIF1<0是奇函數(shù)，所以SKIPIF1<0①；因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0②．令SKIPIF1<0，由①得：SKIPIF1<0，由②得：SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，由①得：SKIPIF1<0，所以SKIPIF1<0．思路一：從定義入手．SKIPIF1<0SKIPIF1<0SKIPIF1<0所以SKIPIF1<0．[方法二]：因?yàn)镾KIPIF1<0是奇函數(shù)，所以SKIPIF1<0①；因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0②．令SKIPIF1<0，由①得：SKIPIF1<0，由②得：SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，由①得：SKIPIF1<0，所以SKIPIF1<0．思路二：從周期性入手由兩個(gè)對(duì)稱(chēng)性可知，函數(shù)SKIPIF1<0的周期SKIPIF1<0．所以SKIPIF1<0．故選：D．【點(diǎn)睛】在解決函數(shù)性質(zhì)類(lèi)問(wèn)題的時(shí)候，我們通常可以借助一些二級(jí)結(jié)論，求出其周期性進(jìn)而達(dá)到簡(jiǎn)便計(jì)算的效果．二、多選題10．（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【分析】方法一：轉(zhuǎn)化題設(shè)條件為函數(shù)的對(duì)稱(chēng)性，結(jié)合原函數(shù)與導(dǎo)函數(shù)圖象的關(guān)系，根據(jù)函數(shù)的性質(zhì)逐項(xiàng)判斷即可得解.【詳解】[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.【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．三、填空題11．（2021·全國(guó)·統(tǒng)考高考真題）寫(xiě)出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)SKIPIF1<0_______．①SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；③SKIPIF1<0是奇函數(shù)．【答案】SKIPIF1<0（答案不唯一，SKIPIF1<0均滿(mǎn)足）【分析】根據(jù)冪函數(shù)的性質(zhì)可得所求的SKIPIF1<0.【詳解】取SKIPIF1<0，則SKIPIF1<0，滿(mǎn)足①，SKIPIF1<0，SKIPIF1<0時(shí)有SKIPIF1<0，滿(mǎn)足②，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，又SKIPIF1<0，故SKIPIF1<0是奇函數(shù)，滿(mǎn)足③.故答案為：SKIPIF1<0（答案不唯一，SKIPIF1<0均滿(mǎn)足）四、雙空題12．（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．【分析】根據(jù)奇函數(shù)的定義即可求出．【詳解】[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．題型一函數(shù)的奇偶性策略方法判斷函數(shù)奇偶性的方法(1)定義法：(2)圖象法：(3)性質(zhì)法：在公共定義域內(nèi)有：奇±奇＝奇，偶±偶＝偶，奇×奇＝偶，偶×偶＝偶，奇×偶＝奇．【典例1】判斷下列函數(shù)的奇偶性：(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；(4)SKIPIF1<0．【答案】(1)偶函數(shù)(2)奇函數(shù)(3)奇函數(shù)(4)非奇非偶函數(shù)【分析】（1）利用偶函數(shù)的定義可判斷函數(shù)的奇偶性；（2）利用奇函數(shù)的定義可判斷函數(shù)的奇偶性；（3）利用奇函數(shù)的定義可判斷函數(shù)的奇偶性；（4）利用反例可判斷該函數(shù)為非奇非偶函數(shù).【詳解】（1）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，它關(guān)于原點(diǎn)對(duì)稱(chēng).SKIPIF1<0，故SKIPIF1<0為偶函數(shù).（2）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，它關(guān)于原點(diǎn)對(duì)稱(chēng).SKIPIF1<0，故SKIPIF1<0為奇函數(shù).（3）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，它關(guān)于原點(diǎn)對(duì)稱(chēng).SKIPIF1<0，故SKIPIF1<0為奇函數(shù).（4）SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0為非奇非偶函數(shù).【題型訓(xùn)練】一、單選題1．函數(shù)SKIPIF1<0的奇偶性是（）A．是奇函數(shù)，不是偶函數(shù)B．是偶函數(shù)，不是奇函數(shù)C．既是奇函數(shù)，也是偶函數(shù)D．非奇非偶函數(shù)【答案】A【分析】由奇偶性定義直接判斷即可.【詳解】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0是奇函數(shù)，不是偶函數(shù).故選：A.2．已知奇函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由SKIPIF1<0得SKIPIF1<0，代入得SKIPIF1<0，根據(jù)奇函數(shù)即可求解.【詳解】當(dāng)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0為奇函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故選：A.3．若函數(shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0（

）A．2 B．1 C．0 D．SKIPIF1<0【答案】C【分析】由SKIPIF1<0為奇函數(shù)求得SKIPIF1<0，即可由分段函數(shù)求值.【詳解】函數(shù)SKIPIF1<0為奇函數(shù)，設(shè)SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0.故選：C.4．函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B．C． D．【答案】C【分析】根據(jù)函數(shù)的奇偶性排除AB，再由特殊值排除D即可得解.【詳解】因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱(chēng)，所以SKIPIF1<0，即函數(shù)為奇函數(shù)，排除AB，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除D.故選：C二、填空題5．函數(shù)SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0時(shí)，SKIPIF1<0___________．【答案】SKIPIF1<0【分析】由偶函數(shù)的定義求解．【詳解】SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0是偶函數(shù)，∴SKIPIF1<0，故答案為：SKIPIF1<0．6．SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0__________.【答案】4【分析】令SKIPIF1<0，可得SKIPIF1<0為奇函數(shù)，再根據(jù)奇函數(shù)的性質(zhì)求解.【詳解】令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0為奇函數(shù)，由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0.故答案為：4.7．已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的解集是__________.【答案】SKIPIF1<0【分析】利用奇偶性求出函數(shù)SKIPIF1<0的解析式SKIPIF1<0，分類(lèi)討論即可求解.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，要解不等式SKIPIF1<0，只需SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，綜上，不等式的解集為SKIPIF1<0.故答案為：SKIPIF1<0.三、解答題8．已知函數(shù)SKIPIF1<0(1)求函數(shù)SKIPIF1<0解析式；(2)判斷函數(shù)SKIPIF1<0的奇偶性并加以證明【答案】(1)SKIPIF1<0(2)奇函數(shù)，證明見(jiàn)解析【分析】（1）利用換元法，令SKIPIF1<0，得SKIPIF1<0，從而可得SKIPIF1<0；（2）先求函數(shù)定義域，利用奇偶性的定義進(jìn)行證明.【詳解】（1）令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.（2）奇函數(shù)；證明：定義域?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0為奇函數(shù).9．已知函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的值；(2)令SKIPIF1<0，求證：SKIPIF1<0為奇函數(shù)；(3)若銳角SKIPIF1<0滿(mǎn)足SKIPIF1<0，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)證明見(jiàn)解析(3)SKIPIF1<0【分析】（1）將SKIPIF1<0和SKIPIF1<0分別代入解析式求解即可；（2）根據(jù)奇偶性的定義證明即可；（3）根據(jù)奇偶性將不等式化為SKIPIF1<0，利用單調(diào)性定義可證得SKIPIF1<0為SKIPIF1<0上的增函數(shù)，由此可得SKIPIF1<0，結(jié)合三角函數(shù)知識(shí)可求得結(jié)果.【詳解】（1）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.（2）SKIPIF1<0，則SKIPIF1<0的定義域?yàn)镾KIPIF1<0；SKIPIF1<0，SKIPIF1<0為奇函數(shù).（3）由SKIPIF1<0得：SKIPIF1<0；SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0為SKIPIF1<0上的增函數(shù)，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0.題型二函數(shù)奇偶性的應(yīng)用策略方法已知函數(shù)奇偶性可以解決的三個(gè)問(wèn)題【典例1】若函數(shù)SKIPIF1<0是定義在R上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．5 D．7【答案】C【分析】求出SKIPIF1<0時(shí)的解析式后，代入SKIPIF1<0可求出結(jié)果.【詳解】因?yàn)镾KIPIF1<0為奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0.故選：C【典例2】若函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0、SKIPIF1<0的值是（

）A．SKIPIF1<0 B．SKIPIF1<0不能確定，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0不能確定 D．SKIPIF1<0【答案】D【分析】根據(jù)定義域關(guān)于原點(diǎn)對(duì)稱(chēng)，求得SKIPIF1<0，再根據(jù)SKIPIF1<0，求得SKIPIF1<0的值，即可求解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0是偶函數(shù)，可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，又由SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.故選：D.【典例3】偶函數(shù)SKIPIF1<0滿(mǎn)足：SKIPIF1<0，且在區(qū)間SKIPIF1<0與SKIPIF1<0上分別遞減和遞增，使SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)題中所給條件，可畫(huà)出符合全部條件的函數(shù)圖象輔助做題.【詳解】根據(jù)題目條件，想象函數(shù)圖象如下:因?yàn)镾KIPIF1<0，SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0和SKIPIF1<0時(shí)，SKIPIF1<0，故選：B.【題型訓(xùn)練】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)SKIPIF1<0為奇函數(shù)，則實(shí)數(shù)SKIPIF1<0的值為（

）A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)題意可得SKIPIF1<0，計(jì)算可得SKIPIF1<0，經(jīng)檢驗(yàn)均符合題意，即可得解.【詳解】由SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，符合題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0符合題意，故選：D2．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用偶函數(shù)的性質(zhì)直接求解即可.【詳解】由已知得，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0為偶函數(shù)，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，故選：SKIPIF1<0.3．（2023·安徽·校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0＋1 D．SKIPIF1<0【答案】B【分析】由定義在SKIPIF1<0上的奇函數(shù)有SKIPIF1<0，求出SKIPIF1<0的值，再由SKIPIF1<0可得出答案.【詳解】函數(shù)SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，則SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0故選：B4．（2023·全國(guó)·高三專(zhuān)題練習(xí)）定義在SKIPIF1<0上的偶函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，若SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)偶函數(shù)及單調(diào)性解不等式即可.【詳解】由題意，SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0.故選:D.5．（2023春·貴州黔東南·高三?？茧A段練習(xí)）已知偶函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用偶函數(shù)的對(duì)稱(chēng)性可得SKIPIF1<0，即可求解集.【詳解】由偶函數(shù)的對(duì)稱(chēng)性知：SKIPIF1<0在SKIPIF1<0上遞增，則在SKIPIF1<0上遞減，所以SKIPIF1<0，故SKIPIF1<0，可得SKIPIF1<0，所以不等式解集為SKIPIF1<0.故選：D6．（2023·湖南長(zhǎng)沙·湖南師大附中?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，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】C【分析】依題意作函數(shù)圖像，根據(jù)單調(diào)性和奇偶性求解.【詳解】依題意，函數(shù)的大致圖像如下圖：因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，則當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式SKIPIF1<0化為SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，即原不等式的解集為SKIPIF1<0；故選：C.二、多選題7．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是偶函數(shù)，在區(qū)間SKIPIF1<0上是單調(diào)函數(shù)，且SKIPIF1<0，則（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【分析】根據(jù)函數(shù)的單調(diào)性和奇偶性直接求解.【詳解】函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)函數(shù)，又SKIPIF1<0，且SKIPIF1<0，故此函數(shù)在區(qū)間SKIPIF1<0上是減函數(shù)．由已知條件及偶函數(shù)性質(zhì)，知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)．對(duì)于A，SKIPIF1<0，故SKIPIF1<0，故A錯(cuò)誤；對(duì)于B，SKIPIF1<0，故SKIPIF1<0，故B正確；對(duì)于C，SKIPIF1<0，故C錯(cuò)誤；對(duì)于D，SKIPIF1<0，故D正確．故選:BD.8．（2023·山東菏澤·山東省東明縣第一中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0為奇函數(shù)，且對(duì)SKIPIF1<0，SKIPIF1<0恒成立，則（

）A．SKIPIF1<0為奇函數(shù) B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【分析】根據(jù)函數(shù)定義換算可得SKIPIF1<0為偶函數(shù)，根據(jù)偶函數(shù)和奇函數(shù)性質(zhì)可知SKIPIF1<0為周期函數(shù)，再根據(jù)函數(shù)周期性和函數(shù)特殊值即可得出選項(xiàng).【詳解】因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，故SKIPIF1<0又SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0為偶函數(shù)，A錯(cuò)誤；SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，B正確；SKIPIF1<0，又SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，所以SKIPIF1<0，C正確；又SKIPIF1<0，所以SKIPIF1<0是以4為周期的函數(shù)，SKIPIF1<0，D正確．故選：BCD．三、填空題9．（2023·廣東潮州·統(tǒng)考二模）已知函數(shù)SKIPIF1<0（其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)，SKIPIF1<0）是奇函數(shù)，則實(shí)數(shù)SKIPIF1<0的值為_(kāi)_____.【答案】SKIPIF1<0【分析】利用奇函數(shù)的性質(zhì)可得出SKIPIF1<0，結(jié)合對(duì)數(shù)運(yùn)算可得出實(shí)數(shù)SKIPIF1<0的值.【詳解】對(duì)于函數(shù)SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.10．（2023·河南周口·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，則不等式SKIPIF1<0的解集為_(kāi)_____.【答案】SKIPIF1<0【分析】由題意和偶函數(shù)的性質(zhì)可知函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)，在SKIPIF1<0上為增函數(shù)，結(jié)合SKIPIF1<0，分類(lèi)討論當(dāng)SKIPIF1<0、SKIPIF1<0時(shí)，利用函數(shù)的單調(diào)性解不等式即可.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0是定義在R上的偶函數(shù)，且在SKIPIF1<0上單調(diào)遞減所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，解得SKIPIF1<0，綜上，不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0.11．（2023春·江蘇南通·高三海安高級(jí)中學(xué)?？茧A段練習(xí)）定義在SKIPIF1<0上的函數(shù)SKIPIF1<0，滿(mǎn)足SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，若SKIPIF1<0，則SKIPIF1<0__________.【答案】1【分析】根據(jù)SKIPIF1<0為偶函數(shù)、SKIPIF1<0為奇函數(shù)的性質(zhì)，利用賦值法可得答案.【詳解】若SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故答案為：1.12．（2023春·福建廈門(mén)·高三廈門(mén)一中?？计谥校┮阎瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，若SKIPIF1<0為奇函數(shù)，且SKIPIF1<0，則SKIPIF1<0_________.【答案】SKIPIF1<0【分析】推導(dǎo)出函數(shù)SKIPIF1<0為周期函數(shù)，確定該函數(shù)的周期，計(jì)算出SKIPIF1<0的值，結(jié)合SKIPIF1<0以及周期性可求得SKIPIF1<0的值.【詳解】因?yàn)镾KIPIF1<0為奇函數(shù)，則SKIPIF1<0，所以，SKIPIF1<0，在等式SKIPIF1<0中，令SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，又因?yàn)镾KIPIF1<0，則SKIPIF1<0，①所以，SKIPIF1<0，②由①②可得SKIPIF1<0，即SKIPIF1<0，所以，函數(shù)SKIPIF1<0為周期函數(shù)，且該函數(shù)的周期為SKIPIF1<0，所以，SKIPIF1<0