高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專(zhuān)題第2講 函數(shù)的對(duì)稱(chēng)性與周期性（含解析）

第2講函數(shù)的對(duì)稱(chēng)性與周期性【考點(diǎn)分析】1.函數(shù)的對(duì)稱(chēng)性、周期性是高考命題熱點(diǎn)，近兩年新高考都考了一道選擇題，分值5分，知識(shí)點(diǎn)比較靈活，需要全面掌握常見(jiàn)對(duì)稱(chēng)性，周期性的結(jié)論考點(diǎn)一：函數(shù)常見(jiàn)對(duì)稱(chēng)性結(jié)論①若函數(shù)SKIPIF1<0對(duì)于任意的SKIPIF1<0均滿足SKIPIF1<0，則函數(shù)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)．②若函數(shù)SKIPIF1<0對(duì)于任意的SKIPIF1<0均滿足SKIPIF1<0則SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)．考點(diǎn)二：函數(shù)常見(jiàn)周期性結(jié)論若函數(shù)對(duì)于任意的SKIPIF1<0都滿足SKIPIF1<0，則SKIPIF1<0為SKIPIF1<0的一個(gè)周期，且SKIPIF1<0幾個(gè)常見(jiàn)周期性結(jié)論=1\*GB3①若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0．=2\*GB3②若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0．=3\*GB3③若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0．④若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0．⑤若函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0，SKIPIF1<0都對(duì)稱(chēng)，則SKIPIF1<0為周期函數(shù)且SKIPIF1<0是它的一個(gè)周期．⑥函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于兩點(diǎn)SKIPIF1<0、SKIPIF1<0都對(duì)稱(chēng)，則函數(shù)SKIPIF1<0是以SKIPIF1<0為周⑦函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于SKIPIF1<0和直線SKIPIF1<0都對(duì)稱(chēng)，則函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)．⑧若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)．【題型目錄】題型一：利用周期性求函數(shù)值題型二：利用周期性求函數(shù)解析式題型三：根據(jù)函數(shù)的對(duì)稱(chēng)性、周期性、奇偶性寫(xiě)函數(shù)題型四：根據(jù)函數(shù)的對(duì)稱(chēng)性、奇偶性、周期性綜合運(yùn)用【典型例題】題型一：利用周期性求函數(shù)值【例1】設(shè)SKIPIF1<0是定義在SKIPIF1<0上周期為2的函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其中SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0的值是．答案：1解析：SKIPIF1<0SKIPIF1<0是定義在SKIPIF1<0上周期為2的函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0【例2】設(shè)SKIPIF1<0為定義在SKIPIF1<0上的奇函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0__________答案：SKIPIF1<0解析：SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0是周期為4的函數(shù)，所以SKIPIF1<0【例3】定義在SKIPIF1<0上的函數(shù)SKIPIF1<0對(duì)任意SKIPIF1<0，都有SKIPIF1<0，則SKIPIF1<0等于A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0答案：D解析：SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是周期為4的函數(shù)，SKIPIF1<0【例4】（重慶南開(kāi)高一上期中）已知定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0,且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<0SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0【例5】（2022·云南昭通·高一期末）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的周期函數(shù)，且周期為2，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】利用函數(shù)的周期性，則SKIPIF1<0，又根據(jù)函數(shù)在SKIPIF1<0的解析式，求解SKIPIF1<0的值，即可得SKIPIF1<0的值.【詳解】解：由題可知SKIPIF1<0所以SKIPIF1<0又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0.故選：C.【題型專(zhuān)練】1.（2021·山東·臨沂市蘭山區(qū)教學(xué)研究室高三開(kāi)學(xué)考試）已知SKIPIF1<0是R上的奇函數(shù)，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．3 B．SKIPIF1<0 C．255 D．SKIPIF1<0【答案】B【分析】根據(jù)題意可知SKIPIF1<0是周期函數(shù)，根據(jù)周期以及奇函數(shù)即可求解.【詳解】由SKIPIF1<0可得，SKIPIF1<0，故SKIPIF1<0是以4為周期的周期函數(shù)，故SKIPIF1<0，故選：B2.（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0，若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．0 B．1C．6 D．216【答案】C【分析】由SKIPIF1<0可得函數(shù)周期為6，進(jìn)而SKIPIF1<0，最后求出答案.【詳解】根據(jù)題意，偶函數(shù)SKIPIF1<0滿足SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0是周期為6的周期函數(shù)，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0故選：C3.（重慶南開(kāi)高一上期末）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，SKIPIF1<0.若對(duì)任意實(shí)數(shù)SKIPIF1<0，SKIPIF1<0都有SKIPIF1<0，則SKIPIF1<0（）ASKIPIF1<0B.-1C.0D.1答案：D解析：由題意知，令SKIPIF1<0，可得SKIPIF1<0，因SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<04．（2022·云南紅河·高一期末）已知SKIPIF1<0是定義在R上的奇函數(shù)，SKIPIF1<0，都有SKIPIF1<0，若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．1 D．2【答案】C【分析】SKIPIF1<0是定義在R上的奇函數(shù)得SKIPIF1<0，有SKIPIF1<0得到SKIPIF1<0是周期函數(shù)，利用函數(shù)周期性可得答案.【詳解】SKIPIF1<0是定義在R上的奇函數(shù)，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，都有SKIPIF1<0，SKIPIF1<0是周期為4的周期函數(shù)，SKIPIF1<0.故選：C.5．（2022·黑龍江·大慶中學(xué)高二期末）SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且滿足SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0______．【答案】SKIPIF1<0【分析】依題意可得SKIPIF1<0，即可得到SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)，再根據(jù)對(duì)數(shù)的運(yùn)算及奇函數(shù)的性質(zhì)計(jì)算可得.【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)，又SKIPIF1<0所以SKIPIF1<0，又SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，所以SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0題型二：利用周期性求函數(shù)解析式【例1】已知定義在實(shí)數(shù)集R上的函數(shù)SKIPIF1<0滿足：（1）SKIPIF1<0；（2）SKIPIF1<0；（3）當(dāng)SKIPIF1<0時(shí)解析式為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，求函數(shù)的解析式。答案：SKIPIF1<0解析：SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，又因SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，因SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此SKIPIF1<0因此當(dāng)SKIPIF1<0時(shí)，函數(shù)的解析式為SKIPIF1<0【例2】（2022·全國(guó)·高一專(zhuān)題練習(xí)）已知SKIPIF1<0是定義在SKIPIF1<0上周期為SKIPIF1<0的函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，那么當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0______.【答案】SKIPIF1<0【分析】根據(jù)周期性求函數(shù)解析式即可.【詳解】解：因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0,SKIPIF1<0是定義在SKIPIF1<0上周期為SKIPIF1<0的函數(shù)所以，SKIPIF1<0，故答案為：SKIPIF1<0【例3】（2021·山東師范大學(xué)附中高三期中）設(shè)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且對(duì)任意實(shí)數(shù)SKIPIF1<0，恒有SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的解析式；(2)計(jì)算SKIPIF1<0.【答案】(1)SKIPIF1<0，(2)SKIPIF1<0【分析】（1）利用奇函數(shù)和SKIPIF1<0判斷出SKIPIF1<0為周期為4的函數(shù)，用代入法求出解析式；（2）利用函數(shù)的周期即可求值.(1)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是周期為4的周期函數(shù).當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由已知得SKIPIF1<0.又SKIPIF1<0是奇函數(shù)，SKIPIF1<0，SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0是周期為4的周期函數(shù)，SKIPIF1<0，從而求得SKIPIF1<0時(shí)，SKIPIF1<0.(2)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0是周期為4的周期函數(shù)，SKIPIF1<0.又SKIPIF1<0，SKIPIF1<0.【題型專(zhuān)練】1.（2021·上海南匯中學(xué)高三期中）設(shè)SKIPIF1<0是定義在R上以2為周期的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式SKIPIF1<0___________．【答案】SKIPIF1<0【分析】設(shè)SKIPIF1<0是SKIPIF1<0時(shí)函數(shù)圖象上的任意一點(diǎn)，然后利用周期和奇偶性將SKIPIF1<0轉(zhuǎn)化到區(qū)間SKIPIF1<0上，進(jìn)而代入解析式化簡(jiǎn)即可.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的周期為2，設(shè)SKIPIF1<0是SKIPIF1<0時(shí)函數(shù)圖象上的任意一點(diǎn)，則點(diǎn)SKIPIF1<0在SKIPIF1<0時(shí)函數(shù)的圖象上，而函數(shù)是R上的奇函數(shù)，則點(diǎn)SKIPIF1<0在SKIPIF1<0時(shí)的圖象上，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上的解析式SKIPIF1<0.故答案為：SKIPIF1<0.2.（2021·吉林·梅河口市第五中學(xué)高三階段練習(xí)（文））函數(shù)SKIPIF1<0滿足是SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為_(kāi)__________.【答案】SKIPIF1<0##SKIPIF1<0【分析】由題設(shè)遞推關(guān)系可得SKIPIF1<0，令SKIPIF1<0結(jié)合已知區(qū)間解析式即可求SKIPIF1<0時(shí)SKIPIF1<0的解析式，再應(yīng)用二次函數(shù)的性質(zhì)求最小值.【詳解】由題設(shè)，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0上，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<03.（2021·江蘇·高一專(zhuān)題練習(xí)）設(shè)SKIPIF1<0是定義在SKIPIF1<0上以2為周期的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在[4，6]上的解析式是__________【答案】SKIPIF1<0【分析】根據(jù)函數(shù)的周期及函數(shù)為奇函數(shù)，分段求解函數(shù)的解析式即可.【詳解】因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上以2為周期的奇函數(shù)且SKIPIF1<0時(shí)，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0,故SKIPIF1<0.綜上可得，函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式是SKIPIF1<0，故答案為：SKIPIF1<04.（2021·北京市十一學(xué)校高一期中）若定義在R上的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0時(shí)SKIPIF1<0，則：（1）SKIPIF1<0__________；（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0_________.【答案】

SKIPIF1<0

SKIPIF1<0【分析】（1）由題可得SKIPIF1<0，再結(jié)合條件可求；（2）由題可求當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，再結(jié)合函數(shù)的周期性即求.【詳解】∵定義在R上的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即函數(shù)SKIPIF1<0是以4為周期的周期函數(shù)，又SKIPIF1<0時(shí)SKIPIF1<0，∴SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0.故答案為：（1）SKIPIF1<0；（2）SKIPIF1<0題型三：根據(jù)函數(shù)的對(duì)稱(chēng)性、周期性、奇偶性寫(xiě)函數(shù)【例1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）寫(xiě)出一個(gè)最小正周期為3的偶函數(shù)SKIPIF1<0___________.【答案】SKIPIF1<0（答案不唯一）【分析】利用余弦函數(shù)的性質(zhì)，結(jié)合已知函數(shù)性質(zhì)寫(xiě)出滿足要求的函數(shù)解析式即可.【詳解】由余弦函數(shù)性質(zhì)知：SKIPIF1<0為偶函數(shù)且SKIPIF1<0為常數(shù)，又最小正周期為3，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0滿足要求.故答案為：SKIPIF1<0(答案不唯一)【例2】（2022·江蘇·金陵中學(xué)高三學(xué)業(yè)考試）寫(xiě)出一個(gè)滿足以下三個(gè)條件的函數(shù)：SKIPIF1<0______．①定義域?yàn)镽；②SKIPIF1<0不是周期函數(shù)；③SKIPIF1<0是周期為SKIPIF1<0的函數(shù)．【答案】SKIPIF1<0（答案不唯一）【分析】由SKIPIF1<0的周期為SKIPIF1<0，結(jié)合正余弦函數(shù)的性質(zhì)確定SKIPIF1<0的解析式形式，即可得符合要求的函數(shù)式.【詳解】SKIPIF1<0的解析式形式：SKIPIF1<0或SKIPIF1<0均可．如：SKIPIF1<0定義域?yàn)镽，不是周期函數(shù)，且SKIPIF1<0是周期為SKIPIF1<0的函數(shù).故答案為：SKIPIF1<0（答案不唯一）【例3】（2022·全國(guó)·高三專(zhuān)題練習(xí)）寫(xiě)出一個(gè)同時(shí)滿足下列性質(zhì)①②③的函數(shù)SKIPIF1<0：__________.①定義域?yàn)镾KIPIF1<0；②SKIPIF1<0為偶函數(shù)；③SKIPIF1<0為奇函數(shù).【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)題意和函數(shù)的奇偶性和周期性可知SKIPIF1<0是關(guān)于SKIPIF1<0軸對(duì)稱(chēng)、關(guān)于SKIPIF1<0中心對(duì)稱(chēng)、以4為周期的函數(shù)，進(jìn)而直接得出結(jié)果.【詳解】由SKIPIF1<0為偶函數(shù)，知SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)；由SKIPIF1<0為奇函數(shù)，知SKIPIF1<0關(guān)于SKIPIF1<0中心對(duì)稱(chēng)，所以SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)；所以SKIPIF1<0，則SKIPIF1<0以4為周期，故可取SKIPIF1<0.故答案為：SKIPIF1<0.【題型專(zhuān)練】1（2022·廣東茂名·二模）請(qǐng)寫(xiě)出一個(gè)函數(shù)SKIPIF1<0_______，使之同時(shí)具有以下性質(zhì)：①圖象關(guān)于y軸對(duì)稱(chēng)；②SKIPIF1<0，SKIPIF1<0．【答案】SKIPIF1<0(答案不唯一)【分析】根據(jù)題設(shè)函數(shù)性質(zhì)的描述，只需寫(xiě)出一個(gè)周期為4的偶函數(shù)，結(jié)合余弦函數(shù)的性質(zhì)即可寫(xiě)出函數(shù)解析式.【詳解】由題設(shè)，寫(xiě)出一個(gè)周期為4的偶函數(shù)即可，所以SKIPIF1<0滿足題設(shè)要求.故答案為：SKIPIF1<0(答案不唯一)2.（2022·北京通州·高三期末）最小正周期為2的函數(shù)的解析式可以是______．（寫(xiě)出一個(gè)即可）【答案】SKIPIF1<0【分析】根據(jù)正弦型三角函數(shù)的周期公式即可找出【詳解】根據(jù)正弦型三角函數(shù)的周期公式，最小正周期為2的函數(shù)的解析式可以是SKIPIF1<0．故答案為：SKIPIF1<0．3.（2022·全國(guó)·高三專(zhuān)題練習(xí)（理））函數(shù)SKIPIF1<0滿足以下條件：①SKIPIF1<0的定義域?yàn)镾KIPIF1<0，其圖像是一條連續(xù)不斷的曲線；②SKIPIF1<0，SKIPIF1<0；③當(dāng)SKIPIF1<0且SKIPIF1<0，SKIPIF1<0；④SKIPIF1<0恰有兩個(gè)零點(diǎn)，請(qǐng)寫(xiě)出函數(shù)SKIPIF1<0的一個(gè)解析式________【答案】SKIPIF1<0（答案不唯一）【分析】由題意可得函數(shù)SKIPIF1<0是偶函數(shù)，且在SKIPIF1<0上為增函數(shù)，函數(shù)圖象與SKIPIF1<0軸只有2個(gè)交點(diǎn)，由此可得函數(shù)解析式【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，因?yàn)楫?dāng)SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0恰有兩個(gè)零點(diǎn)，所以SKIPIF1<0圖象與SKIPIF1<0軸只有2個(gè)交點(diǎn)，所以函數(shù)SKIPIF1<0的一個(gè)解析式可以為SKIPIF1<0，故答案為：SKIPIF1<0（答案不唯一）題型四：根據(jù)函數(shù)的對(duì)稱(chēng)性、奇偶性、周期性綜合運(yùn)用【例1】（2022·貴州銅仁·高二期末（理））已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且滿足：SKIPIF1<0，又SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的值為（

）A．4 B．SKIPIF1<0 C．0 D．2【答案】C【分析】由SKIPIF1<0，可得SKIPIF1<0，再根據(jù)條件得到周期后即可求解.【詳解】由SKIPIF1<0，可知函數(shù)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，即有SKIPIF1<0；由SKIPIF1<0為偶函數(shù)，可知函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，即有SKIPIF1<0.于是有SKIPIF1<0，從而可得SKIPIF1<0，因此可得函數(shù)SKIPIF1<0的周期為4.所以SKIPIF1<0，SKIPIF1<0.再由SKIPIF1<0，令SKIPIF1<0，有SKIPIF1<0，即SKIPIF1<0.所以SKIPIF1<0.故選：C【例2】（2022·陜西·長(zhǎng)安一中高一期末）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0為奇函數(shù)，SKIPIF1<0為偶函數(shù)，則函數(shù)SKIPIF1<0的周期是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由奇函數(shù)性質(zhì)可得SKIPIF1<0，由偶函數(shù)性質(zhì)可得SKIPIF1<0，化簡(jiǎn)整理可得SKIPIF1<0，即可求出周期.【詳解】因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的周期是4.故選：C.【例3】（2022·湖南·長(zhǎng)沙一中高三開(kāi)學(xué)考試）已知SKIPIF1<0是定義在R上的奇函數(shù)，SKIPIF1<0為偶函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．SKIPIF1<0 D．1【答案】D【分析】根據(jù)奇偶性的性質(zhì)化簡(jiǎn)可得SKIPIF1<0是以4為周期的函數(shù)，即可求出.【詳解】因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，故可得SKIPIF1<0，又SKIPIF1<0為偶函數(shù)，故可得SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0以4為周期，故SKIPIF1<0.故選：D.【例4】（2022·山東日照·高二期末）已知SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，SKIPIF1<0是定義域?yàn)镾KIPIF1<0的偶函數(shù)，且SKIPIF1<0與SKIPIF1<0的圖像關(guān)于y軸對(duì)稱(chēng)，則（

）A．SKIPIF1<0是奇函數(shù) B．SKIPIF1<0是偶函數(shù)C．2是SKIPIF1<0一個(gè)周期 D．SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)【答案】A【分析】根據(jù)函數(shù)奇偶性，對(duì)稱(chēng)性、周期性的定義一一判斷即可；【詳解】解：根據(jù)題意，SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，則SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0成中心對(duì)稱(chēng)，SKIPIF1<0是定義域?yàn)镾KIPIF1<0的偶函數(shù)，則SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0與SKIPIF1<0的圖像關(guān)于y軸對(duì)稱(chēng)，則SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0關(guān)于原點(diǎn)中心對(duì)稱(chēng)，故SKIPIF1<0是奇函數(shù)，故A正確.SKIPIF1<0是奇函數(shù)，且SKIPIF1<0與SKIPIF1<0的圖像關(guān)于y軸對(duì)稱(chēng)，故SKIPIF1<0是奇函數(shù)，故B錯(cuò)誤.SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，則SKIPIF1<0，①SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，故SKIPIF1<0，可得SKIPIF1<0，聯(lián)立①得SKIPIF1<0，故SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0，函數(shù)SKIPIF1<0是周期為4的周期函數(shù)，由題意可得出4是函數(shù)SKIPIF1<0的周期，故C錯(cuò)誤.因?yàn)?是函數(shù)SKIPIF1<0的周期，SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，所以SKIPIF1<0是SKIPIF1<0的中心對(duì)稱(chēng)，SKIPIF1<0關(guān)于y軸對(duì)稱(chēng)為SKIPIF1<0，為SKIPIF1<0的對(duì)稱(chēng)中心，故D錯(cuò)誤.故選：A【例5】已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足條件SKIPIF1<0，且函數(shù)SKIPIF1<0為奇函數(shù)，下列有關(guān)命題的說(shuō)法錯(cuò)誤的是（）A．函數(shù)SKIPIF1<0是周期函數(shù) B．函數(shù)SKIPIF1<0為SKIPIF1<0上的偶函數(shù) C．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)函數(shù) D．SKIPIF1<0為SKIPIF1<0上的單調(diào)函數(shù)答案：D解析：SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0是周期為3的函數(shù)，故A正確；因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，令SKIPIF1<0可得，SKIPIF1<0，即SKIPIF1<0，又因SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0上的偶函數(shù)，所以B對(duì)，D錯(cuò)，因SKIPIF1<0為奇函數(shù)，所以它關(guān)于原點(diǎn)對(duì)稱(chēng)，故把SKIPIF1<0向左平移SKIPIF1<0單位，得到SKIPIF1<0的圖象，所以SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以C對(duì)。【例6】（2021新高考2卷8）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<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0答案：B解析：SKIPIF1<0SKIPIF1<0是偶函數(shù)，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，又因SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，又因SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0【例7】若函數(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】【分析】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】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．【題型專(zhuān)練】1.（2022·四川雅安·高二期末（文））已知函數(shù)SKIPIF1<0是SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的值為（

）A．1 B．2 C．SKIPIF1<0 D．0【答案】A【分析】由偶函數(shù)可得SKIPIF1<0，由SKIPIF1<0可得對(duì)稱(chēng)性，再化簡(jiǎn)整理可得周期SKIPIF1<0，進(jìn)而根據(jù)性質(zhì)轉(zhuǎn)換SKIPIF1<0到SKIPIF1<0，再代入解析式求解即可.【詳解】由題，因?yàn)榕己瘮?shù)，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0,即SKIPIF1<0，所以SKIPIF1<0是周期函數(shù)，SKIPIF1<0，故SKIPIF1<0故選：A2．（2022·河南新鄉(xiāng)·高二期末（理））已知SKIPIF1<0是定義在R上的奇函數(shù)，且滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．－8 B．－4 C．0 D．4【答案】B【分析】結(jié)合條件證得SKIPIF1<0的周期為8，即可求出結(jié)果.【詳解】因?yàn)镾KIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的周期為8，所以SKIPIF1<0，故SKIPIF1<0．故選：B.3．（2022·湖南·高二期末）已知定義域是R的函數(shù)SKIPIF1<0滿足：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為偶函數(shù)，SKIPIF1<0，則SKIPIF1<0（

）A．1 B．-1 C．2 D．-3【答案】B【分析】根據(jù)對(duì)稱(chēng)性可得函數(shù)具有周期性，根據(jù)周期可將SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，又由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0的周期為4，所以SKIPIF1<0．故選:B．4.函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，若SKIPIF1<0與SKIPIF1<0都是奇函數(shù)，則（）A.SKIPIF1<0是偶函數(shù)B.SKIPIF1<0是奇函數(shù)C.SKIPIF1<0D.SKIPIF1<0是奇函數(shù)答案：D解析：SKIPIF1<0SKIPIF1<0是奇函數(shù)，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)。5.（2021全國(guó)卷甲卷理科12）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，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（）SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0答案：D解析：SKIPIF1<0SKIPIF1<0是奇函數(shù)，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0SKIPIF1<0是偶函數(shù)，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，又因SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，因SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0，因SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又因SKIPIF1<0，解得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<06.已知SKIPIF1<0是定義在SKIPIF1<0上的函數(shù)，且對(duì)任意SKIPIF1<0都有SKIPIF1<0，若函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，且SKIPIF1<0，則SKIPIF1<0答案：SKIPIF1<0解析：SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以關(guān)于SKIPIF1<0對(duì)稱(chēng)，因?yàn)镾KIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，且SKIPIF1<0為奇函數(shù)，所以SKIPIF1<07.（2020?岳麓區(qū)校級(jí)模擬）若對(duì)任意的SKIPIF1<0，都有SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的值為．答案：SKIPIF1<0解析：SKIPIF1<0若對(duì)任意的SKIPIF1<0，都有SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<08.（2022·河北深州市中學(xué)高三階段練習(xí)多選）已知函數(shù)SKIPIF1<0對(duì)SKIPIF1<0，都有SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱(chēng)B．SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)C．SKIPIF1<0D．SKIPIF1<0【答案】ABC【分析】A選項(xiàng)根據(jù)題目條件立即得出，BCD選項(xiàng)通過(guò)已知條件合理的進(jìn)行“取代”，推出函數(shù)周期后便容易得出結(jié)果.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，A選項(xiàng)正確；又SKIPIF1<0，令SKIPIF1<0去取代SKIPIF1<0，所以SKIPIF1<0，再令SKIPIF1<0取代SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的周期為4，由SKIPIF1<0可得：SKIPIF1<0，所以SKIPIF1<0的圖像關(guān)于SKIPIF1<0對(duì)稱(chēng)，結(jié)合SKIPIF1<0的周期為4，所以SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，故B正確；定義在SKIPIF1<0上的奇函數(shù)滿足SKIPIF1<0，令SKIPIF1<0中SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，故C正確；SKIPIF1<0，故D不正確.故選：ABC.9.（2022·黑龍江齊齊哈爾·高二期末多選）已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且函數(shù)SKIPIF1<0為偶函數(shù)，則下列結(jié)論正確的是（

）A．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的零點(diǎn)有6個(gè)C