新高考數(shù)學(xué)一輪復(fù)習(xí)講練測(cè)專題5.3三角函數(shù)的圖象與性質(zhì)（練）解析版

專題5.3三角函數(shù)的圖象與性質(zhì)練基礎(chǔ)練基礎(chǔ)1．（2021·北京市大興區(qū)精華培訓(xùn)學(xué)校高三三模）下列函數(shù)中，既是奇函數(shù)又以SKIPIF1<0為最小正周期的函數(shù)是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由三角函數(shù)的奇偶性和周期性判斷即可得出答案.【詳解】解：A選項(xiàng)：SKIPIF1<0是周期為SKIPIF1<0的偶函數(shù)，故A不正確；B選項(xiàng)：SKIPIF1<0是周期為SKIPIF1<0的奇函數(shù)，故B正確；C選項(xiàng)：SKIPIF1<0，周期為SKIPIF1<0且非奇非偶函數(shù)，故C不正確；D選項(xiàng)：SKIPIF1<0是周期為SKIPIF1<0的奇函數(shù)，故D不正確.故選：B.2．（2021·海南高三其他模擬）下列函數(shù)中，既是偶函數(shù)又存在零點(diǎn)的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】根據(jù)題意，依次分析選項(xiàng)中函數(shù)的奇偶性以及是否存在零點(diǎn)，綜合即可得答案．【詳解】解：根據(jù)題意，依次分析選項(xiàng)：對(duì)于SKIPIF1<0，SKIPIF1<0，為對(duì)數(shù)函數(shù)，不是奇函數(shù)，不符合題意，對(duì)于SKIPIF1<0，SKIPIF1<0，為二次函數(shù)，是偶函數(shù)，但不存在零點(diǎn)，不符合題意，對(duì)于SKIPIF1<0，SKIPIF1<0，為正弦函數(shù)，是奇函數(shù)，不符合題意，對(duì)于SKIPIF1<0，SKIPIF1<0，為余弦函數(shù)，既是偶函數(shù)又存在零點(diǎn)，符合題意，故選：SKIPIF1<0．3．（2021·浙江高三其他模擬）函數(shù)y=SKIPIF1<0在[-2，2]上的圖像可能是（）A． B．C． D．【答案】B【解析】利用同角三角函數(shù)的商數(shù)關(guān)系并注意利用正切函數(shù)的性質(zhì)求得函數(shù)的定義域，可以化簡(jiǎn)得到SKIPIF1<0，考察當(dāng)SKIPIF1<0趨近于0時(shí)，函數(shù)的變化趨勢(shì)，可以排除A,考察端點(diǎn)值的正負(fù)可以評(píng)出CD.【詳解】SKIPIF1<0,當(dāng)SKIPIF1<0趨近于0時(shí)，函數(shù)值趨近于SKIPIF1<0,故排除A;SKIPIF1<0,故排除CD,故選：B4．（2021·全國(guó)高三其他模擬（理））函數(shù)y=tan(3x+SKIPIF1<0)的一個(gè)對(duì)稱中心是（）A．(0，0) B．(SKIPIF1<0，0)C．(SKIPIF1<0，0) D．以上選項(xiàng)都不對(duì)【答案】C【解析】根據(jù)正切函數(shù)y=tanx圖象的對(duì)稱中心是(SKIPIF1<0，0)求出函數(shù)y=tan(3x+SKIPIF1<0)圖象的對(duì)稱中心，即可得到選項(xiàng).【詳解】解：因?yàn)檎泻瘮?shù)y=tanx圖象的對(duì)稱中心是(SKIPIF1<0，0)，k∈Z；令3x+SKIPIF1<0=SKIPIF1<0，解得SKIPIF1<0，k∈Z；所以函數(shù)y=tan(3x+SKIPIF1<0)的圖象的對(duì)稱中心為(SKIPIF1<0，0)，k∈Z；當(dāng)k=3時(shí),C正確，故選：C.5．（2019年高考全國(guó)Ⅱ卷文）若x1=，x2=是函數(shù)f(x)=(>0)兩個(gè)相鄰的極值點(diǎn)，則=（）A．2 B． C．1 D．【答案】A【解析】由題意知，的周期，解得．故選A．6．（2021·臨川一中實(shí)驗(yàn)學(xué)校高三其他模擬（文））若函數(shù)SKIPIF1<0的圖象在區(qū)間SKIPIF1<0上只有一個(gè)對(duì)稱中心，則SKIPIF1<0的取范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】根據(jù)題意可得SKIPIF1<0，即可求出.【詳解】由題可知，SKIPIF1<0在SKIPIF1<0上只有一個(gè)零點(diǎn)，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：A.7.(2019年高考北京卷文)設(shè)函數(shù)f（x）=cosx+bsinx（b為常數(shù)），則“b=0”是“f（x）為偶函數(shù)”的（）A．充分而不必要條件 B．必要而不充分條件

C．充分必要條件 D．既不充分也不必要條件【答案】C【解析】時(shí)，，為偶函數(shù)；為偶函數(shù)時(shí)，對(duì)任意的恒成立，即，，得對(duì)任意的恒成立，從而.從而“”是“為偶函數(shù)”的充分必要條件，故選C.8．（2021·青海西寧市·高三二模（文））函數(shù)SKIPIF1<0圖象的一個(gè)對(duì)稱中心為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】根據(jù)余弦函數(shù)的對(duì)稱中心整體代換求解即可.【詳解】令SKIPIF1<0，可得SKIPIF1<0．所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0滿足條件，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0滿足條件；故選：D9．（2021·全國(guó)高一專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，則下列結(jié)論錯(cuò)誤的是（）A．SKIPIF1<0的最小正周期為SKIPIF1<0 B．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0在SKIPIF1<0單調(diào)遞減 D．SKIPIF1<0的一個(gè)零點(diǎn)為SKIPIF1<0【答案】C【解析】根據(jù)解析式結(jié)合余弦函數(shù)的性質(zhì)依次判斷每個(gè)選項(xiàng)的正誤即可.【詳解】SKIPIF1<0函數(shù)SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0，故A正確；SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故B正確；當(dāng)SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0沒有單調(diào)性，故C錯(cuò)誤；SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0的一個(gè)零點(diǎn)為SKIPIF1<0，故D正確.綜上，錯(cuò)誤的選項(xiàng)為C.故選：C.10．（2017·全國(guó)高考真題（理））函數(shù)fx=s【答案】1【解析】化簡(jiǎn)三角函數(shù)的解析式，則fx=1?cos2x+3cosx?34=?練提升TIDHNEG練提升TIDHNEG1．（2021·河南高二月考（文））已知函數(shù)SKIPIF1<0的相鄰的兩個(gè)零點(diǎn)之間的距離是SKIPIF1<0，且直線SKIPIF1<0是SKIPIF1<0圖象的一條對(duì)稱軸，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由相鄰兩個(gè)零點(diǎn)的距離確定周期求出SKIPIF1<0，再由對(duì)稱軸確定SKIPIF1<0，代入SKIPIF1<0可求出結(jié)果.【詳解】解：因?yàn)橄噜彽膬蓚€(gè)零點(diǎn)之間的距離是SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0.故選：D.2.（2020·山東濰坊?高一期末）若函數(shù)的最小正周期為，則（）A． B．C． D．【答案】C【解析】由題意，函數(shù)的最小正周期為，可得，解得，即，令，即，當(dāng)時(shí)，，即函數(shù)在上單調(diào)遞增，又由，又由，所以.故選：C.3．（2021·廣東佛山市·高三二模）設(shè)SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】由條件即SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0；反之不成立，可舉反例．再由充分必要條件的判定得答案．【詳解】由SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，由正弦函數(shù)SKIPIF1<0的單調(diào)性可得SKIPIF1<0，即由SKIPIF1<0可以得到SKIPIF1<0.反之不成立，例如當(dāng)SKIPIF1<0時(shí)，也有SKIPIF1<0成立，但SKIPIF1<0不成立.故“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件故選：A4．（2021·四川省華鎣中學(xué)高三其他模擬（理））已知函數(shù)SKIPIF1<0的最大值為2，其圖象相鄰兩條對(duì)稱軸之間的距離為SKIPIF1<0且SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，則下列判斷不正確的是（）A．要得到函數(shù)SKIPIF1<0的圖象，只需將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位B．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減【答案】C【解析】根據(jù)最大值為2，可得A，根據(jù)正弦型函數(shù)的周期性，可求得SKIPIF1<0，根據(jù)對(duì)稱性，可求得SKIPIF1<0，即可得SKIPIF1<0解析式，根據(jù)正弦型函數(shù)的單調(diào)性、值域的求法，逐一分析選項(xiàng)，即可得答案.【詳解】由題意得A=2，因?yàn)槠鋱D象相鄰兩條對(duì)稱軸之間的距離為SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0為對(duì)稱中心，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，令k=0，可得SKIPIF1<0，所以SKIPIF1<0.對(duì)于A：將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，可得SKIPIF1<0，故A正確；對(duì)于B：令SKIPIF1<0，解得SKIPIF1<0，令k=1，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故B正確；對(duì)于C：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故C錯(cuò)誤；對(duì)于D：令SKIPIF1<0，解得SKIPIF1<0，令k=0，可得一個(gè)單調(diào)減區(qū)間為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D正確.故選：C5．（2021·玉林市第十一中學(xué)高三其他模擬（文））已知函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得y=g(x)的圖象，若函數(shù)g(x)的圖象與直線SKIPIF1<0在SKIPIF1<0上恰有兩個(gè)交點(diǎn)，則a的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由函數(shù)的平移可得SKIPIF1<0，結(jié)合三角函數(shù)的圖象與性質(zhì)可得SKIPIF1<0滿足的不等式，即可得解.【詳解】由題意，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)g(x)的圖象與直線SKIPIF1<0在SKIPIF1<0上恰有兩個(gè)交點(diǎn)，則SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故選：B.6．（2020·北京四中高三其他模擬）函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0（）A．6 B．5 C．4 D．3【答案】A【解析】根據(jù)正切函數(shù)的圖象求出A、B兩點(diǎn)的坐標(biāo)，再求出向量的坐標(biāo)，根據(jù)向量數(shù)量積的坐標(biāo)運(yùn)算求出結(jié)果．【詳解】由圖象得,令SKIPIF1<0=0,即SKIPIF1<0=kπ，SKIPIF1<0k=0時(shí)解得x=2，令SKIPIF1<0=1,即SKIPIF1<0，解得x=3，∴A(2,0),B(3,1)，∴SKIPIF1<0，∴SKIPIF1<0.故選：A.7．（2020·全國(guó)高三其他模擬（文））若函數(shù)SKIPIF1<0圖象上的相鄰一個(gè)最高點(diǎn)和一個(gè)最低點(diǎn)恰好都在圓SKIPIF1<0上，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】首先由題意判斷該正弦型函數(shù)的大概圖象及相鄰最高點(diǎn)和最低點(diǎn)與圓的交點(diǎn)情況.從而解得n的取值，再代入SKIPIF1<0求解.【詳解】解：設(shè)兩交點(diǎn)坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又函數(shù)SKIPIF1<0為奇函數(shù)，∴SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)取得最大值SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，由題，函數(shù)SKIPIF1<0圖象上的相鄰一個(gè)最高點(diǎn)和一個(gè)最低點(diǎn)恰好都在圓SKIPIF1<0上，∴SKIPIF1<0，則SKIPIF1<0.故選：A.8．【多選題】（2021·全國(guó)高三其他模擬）已知函數(shù)SKIPIF1<0圖象的一條對(duì)稱軸為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞減，則以下說(shuō)法正確的是（）A．SKIPIF1<0是其中一個(gè)對(duì)稱中心 B．SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0單増 D．SKIPIF1<0【答案】AD【解析】先根據(jù)條件求解函數(shù)的解析式，然后根據(jù)選項(xiàng)驗(yàn)證可得答案.【詳解】∵f(x)關(guān)SKIPIF1<0對(duì)稱，SKIPIF1<0，f(x)在SKIPIF1<0單調(diào)遞減，SKIPIF1<0，B錯(cuò)誤；SKIPIF1<0令SKIPIF1<0，可得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0即SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，A正確；令SKIPIF1<0得SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0單調(diào)遞増，即C錯(cuò)誤；SKIPIF1<0，D正確，故選：AD.9．【多選題】（2021·重慶市蜀都中學(xué)校高三月考）已知函數(shù)SKIPIF1<0滿足SKIPIF1<0，有SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則下列說(shuō)法正確的是（）A．SKIPIF1<0B．SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增C．SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱D．SKIPIF1<0時(shí)，方程SKIPIF1<0的所有根的和為SKIPIF1<0【答案】CD【解析】利用已知條件可知SKIPIF1<0在SKIPIF1<0上為奇函數(shù)且單調(diào)遞減，關(guān)于SKIPIF1<0、SKIPIF1<0，SKIPIF1<0對(duì)稱，且周期為4，即可判斷各選項(xiàng)的正誤.【詳解】由題設(shè)知：SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為奇函數(shù)且單調(diào)遞減，又SKIPIF1<0，即關(guān)于SKIPIF1<0、SKIPIF1<0，SKIPIF1<0對(duì)稱，且最小周期為4，A：SKIPIF1<0，錯(cuò)誤；B：SKIPIF1<0等價(jià)于SKIPIF1<0，由上易知：SKIPIF1<0上遞減，SKIPIF1<0上遞增，故SKIPIF1<0不單調(diào)，錯(cuò)誤；C：由上知：SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱且SKIPIF1<0，所以SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，正確；D：由題意，只需確定SKIPIF1<0與SKIPIF1<0在SKIPIF1<0的交點(diǎn)，判斷交點(diǎn)橫坐標(biāo)的對(duì)稱情況即可求和，如下圖示，∴共有6個(gè)交點(diǎn)且關(guān)于SKIPIF1<0對(duì)稱，則SKIPIF1<0，∴所有根的和為SKIPIF1<0，正確.故選：CD10.（2021·浙江杭州市·杭州高級(jí)中學(xué)高三其他模擬）設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，最小值為SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上最大值為________．【答案】1【解析】依題意可得函數(shù)在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0，所以SKIPIF1<0，即可求出函數(shù)的最大值；【詳解】解：函數(shù)SKIPIF1<0的周期為6，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0當(dāng)SKIPIF1<0時(shí)取最大值1故答案為：SKIPIF1<0練真題TIDHNEG練真題TIDHNEG1．（2021·全國(guó)高考真題（理））已知命題SKIPIF1<0﹔命題SKIPIF1<0﹐SKIPIF1<0，則下列命題中為真命題的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由正弦函數(shù)的有界性確定命題SKIPIF1<0的真假性，由指數(shù)函數(shù)的知識(shí)確定命題SKIPIF1<0的真假性，由此確定正確選項(xiàng).【詳解】由于SKIPIF1<0，所以命題SKIPIF1<0為真命題；由于SKIPIF1<0，所以SKIPIF1<0，所以命題SKIPIF1<0為真命題；所以SKIPIF1<0為真命題，SKIPIF1<0、SKIPIF1<0、SKIPIF1<0為假命題.故選：A．2.（2021·全國(guó)高考真題）下列區(qū)間中，函數(shù)SKIPIF1<0單調(diào)遞增的區(qū)間是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】解不等式SKIPIF1<0，利用賦值法可得出結(jié)論.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，對(duì)于函數(shù)SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，取SKIPIF1<0，可得函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞增區(qū)間為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，A選項(xiàng)滿足條件，B不滿足條件；取SKIPIF1<0，可得函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，CD選項(xiàng)均不滿足條件.故選：A.3.（2019年