新高考數(shù)學(xué)一輪復(fù)習(xí)講練測(cè)第4章第03講 三角函數(shù)的圖象與性質(zhì)（練習(xí)）（解析版）

第03講三角函數(shù)的圖象與性質(zhì)（模擬精練+真題演練）1．（2023·陜西西安·陜西師大附中校考模擬預(yù)測(cè)）將函數(shù)SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖像，則下列正確的是（

）A．直線SKIPIF1<0是SKIPIF1<0圖像的一條對(duì)稱軸 B．SKIPIF1<0的最小正周期為SKIPIF1<0C．SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱 D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增【答案】C【解析】由SKIPIF1<0SKIPIF1<0，則SKIPIF1<0圖像向右平移SKIPIF1<0個(gè)單位長(zhǎng)度可得，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0不是SKIPIF1<0圖像的一條對(duì)稱軸，A錯(cuò)；由SKIPIF1<0，得SKIPIF1<0的最小正周期為SKIPIF1<0，B錯(cuò)；由SKIPIF1<0，所以點(diǎn)SKIPIF1<0是SKIPIF1<0圖像的一個(gè)對(duì)稱中心，C正確；由SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有增有減，D錯(cuò).故選：C2．（2023·四川成都·石室中學(xué)校考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0圖象的對(duì)稱軸可以是（

）A．直線SKIPIF1<0 B．直線SKIPIF1<0C．直線SKIPIF1<0 D．直線SKIPIF1<0【答案】A【解析】SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的對(duì)稱軸為直線SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故選：A.3．（2023·河南·襄城高中校聯(lián)考三模）將函數(shù)SKIPIF1<0的圖象上所有點(diǎn)向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，然后橫坐標(biāo)伸長(zhǎng)為原來(lái)的2倍，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象，則SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】將SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0，縱坐標(biāo)不變得到SKIPIF1<0的圖象，再將SKIPIF1<0圖象上所有點(diǎn)向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0的圖象．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0．故選：C.4．（2023·重慶·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若對(duì)于任意實(shí)數(shù)x，都有SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．4 D．8【答案】C【解析】因?yàn)閷?duì)于任意實(shí)數(shù)x，都有SKIPIF1<0，則有函數(shù)SKIPIF1<0圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，因此SKIPIF1<0，解得SKIPIF1<0，而SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值4.故選：C5．（2023·河南·校聯(lián)考模擬預(yù)測(cè)）某次實(shí)驗(yàn)得交變電流SKIPIF1<0（單位：A）隨時(shí)間SKIPIF1<0（單位：s）變化的函數(shù)解析式為SKIPIF1<0，其中SKIPIF1<0且SKIPIF1<0，其圖象如圖所示，則下列說(shuō)法錯(cuò)誤的是（

）

A．SKIPIF1<0 B．SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0【答案】D【解析】由題知SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，因此SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此ABC正確，D錯(cuò)誤，故選：D.6．（2023·北京西城·北師大實(shí)驗(yàn)中學(xué)?？既＃┰谙铝兴膫€(gè)函數(shù)中，在定義域內(nèi)單調(diào)遞增的有（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】A.SKIPIF1<0的增區(qū)間為SKIPIF1<0，在整個(gè)定義域上不單調(diào)，故錯(cuò)誤；B.SKIPIF1<0的增區(qū)間是SKIPIF1<0，在整個(gè)定義域上不單調(diào)，故錯(cuò)誤；C.SKIPIF1<0在R上遞增，故正確；D.SKIPIF1<0的增區(qū)間是SKIPIF1<0，在整個(gè)定義域上不單調(diào)，故錯(cuò)誤；故選：C7．（2023·北京大興·校考三模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)下列變換可以與SKIPIF1<0的圖象重合的是（

）A．向左平移SKIPIF1<0個(gè)單位 B．向左平移SKIPIF1<0個(gè)單位C．向右平移SKIPIF1<0個(gè)單位 D．向右平移SKIPIF1<0個(gè)單位【答案】D【解析】因?yàn)镾KIPIF1<0，所以將SKIPIF1<0向右平移SKIPIF1<0個(gè)單位得到SKIPIF1<0.故選：D8．（2023·陜西咸陽(yáng)·武功縣普集高級(jí)中學(xué)校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則關(guān)于SKIPIF1<0的下列結(jié)論不正確的是（

）A．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱B．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)遞減函數(shù)D．將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位即可得到SKIPIF1<0的圖象【答案】D【解析】∵SKIPIF1<0，∴SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故A正確；∵SKIPIF1<0，∴SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，故B正確；令SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)，根據(jù)復(fù)合函數(shù)的單調(diào)性知，SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)遞減函數(shù)，故C正確；∵SKIPIF1<0，∴將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位即可得到SKIPIF1<0的圖象，而SKIPIF1<0時(shí)，SKIPIF1<0，故D錯(cuò)誤，故選：D.9．（多選題）（2023·福建漳州·統(tǒng)考模擬預(yù)測(cè)）把函數(shù)SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變，再把所得曲線向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，則（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減B．SKIPIF1<0在SKIPIF1<0上有2個(gè)零點(diǎn)C．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱D．SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0【答案】BC【解析】把函數(shù)SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變，可得到SKIPIF1<0的圖象；再把所得曲線向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，故A錯(cuò)誤；令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有2個(gè)零點(diǎn)，故B正確；因?yàn)镾KIPIF1<0，為SKIPIF1<0的最大值，所以直線SKIPIF1<0是SKIPIF1<0的圖象的一條對(duì)稱軸，故C正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故D錯(cuò)誤.故選：BC10．（多選題）（2023·江蘇鹽城·鹽城市伍佑中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0）個(gè)單位長(zhǎng)度后對(duì)應(yīng)的函數(shù)為SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上單調(diào)，則SKIPIF1<0的可?。?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】CD【解析】依題意，SKIPIF1<0，于是SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增時(shí)，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，不存在整數(shù)SKIPIF1<0使得SKIPIF1<0取得ABCD選項(xiàng)中的值；當(dāng)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減時(shí)，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，CD符合，不存在整數(shù)SKIPIF1<0使得SKIPIF1<0取得AB選項(xiàng)中的值.故選：CD11．（多選題）（2023·廣東佛山·校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的初相為SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱B．函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞減區(qū)間為SKIPIF1<0C．若把函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象，則SKIPIF1<0為偶函數(shù)D．若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)镾KIPIF1<0【答案】AB【解析】由題意知SKIPIF1<0，所以SKIPIF1<0．對(duì)于選項(xiàng)A，SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故A項(xiàng)正確；對(duì)于選項(xiàng)B，由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞減區(qū)間為SKIPIF1<0，故B項(xiàng)正確；對(duì)于選項(xiàng)C，SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象，所以SKIPIF1<0為奇函數(shù)，故C項(xiàng)錯(cuò)誤；對(duì)于選項(xiàng)D，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即：SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)镾KIPIF1<0，故D項(xiàng)錯(cuò)誤．故選：AB.12．（多選題）（2023·湖南衡陽(yáng)·衡陽(yáng)市八中?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，其圖象相鄰對(duì)稱軸間的距離為SKIPIF1<0，點(diǎn)SKIPIF1<0是其中的一個(gè)對(duì)稱中心，則下列結(jié)論正確的是（

）A．函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0B．函數(shù)SKIPIF1<0圖象的一條對(duì)稱軸方程是SKIPIF1<0C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D．將函數(shù)SKIPIF1<0圖象上所有點(diǎn)橫坐標(biāo)伸長(zhǎng)原來(lái)的2倍，縱坐標(biāo)縮短原來(lái)的一半，再把得到的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得到正弦函數(shù)SKIPIF1<0的圖象【答案】ACD【解析】因?yàn)楹瘮?shù)SKIPIF1<0圖象相鄰對(duì)稱軸間的距離為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0正確；因?yàn)镾KIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，且點(diǎn)SKIPIF1<0是對(duì)稱中心，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.令SKIPIF1<0,解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，所以SKIPIF1<0錯(cuò)誤；令SKIPIF1<0,解得SKIPIF1<0，函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為：SKIPIF1<0，所以C正確；函數(shù)SKIPIF1<0圖象上所有點(diǎn)橫坐標(biāo)伸長(zhǎng)原來(lái)的2倍，縱坐標(biāo)縮短原來(lái)的一半，得到SKIPIF1<0的圖象，再把得到的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得函數(shù)SKIPIF1<0，所以SKIPIF1<0正確.故選:ACD.13．（2023·河北滄州·?？寄M預(yù)測(cè)）若函數(shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0的最小值為_(kāi)_____.【答案】SKIPIF1<0【解析】因?yàn)楹瘮?shù)SKIPIF1<0為R上的奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<014．（2023·陜西咸陽(yáng)·武功縣普集高級(jí)中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0，當(dāng)SKIPIF1<0（其中SKIPIF1<0）時(shí)，SKIPIF1<0有且只有一個(gè)解，則SKIPIF1<0的取值范圍是____________.【答案】SKIPIF1<0【解析】由于SKIPIF1<0，所以SKIPIF1<0有且只有一個(gè)解，即SKIPIF1<0有且只有一個(gè)解，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由題意知SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0的取值范圍是為SKIPIF1<0，故答案為：SKIPIF1<015．（2023·江蘇鎮(zhèn)江·江蘇省鎮(zhèn)江中學(xué)?？既＃?xiě)出一個(gè)同時(shí)具有下列性質(zhì)①②③，且定義域?yàn)閷?shí)數(shù)集SKIPIF1<0的函數(shù)SKIPIF1<0__________.①最小正周期為2；②SKIPIF1<0；③無(wú)零點(diǎn).【答案】SKIPIF1<0（答案不唯一）【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，最小正周期為SKIPIF1<0，SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0無(wú)零點(diǎn)，綜上，SKIPIF1<0符合題意故答案為：SKIPIF1<0.16．（2023·上海徐匯·位育中學(xué)?？寄M預(yù)測(cè)）若函數(shù)SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖像，若對(duì)滿足SKIPIF1<0的SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0的最小值為SKIPIF1<0，則SKIPIF1<0________．【答案】SKIPIF1<0【解析】由函數(shù)SKIPIF1<0的圖像向右平移SKIPIF1<0，可得SKIPIF1<0由SKIPIF1<0可知一個(gè)取得最大值一個(gè)取得最小值，不妨設(shè)SKIPIF1<0取得最大值，SKIPIF1<0取得最小值，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．可得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，故答案為：SKIPIF1<0.17．（2023·湖南岳陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示.(1)求SKIPIF1<0的最小正周期及解析式；(2)將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象，求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值和最小值.【解析】（1）由圖象可知SKIPIF1<0的最大值為1，最小值-1，故SKIPIF1<0；又SKIPIF1<0∴SKIPIF1<0，將點(diǎn)SKIPIF1<0代入SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0故答案為：SKIPIF1<0，SKIPIF1<0.（2）由SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0∵SKIPIF1<0∴SKIPIF1<0∴當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0故答案為：SKIPIF1<018．（2023·黑龍江哈爾濱·哈爾濱市第六中學(xué)校?？既＃┮阎瘮?shù)SKIPIF1<0，其圖象的一條對(duì)稱軸與相鄰對(duì)稱中心的橫坐標(biāo)相差SKIPIF1<0，______，從以下兩個(gè)條件中任選一個(gè)補(bǔ)充在空白橫線中.①函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到的圖象關(guān)于y軸對(duì)稱且SKIPIF1<0；②函數(shù)SKIPIF1<0的圖象的一個(gè)對(duì)稱中心為SKIPIF1<0且SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的解析式；(2)將函數(shù)SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，求t的取值范圍.【解析】（1）由題意可得SKIPIF1<0SKIPIF1<0，SKIPIF1<0，由于其圖象的一條對(duì)稱軸與相鄰對(duì)稱中心的橫坐標(biāo)相差SKIPIF1<0，故SKIPIF1<0,故SKIPIF1<0若選①，函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到的圖象對(duì)應(yīng)的函數(shù)為SKIPIF1<0，由題意知該函數(shù)為偶函數(shù)，故SKIPIF1<0，由于SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0；若選②，函數(shù)SKIPIF1<0的圖象的一個(gè)對(duì)稱中心為SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，由于SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0；（2）由題意可得SKIPIF1<0，由于SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，故SKIPIF1<0，即SKIPIF1<0.19．（2023·湖南常德·常德市一中?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，其中SKIPIF1<0為正整數(shù).(1)求函數(shù)SKIPIF1<0的解析式；(2)將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到函數(shù)SKIPIF1<0的圖象，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間.【解析】（1）由SKIPIF1<0，得SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，于是SKIPIF1<0，解得SKIPIF1<0，而SKIPIF1<0為正整數(shù)，因此SKIPIF1<0，所以SKIPIF1<0.（2）由（1）知，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，即有SKIPIF1<0，因此SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0.20．（2023·黑龍江哈爾濱·哈師大附中?？寄M預(yù)測(cè)）將函數(shù)SKIPIF1<0的圖象先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將所得函圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0（ω＞0）倍（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象．(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上沒(méi)有零點(diǎn)，求ω的取值范圍．【解析】（1）函數(shù)SKIPIF1<0的圖象先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，則解析式變?yōu)椋篠KIPIF1<0，再將所得函圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0（ω＞0）倍（縱坐標(biāo)不變），則解析式變?yōu)镾KIPIF1<0.則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0.∴SKIPIF1<0，∴SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0．（2）SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，要使SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)，則SKIPIF1<0，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0舍去．綜上：SKIPIF1<0的取值范圍為SKIPIF1<0．1．（2023?天津）已知函數(shù)SKIPIF1<0的一條對(duì)稱軸為直線SKIPIF1<0，一個(gè)周期為4，則SKIPIF1<0的解析式可能為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】SKIPIF1<0：若SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0不是對(duì)稱軸，不符合題意；SKIPIF1<0：若SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0是一條對(duì)稱軸，SKIPIF1<0符合題意；SKIPIF1<0，則SKIPIF1<0，不符合題意；SKIPIF1<0，則SKIPIF1<0，不符合題意．故選：SKIPIF1<0．2．（2022?天津）已知SKIPIF1<0，關(guān)于該函數(shù)有下列四個(gè)說(shuō)法：①SKIPIF1<0的最小正周期為SKIPIF1<0；②SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增；③當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0的取值范圍為SKIPIF1<0，SKIPIF1<0；④SKIPIF1<0的圖象可由SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到．以上四個(gè)說(shuō)法中，正確的個(gè)數(shù)為SKIPIF1<0SKIPIF1<0A．1 B．2 C．3 D．4【答案】SKIPIF1<0【解析】對(duì)于SKIPIF1<0，它的最小正周期為SKIPIF1<0，故①錯(cuò)誤；在SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，故②正確；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的取值范圍為SKIPIF1<0，SKIPIF1<0，故③錯(cuò)誤；SKIPIF1<0的圖象可由SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到，故④錯(cuò)誤，故選：SKIPIF1<0．3．（2022?浙江）為了得到函數(shù)SKIPIF1<0的圖象，只要把函數(shù)SKIPIF1<0圖象上所有的點(diǎn)SKIPIF1<0SKIPIF1<0A．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度 C．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 D．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度【答案】SKIPIF1<0【解析】把SKIPIF1<0圖象上所有的點(diǎn)向右平移SKIPIF1<0個(gè)單位可得SKIPIF1<0的圖象．故選：SKIPIF1<0．4．（2022?新高考Ⅰ）記函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0．若SKIPIF1<0，且SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0，SKIPIF1<0中心對(duì)稱，則SKIPIF1<0SKIPIF1<0A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】SKIPIF1<0【解析】函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0，SKIPIF1<0中心對(duì)稱，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．SKIPIF1<0SKIPIF1<0，SKIPIF1<0，取SKIPIF1<0，可得SKIPIF1<0．SKIPIF1<0，則SKIPIF1<0．故選：SKIPIF1<0．5．（2022?甲卷）將函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到曲線SKIPIF1<0，若SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱，則SKIPIF1<0的最小值是SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】將函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到曲線SKIPIF1<0，則SKIPIF1<0對(duì)應(yīng)函數(shù)為SKIPIF1<0，SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，則令SKIPIF1<0，可得SKIPIF1<0的最小值是SKIPIF1<0，故選：SKIPIF1<0．6．（2022?甲卷）設(shè)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0恰有三個(gè)極值點(diǎn)、兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是SKIPIF1<0SKIPIF1<0A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí)，不能滿足在區(qū)間SKIPIF1<0極值點(diǎn)比零點(diǎn)多，所以SKIPIF1<0；函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0恰有三個(gè)極值點(diǎn)、兩個(gè)零點(diǎn)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，求得SKIPIF1<0，故選：SKIPIF1<0．7．（多選題）（2022?新高考Ⅱ）已知函數(shù)SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0，SKIPIF1<0中心對(duì)稱，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減 B．SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0有兩個(gè)極值點(diǎn) C．直線SKIPIF1<0是曲線SKIPIF1<0的對(duì)稱軸 D．直線SKIPIF1<0是曲線SKIPIF1<0的切線【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0，SKIPIF1<0對(duì)稱，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0正確；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，根據(jù)函數(shù)的單調(diào)性，故函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0只有一個(gè)極值點(diǎn)，故SKIPIF1<0錯(cuò)誤；令SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0顯然錯(cuò)誤；SKIPIF1<0，求導(dǎo)可得，SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為SKIPIF1<0，故切線方程為SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0正確．直線SKIPIF1<0顯然與SKIPIF1<0相切，故直線SKIPIF1<0顯然是曲線的切線，故SKIPIF1<0正確．故選：SKIPIF1<0．8．（2023?新高考Ⅱ）已知函數(shù)SKIPIF1<0，如圖，SKIPIF1<0，SKIPIF1<0是直線SKIPIF1<0與曲線SKIPIF1<0的兩個(gè)交點(diǎn)，若SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0【解析】由題意：設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0的圖象可知：SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，觀察圖象，可知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0滿足條件，SKIPIF1<0．故答案為：SKIPIF1<0．9．（2023?新高考Ⅰ）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0有且僅有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0，SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0，函數(shù)的周期為SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0有且僅有3個(gè)零點(diǎn)，可得SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0，SKIPIF1<0．10．（2022?上海）函數(shù)SKIPIF1<0的周期為．【答案】SKIPIF1<0【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0