高考數(shù)學(xué)二輪復(fù)習(xí)壓軸題專題13 三角函數(shù)中參數(shù)ω的取值范圍問(wèn)題（教師版）

專題13三角函數(shù)中參數(shù)ω的取值范圍問(wèn)題目錄TOC\o"1-1"\h\u①ω的取值范圍與單調(diào)性結(jié)合 1②ω的取值范圍與對(duì)稱性相結(jié)合 4③ω的取值范圍與三角函數(shù)的最值相結(jié)合 6④ω的取值范圍與三角函數(shù)的零點(diǎn)相結(jié)合 9⑤ω的取值范圍與三角函數(shù)的極值相結(jié)合 15①ω的取值范圍與單調(diào)性結(jié)合1．（2023春·海南海口·高一?？谝恢行？计谥校⒑瘮?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，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的值可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．4【答案】A【詳解】由已知可得，SKIPIF1<0.因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的值可能為SKIPIF1<0，故選：A2．（2023春·湖北武漢·高三武漢市黃陂區(qū)第一中學(xué)校考階段練習(xí)）將函數(shù)SKIPIF1<0（SKIPIF1<0）的圖像向左平移SKIPIF1<0個(gè)單位，得到函數(shù)SKIPIF1<0的圖像，若函數(shù)SKIPIF1<0）的一個(gè)極值點(diǎn)是SKIPIF1<0，且在SKIPIF1<0上單調(diào)遞增，則ω的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題意得：SKIPIF1<0，又函數(shù)SKIPIF1<0）的一個(gè)極值點(diǎn)是SKIPIF1<0，即SKIPIF1<0是函數(shù)SKIPIF1<0一條對(duì)稱軸，所以SKIPIF1<0，則SKIPIF1<0（SKIPIF1<0），函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則函數(shù)SKIPIF1<0的周期SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故選：A.3．（2023·全國(guó)·高三專題練習(xí)）將函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖像，若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．3 D．4【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖像，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以有SKIPIF1<0，因此SKIPIF1<0的最小值為SKIPIF1<0.故選：A.4．（2023·全國(guó)·高三專題練習(xí)）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后，得到函數(shù)SKIPIF1<0的圖象，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)遞減函數(shù)，則實(shí)數(shù)SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由題意，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)遞減函數(shù)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的最大值為SKIPIF1<0.故選：C.5．（2023春·河南鄭州·高三鄭州四中?？茧A段練習(xí)）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，若SKIPIF1<0在SKIPIF1<0上為增函數(shù)，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．3【答案】C【詳解】由題意可得SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0.∵SKIPIF1<0在SKIPIF1<0上為增函數(shù)，∴SKIPIF1<0,解得SKIPIF1<0.∴SKIPIF1<0的最大值為2.故選:C.②ω的取值范圍與對(duì)稱性相結(jié)合1．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，將SKIPIF1<0的圖象先向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，然后再向下平移1個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0圖象關(guān)于SKIPIF1<0對(duì)稱，則SKIPIF1<0為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，SKIPIF1<0的圖象先向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，然后再向下平移1個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0.故選：A2．（2023·四川瀘州·四川省瀘縣第一中學(xué)?？寄M預(yù)測(cè)）將函數(shù)SKIPIF1<0(其中SKIPIF1<0)的圖像向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得圖像關(guān)于SKIPIF1<0對(duì)稱，則SKIPIF1<0的最小值是A．6 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，可得SKIPIF1<0所得圖象關(guān)于SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，即SKIPIF1<0由于SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)取得最小值SKIPIF1<0.故選：D3．（2023·全國(guó)·模擬預(yù)測(cè)）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象.若函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，則SKIPIF1<0的最小值為.【答案】SKIPIF1<0【詳解】由題可得SKIPIF1<0，SKIPIF1<0SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，所以SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.4．（2023·廣東深圳·?？家荒＃⒑瘮?shù)SKIPIF1<0的圖像上所有點(diǎn)的縱坐標(biāo)保持不變，橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍后，所得函數(shù)SKIPIF1<0的圖像在區(qū)間SKIPIF1<0上有且僅有兩條對(duì)稱軸和兩個(gè)對(duì)稱中心，則SKIPIF1<0的值為．【答案】2【詳解】由題可知SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0．所以SKIPIF1<0的圖像大致如圖所示，要使SKIPIF1<0的圖像在區(qū)間SKIPIF1<0上有且僅有兩條對(duì)稱軸和兩個(gè)對(duì)稱中心，則SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0．故答案為：25．（2023·全國(guó)·高三專題練習(xí)）將函數(shù)SKIPIF1<0(SKIPIF1<0)的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度,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．【答案】SKIPIF1<0/0.5【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0設(shè)曲線SKIPIF1<0所對(duì)應(yīng)的函數(shù)為SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對(duì)稱,SKIPIF1<0SKIPIF1<0,SKIPIF1<0,解得:SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0的最小值是SKIPIF1<0.故答案為:SKIPIF1<0.③ω的取值范圍與三角函數(shù)的最值相結(jié)合1．（2023春·北京東城·高一北京二中校考階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，將函數(shù)SKIPIF1<0圖像上所有點(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的圖像，若對(duì)于任意的實(shí)數(shù)SKIPIF1<0，SKIPIF1<0恒成立，則SKIPIF1<0的最小值等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】將函數(shù)SKIPIF1<0圖像上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍（縱坐標(biāo)不變），則可得SKIPIF1<0，且對(duì)于任意的實(shí)數(shù)SKIPIF1<0，SKIPIF1<0恒成立，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故選：C2．（2023春·陜西西安·高一高新一中校考階段練習(xí)）將函數(shù)SKIPIF1<0先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將圖象上各點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍（縱坐標(biāo)不變），則所得函數(shù)SKIPIF1<0圖象，若SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0，則SKIPIF1<0的最小值等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】函數(shù)SKIPIF1<0先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得函數(shù)SKIPIF1<0，再將圖象上各點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍（縱坐標(biāo)不變），得函數(shù)SKIPIF1<0，∵SKIPIF1<0，則SKIPIF1<0∴由題意得：SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的最小值等于SKIPIF1<0，故選：B．3．（2023秋·廣東廣州·高三廣州市禺山高級(jí)中學(xué)校考階段練習(xí)）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將各點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0，得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，則SKIPIF1<0范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得SKIPIF1<0的圖象；再將各點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0，得到函數(shù)SKIPIF1<0的圖象．若SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，此時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，求得SKIPIF1<0，故選：A．4．（2023春·安徽亳州·高一亳州二中?？计谀┮阎瘮?shù)SKIPIF1<0圖象的縱坐標(biāo)不變、橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍后，得到的函數(shù)在SKIPIF1<0上恰有5個(gè)不同的SKIPIF1<0值，使其取到最值，則正實(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【詳解】解：∵函數(shù)SKIPIF1<0圖象的縱坐標(biāo)不變、橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍后，得到的函數(shù)為SKIPIF1<0在SKIPIF1<0上恰有5個(gè)不同的SKIPIF1<0值，使其取到最值；SKIPIF1<0，∴SKIPIF1<0，則正實(shí)數(shù)SKIPIF1<0，故選：A．5．（2023·貴州貴陽(yáng)·校聯(lián)考模擬預(yù)測(cè)）將函數(shù)SKIPIF1<0向右平移SKIPIF1<0個(gè)周期后所得的圖象在SKIPIF1<0內(nèi)有SKIPIF1<0個(gè)最高點(diǎn)和SKIPIF1<0個(gè)最低點(diǎn)，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，將函數(shù)SKIPIF1<0向右平移SKIPIF1<0后的解析式為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，要使得平移后的圖象有SKIPIF1<0個(gè)最高點(diǎn)和SKIPIF1<0個(gè)最低點(diǎn)，則需：SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0.④ω的取值范圍與三角函數(shù)的零點(diǎn)相結(jié)合1．（2023·貴州畢節(jié)·校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0在SKIPIF1<0上恰有兩個(gè)零點(diǎn)，則下列區(qū)間中，SKIPIF1<0的一個(gè)取值區(qū)間可以為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】依題意得SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上恰有兩個(gè)零點(diǎn)，因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0在SKIPIF1<0上超過(guò)兩個(gè)零點(diǎn)，不符合題意.綜上所述：SKIPIF1<0的取值范圍是SKIPIF1<0.結(jié)合四個(gè)選項(xiàng)可知，C正確.故選：C2．（2023春·浙江·高一校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有且僅有三個(gè)不相等的實(shí)根，則實(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】B【詳解】SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有且僅有三個(gè)不相等的實(shí)根，則SKIPIF1<0，解得SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B．3．（2023·浙江金華·?？既＃┮阎瘮?shù)SKIPIF1<0，若將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有且僅有兩個(gè)不相等的實(shí)根，則實(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】B【詳解】將函數(shù)SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0上有且僅有兩個(gè)不相等的實(shí)根，∴SKIPIF1<0，解得SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0，故選：B.4．（2023·四川內(nèi)江·統(tǒng)考三模）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0是SKIPIF1<0的一個(gè)單調(diào)遞增區(qū)間，且SKIPIF1<0在SKIPIF1<0上有5個(gè)零點(diǎn)，則SKIPIF1<0（

）A．1 B．5 C．9 D．13【答案】B【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，所以SKIPIF1<0，因?yàn)镾KIPIF1<0是SKIPIF1<0的一個(gè)單調(diào)遞增區(qū)間，所以，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上有5個(gè)零點(diǎn)，作出其草圖如圖，所以，由上圖可知，SKIPIF1<0，解得SKIPIF1<0

，所以，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，故選：B5．（2023春·高一單元測(cè)試）已知函數(shù)SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將所得函數(shù)圖象上所有點(diǎn)的縱坐標(biāo)不變，橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍得到函數(shù)SKIPIF1<0的圖象，若函數(shù)SKIPIF1<0在SKIPIF1<0上有且僅有4個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】方法一：由題意，函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得SKIPIF1<0的圖象，再將所得函數(shù)圖象上所有點(diǎn)的縱坐標(biāo)不變，橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0，得SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，SKIPIF1<0﹐解得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，欲使函數(shù)SKIPIF1<0在SKIPIF1<0上有且僅有4個(gè)零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0，方法二：由方法一得SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0.令SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，欲使方程SKIPIF1<0在SKIPIF1<0上有且僅有4個(gè)實(shí)根，則SKIPIF1<0，所以SKIPIF1<0，故選:B.6．（2023秋·天津南開(kāi)·高三南開(kāi)中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0是偶函數(shù).若將曲線SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到曲線SKIPIF1<0，若方程SKIPIF1<0在SKIPIF1<0有且僅有兩個(gè)不相等實(shí)根，則實(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】B【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0是偶函數(shù)，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0．若將曲線SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后，得到曲線SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0，則SKIPIF1<0，若方程SKIPIF1<0在SKIPIF1<0有且僅有兩個(gè)不相等實(shí)根，則有SKIPIF1<0，解得SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B7．（2023春·廣東惠州·高一?？茧A段練習(xí)）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將圖象上每個(gè)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上有且僅有一個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0的圖象再將圖象上每個(gè)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，SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn)，SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0故答案為SKIPIF1<08．（2023·全國(guó)·高一專題練習(xí)）將函數(shù)SKIPIF1<0的圖像先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再把所得函數(shù)圖像的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖像，若函數(shù)SKIPIF1<0在SKIPIF1<0上沒(méi)有零點(diǎn)，則SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】將函數(shù)SKIPIF1<0的圖像先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖像，再把所得函數(shù)圖像的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖像，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.由SKIPIF1<0在SKIPIF1<0上沒(méi)有零點(diǎn)，得SKIPIF1<0SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0.9．（2023·全國(guó)·高三專題練習(xí)）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，若函數(shù)SKIPIF1<0在SKIPIF1<0上有且只有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】由已知得SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有且只有三個(gè)根，分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，接下來(lái)第四個(gè)根為SKIPIF1<0所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0，故答案為：SKIPIF1<010．（2023春·上海虹口·高一上外附中?？计谀┮阎瘮?shù)SKIPIF1<0（其中SKIPIF1<0為常數(shù)，且SKIPIF1<0）有且僅有三個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】因?yàn)楹瘮?shù)SKIPIF1<0（其中SKIPIF1<0為常數(shù)，且SKIPIF1<0）有且僅有三個(gè)零點(diǎn)，故必有一個(gè)零點(diǎn)為x=0，所以SKIPIF1<0.所以問(wèn)題等價(jià)于函數(shù)SKIPIF1<0與直線y=1的圖像在SKIPIF1<0上有3個(gè)交點(diǎn)，如圖所示：所以SKIPIF1<0.故答案為：[2,4).⑤ω的取值范圍與三角函數(shù)的極值相結(jié)合1．（2023春·河南平頂山·高三校聯(lián)考階段練習(xí)）把函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將所得圖象上的所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍（SKIPIF1<0），縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象，若函數(shù)SKIPIF1<0在SKIPIF1<0上有兩個(gè)極值點(diǎn)、兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0，SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0的圖象，再將所得圖象上的所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍，得SKIPIF1<0的圖象．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，若SKIPIF1<0有兩個(gè)極值點(diǎn)，則SKIP