新高考數(shù)學(xué)二輪復(fù)習(xí) 多選題分類提升練習(xí)專題二【三角函數(shù)與解三角形】多選題專練六十題（原卷版）

多選題專練六十題專題二三角函數(shù)與解三角形（學(xué)生版）第一部——高考真題練1．（2000·全國(guó)·高考真題）已知SKIPIF1<0，那么下列命題中成立的是（

）A．若SKIPIF1<0、SKIPIF1<0是第一象限角，則SKIPIF1<0B．若SKIPIF1<0、SKIPIF1<0是第二象限角，則SKIPIF1<0C．若SKIPIF1<0、SKIPIF1<0是第二象限角，則SKIPIF1<0D．若SKIPIF1<0、SKIPIF1<0是第四象限角，則SKIPIF1<02．（2022·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱，則（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減B．SKIPIF1<0在區(qū)間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3．（2021·全國(guó)·統(tǒng)考高考真題）已知SKIPIF1<0為坐標(biāo)原點(diǎn)，點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2020·海南·高考真題）下圖是函數(shù)y=sin(ωx+φ)的部分圖像，則sin(ωx+φ)=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2023·湖北武漢·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，下列結(jié)論中正確的有（

）A．若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的整數(shù)倍B．函數(shù)SKIPIF1<0的圖象可由函數(shù)SKIPIF1<0的圖象上所有點(diǎn)的縱坐標(biāo)不變，橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0，再向左平移SKIPIF1<0單位得到C．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增第二部——基礎(chǔ)模擬題6．（2023·海南省直轄縣級(jí)單位·文昌中學(xué)校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0是SKIPIF1<0的一個(gè)極值點(diǎn)，SKIPIF1<0是與其相鄰的一個(gè)零點(diǎn)，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．直線SKIPIF1<0是函數(shù)SKIPIF1<0的對(duì)稱軸 D．SKIPIF1<07．（2023·河北張家口·統(tǒng)考三模）關(guān)于函數(shù)SKIPIF1<0，下列選項(xiàng)正確的有（

）A．SKIPIF1<0為偶函數(shù)B．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0的最小值為2D．SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個(gè)零點(diǎn)8．（2023·重慶巴南·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0的最小正周期為SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增9．（2023·廣東廣州·廣州市從化區(qū)從化中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)（

）A．SKIPIF1<0是由SKIPIF1<0圖象上的點(diǎn)橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變，再把得到的曲線向左平移SKIPIF1<0得到的B．SKIPIF1<0是由SKIPIF1<0圖象上的點(diǎn)橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變，再把得到的曲線向右平移SKIPIF1<0得到的C．SKIPIF1<0的對(duì)稱中心坐標(biāo)是SKIPIF1<0D．SKIPIF1<0是SKIPIF1<0的一條切線方程.10．（2023·江蘇南京·南京市第一中學(xué)校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，則正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0是函數(shù)SKIPIF1<0的零點(diǎn)C．函數(shù)SKIPIF1<0是非奇非偶函數(shù) D．SKIPIF1<0為SKIPIF1<0圖象的一條對(duì)稱軸11．（2023·山東菏澤·山東省鄄城縣第一中學(xué)校考三模）已知函數(shù)SKIPIF1<0，把函數(shù)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，若SKIPIF1<0時(shí)，方程SKIPIF1<0有實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值可以為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．（2023·廣東茂名·茂名市第一中學(xué)?？既＃㏒KIPIF1<0中，角SKIPIF1<0所對(duì)的邊分別為SKIPIF1<0.以下結(jié)論中正確的有（

）A．若SKIPIF1<0，則SKIPIF1<0必有兩解B．若SKIPIF1<0，則SKIPIF1<0一定為等腰三角形C．若SKIPIF1<0，則SKIPIF1<0一定為直角三角形D．若SKIPIF1<0，且該三角形有兩解，則SKIPIF1<0的范圍是SKIPIF1<013．（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<014．（2023·安徽六安·安徽省舒城中學(xué)?？寄M預(yù)測(cè)）定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足在區(qū)間SKIPIF1<0內(nèi)恰有兩個(gè)零點(diǎn)和一個(gè)極值點(diǎn)，則下列說(shuō)法不正確的是（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后關(guān)于原點(diǎn)對(duì)稱C．SKIPIF1<0圖象的一個(gè)對(duì)稱中心為SKIPIF1<0D．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增15．（2023·廣東東莞·校考三模）已知SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，則下列命題中成立的是（

）A．若SKIPIF1<0，SKIPIF1<0是第一象限角，則SKIPIF1<0B．若SKIPIF1<0，SKIPIF1<0是第二象限角，則SKIPIF1<0C．若SKIPIF1<0，SKIPIF1<0是第三象限角，則SKIPIF1<0D．若SKIPIF1<0，SKIPIF1<0是第四象限角，則SKIPIF1<016．（2023·廣東佛山·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱，則（

）A．SKIPIF1<0的最大值為2B．SKIPIF1<0是偶函數(shù)C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增D．把SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱17．（2023·湖南長(zhǎng)沙·長(zhǎng)沙市實(shí)驗(yàn)中學(xué)校考二模）如圖是函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的部分圖像，則（

）

A．SKIPIF1<0的最小正周期為SKIPIF1<0B．SKIPIF1<0是的函數(shù)SKIPIF1<0的一條對(duì)稱軸C．將函數(shù)SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位后，得到的函數(shù)為奇函數(shù)D．若函數(shù)SKIPIF1<0（SKIPIF1<0）在SKIPIF1<0上有且僅有兩個(gè)零點(diǎn)，則SKIPIF1<018．（2023·福建泉州·泉州五中?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0的最小正周期是SKIPIF1<0B．函數(shù)SKIPIF1<0的遞增區(qū)間是SKIPIF1<0，SKIPIF1<0C．函數(shù)SKIPIF1<0的對(duì)稱中心SKIPIF1<0，SKIPIF1<0D．當(dāng)SKIPIF1<0，函數(shù)SKIPIF1<0的值域是SKIPIF1<019．（2023·安徽安慶·安徽省桐城中學(xué)?？家荒＃╆P(guān)于函數(shù)SKIPIF1<0，則下列結(jié)論正確的有（）A．SKIPIF1<0是奇函數(shù) B．SKIPIF1<0的最小正周期為SKIPIF1<0C．SKIPIF1<0的最大值為SKIPIF1<0 D．SKIPIF1<0在SKIPIF1<0單調(diào)遞增20．（2023·海南?？凇ずＤ先A僑中學(xué)校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0（SKIPIF1<0）的圖象與函數(shù)SKIPIF1<0的圖象的對(duì)稱中心完全相同，且在SKIPIF1<0上，SKIPIF1<0有極小值，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．函數(shù)SKIPIF1<0是偶函數(shù) D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增21．（2023·江蘇無(wú)錫·校聯(lián)考三模）在SKIPIF1<0中，若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<022．（2023·安徽·合肥一中校聯(lián)考模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的最小正周期為πB．SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0的最小值為-1D．SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<023．（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<024．（2023·廣東·校聯(lián)考模擬預(yù)測(cè)）如圖是函數(shù)SKIPIF1<0的部分圖象，則下列結(jié)論正確的是（

）

A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<025．（2023·廣東東莞·統(tǒng)考模擬預(yù)測(cè)）隨著時(shí)代與科技的發(fā)展，信號(hào)處理以各種方式被廣泛應(yīng)用于醫(yī)學(xué)、聲學(xué)、密碼學(xué)、計(jì)算機(jī)科學(xué)、量子力學(xué)等各個(gè)領(lǐng)域.而信號(hào)處理背后的“功臣”就是正弦型函數(shù)，SKIPIF1<0的圖象就可以近似的模擬某種信號(hào)的波形，則下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱B．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．函數(shù)SKIPIF1<0為周期函數(shù)，且最小正周期為SKIPIF1<0D．函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的最大值為326．（2023·江蘇鎮(zhèn)江·江蘇省鎮(zhèn)江中學(xué)?？既＃┮阎瘮?shù)SKIPIF1<0，則（

）A．若SKIPIF1<0在區(qū)間SKIPIF1<0上為增函數(shù)，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0B．若SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0C．若SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)極大值，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0D．若SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)最大值，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<027．（2023·云南·校聯(lián)考模擬預(yù)測(cè)）在如圖所示的平面直角坐標(biāo)系中，銳角SKIPIF1<0，SKIPIF1<0的終邊分別與單位圓交于SKIPIF1<0，SKIPIF1<0兩點(diǎn).則（

）

A．若A點(diǎn)的橫坐標(biāo)為SKIPIF1<0，SKIPIF1<0點(diǎn)的縱坐標(biāo)為SKIPIF1<0，則SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為三邊構(gòu)成的三角形的外接圓的面積為SKIPIF1<028．（2023·安徽安慶·安慶一中?？寄M預(yù)測(cè)）正割（Secant）及余割（Cosecant）這兩個(gè)概念是由伊朗數(shù)學(xué)家?天文學(xué)家阿布爾·威發(fā)首先引入，SKIPIF1<0這兩個(gè)符號(hào)是荷蘭數(shù)學(xué)家基拉德在《三角學(xué)》中首先使用，后經(jīng)歐拉采用得以通行.在三角中，定義正割SKIPIF1<0，余割SKIPIF1<0.已知函數(shù)SKIPIF1<0，給出下列說(shuō)法正確的是（

）A．SKIPIF1<0的定義域?yàn)镾KIPIF1<0；B．SKIPIF1<0的最小正周期為SKIPIF1<0；C．SKIPIF1<0的值域?yàn)镾KIPIF1<0；D．SKIPIF1<0圖象的對(duì)稱軸為直線SKIPIF1<0.29．（2023·廣東深圳·統(tǒng)考模擬預(yù)測(cè)）已知SKIPIF1<0，則下列選項(xiàng)中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱C．SKIPIF1<0關(guān)于SKIPIF1<0中心對(duì)稱 D．SKIPIF1<0的值域?yàn)镾KIPIF1<030．（2023·湖南長(zhǎng)沙·長(zhǎng)沙市實(shí)驗(yàn)中學(xué)?？既＃┮阎瘮?shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0），若函數(shù)SKIPIF1<0的部分圖象如圖所示，則關(guān)于函數(shù)SKIPIF1<0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B．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D．函數(shù)SKIPIF1<0的圖象可由函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到31．（2023·廣東汕頭·金山中學(xué)?？既＃┮阎瘮?shù)SKIPIF1<0SKIPIF1<0，且SKIPIF1<0所有的正零點(diǎn)構(gòu)成一個(gè)公差為SKIPIF1<0的等差數(shù)列，把函數(shù)SKIPIF1<0的圖象沿SKIPIF1<0軸向左平移SKIPIF1<0個(gè)單位，橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍得到函數(shù)SKIPIF1<0的圖象，則下列關(guān)于函數(shù)SKIPIF1<0的結(jié)論正確的是（

）A．函數(shù)SKIPIF1<0是偶函數(shù)B．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．SKIPIF1<0在SKIPIF1<0上是增函數(shù)D．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的值域是SKIPIF1<032．（2023·云南·校聯(lián)考三模）在平面直角坐標(biāo)系SKIPIF1<0中，已知任意角SKIPIF1<0以坐標(biāo)原點(diǎn)SKIPIF1<0為頂點(diǎn)，SKIPIF1<0軸的非負(fù)半軸為始邊，若終邊經(jīng)過(guò)點(diǎn)SKIPIF1<0，且SKIPIF1<0，定義SKIPIF1<0，稱“SKIPIF1<0”為“正余弦函數(shù)”.對(duì)于“正余弦函數(shù)SKIPIF1<0”，下列結(jié)論中正確的是（

）A．將SKIPIF1<0圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到的圖象關(guān)于原點(diǎn)對(duì)稱B．SKIPIF1<0在區(qū)間SKIPIF1<0上的所有零點(diǎn)之和為SKIPIF1<0C．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減D．SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有5個(gè)極大值點(diǎn)33．（2023·遼寧·朝陽(yáng)市第一高級(jí)中學(xué)校聯(lián)考三模）關(guān)于函數(shù)SKIPIF1<0，下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0在SKIPIF1<0上最大值為SKIPIF1<0 B．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<034．（2023·廣東佛山·?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0是SKIPIF1<0的兩個(gè)極值點(diǎn)，且SKIPIF1<0，下列說(shuō)法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0在SKIPIF1<0上的單調(diào)遞增區(qū)間為SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上存在兩個(gè)不相等的根D．若SKIPIF1<0在SKIPIF1<0上恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<035．（2023·廣東廣州·廣州市培正中學(xué)?？寄M預(yù)測(cè)）在銳角SKIPIF1<0中，角SKIPIF1<0所對(duì)的邊為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的可能取值為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<036．（2023·遼寧沈陽(yáng)·沈陽(yáng)二中校考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱，且在區(qū)間SKIPIF1<0內(nèi)恰有三個(gè)極值點(diǎn)，則（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增B．SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有3個(gè)零點(diǎn)C．直線SKIPIF1<0是曲線SKIPIF1<0的對(duì)稱軸D．將SKIPIF1<0圖象向左平移SKIPIF1<0個(gè)單位，所得圖象對(duì)應(yīng)的函數(shù)為奇函數(shù)37．（2023·全國(guó)·鎮(zhèn)海中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0為坐標(biāo)原點(diǎn)，點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列說(shuō)法中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<038．（2023·湖北恩施·校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0部分圖像如下，它過(guò)SKIPIF1<0，SKIPIF1<0兩點(diǎn)，將SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位到SKIPIF1<0的圖像，則下列關(guān)于SKIPIF1<0的成立是（

）

A．圖像關(guān)于y軸對(duì)稱B．圖像關(guān)于SKIPIF1<0中心對(duì)稱C．在SKIPIF1<0上單調(diào)遞增D．在SKIPIF1<0最小值為SKIPIF1<039．（2023·遼寧沈陽(yáng)·東北育才學(xué)校校考模擬預(yù)測(cè)）在△ABC中，已知a＝2b，且SKIPIF1<0，則（

）A．a(chǎn)，c，b成等比數(shù)列B．SKIPIF1<0C．若a＝4，則SKIPIF1<0D．A，B，C成等差數(shù)列40．（2023·廣東潮州·統(tǒng)考模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0，且過(guò)點(diǎn)SKIPIF1<0，則下列正確的有（

）A．SKIPIF1<0在SKIPIF1<0單調(diào)遞減B．SKIPIF1<0的一條對(duì)稱軸為SKIPIF1<0C．SKIPIF1<0的周期為SKIPIF1<0D．把函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)長(zhǎng)度單位得到函數(shù)SKIPIF1<0的解析式為SKIPIF1<0第三部分能力提升模擬題41．（2023·湖南邵陽(yáng)·邵陽(yáng)市第二中學(xué)?？寄M預(yù)測(cè)）函數(shù)SKIPIF1<0的部分圖象如圖所示，則下列結(jié)論正確的是（

）

A．SKIPIF1<0B．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減C．將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位所得函數(shù)為奇函數(shù)D．方程SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有4個(gè)根42．（2023·河北·校聯(lián)考三模）已知SKIPIF1<0，則下列不等式成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<043．（2023·河北·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．SKIPIF1<0圖象的一條對(duì)稱軸為直線SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D．存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0在區(qū)間SKIPIF1<0上恰有2023個(gè)零點(diǎn)44．（2023·黑龍江哈爾濱·哈爾濱市第六中學(xué)校?？既＃┮阎猄KIPIF1<0的三個(gè)內(nèi)角SKIPIF1<0所對(duì)邊的長(zhǎng)分別為SKIPIF1<0，若SKIPIF1<0，則下列正確的是（

）A．SKIPIF1<0的取值范圍是SKIPIF1<0B．若SKIPIF1<0是SKIPIF1<0邊上的一點(diǎn)，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的面積的最大值為SKIPIF1<0C．若SKIPIF1<0是銳角三角形，則SKIPIF1<0的取值范圍是SKIPIF1<0D．若SKIPIF1<0平分SKIPIF1<0交SKIPIF1<0點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<045．（2023·湖南長(zhǎng)沙·雅禮中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0是以SKIPIF1<0為周期的函數(shù)B．直線SKIPIF1<0是曲線SKIPIF1<0的對(duì)稱軸C．函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，最小值為SKIPIF1<0D．若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有2023個(gè)零點(diǎn)，則SKIPIF1<046．（2023·湖北·荊門(mén)市龍泉中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0在SKIPIF1<0上有最大值，則（

）A．SKIPIF1<0的取值范圍為SKIPIF1<0 B．SKIPIF1<0在區(qū)間SKIPIF1<0上有零點(diǎn)C．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減 D．存在兩個(gè)SKIPIF1<0，使得SKIPIF1<047．（2023·山東濰坊·統(tǒng)考模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0其中SKIPIF1<0，SKIPIF1<0．若SKIPIF1<0，SKIPIF1<0，且相鄰兩個(gè)極值點(diǎn)之間的距離大于SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D．SKIPIF1<0在SKIPIF1<0上存在唯一極值點(diǎn)48．（2023·湖南郴州·校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0處取得最大值.下列結(jié)論正確的有（

）A．SKIPIF1<0B．SKIPIF1<0的最小值為SKIPIF1<0C．若函數(shù)SKIPIF1<0在SKIPIF1<0上存在零點(diǎn)，則SKIPIF1<0的最小值為SKIPIF1<0D．函數(shù)SKIPIF1<0在SKIPIF1<0上一定存在零點(diǎn)49．（2023·浙江紹興·統(tǒng)考二模）聲音是由物體振動(dòng)產(chǎn)生的聲波，純音的數(shù)學(xué)模型是函數(shù)SKIPIF1<0，我們聽(tīng)到的聲音是由純音合成的，稱之為復(fù)合音.我們聽(tīng)到的聲音函數(shù)是SKIPIF1<0，記SKIPIF1<0，SKIPIF1<0則下列結(jié)論中正確的為（

）A．SKIPIF1<0在SKIPIF1<0上是增函數(shù) B．SKIPIF1<0的最大值為SKIPIF1<0C．SKIPIF1<0的最小正周期為SKIPIF1<0 D．SKIPIF1<050．（2023·山西陽(yáng)泉·統(tǒng)考三模）設(shè)SKIPIF1<0內(nèi)角A，B，C的對(duì)邊分別為a，b，c．若SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<051．（2022·海南·校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，滿足SKIPIF1<0，且對(duì)任意SKIPIF1<0，都有SKIPIF1<0，當(dāng)SKIPIF1<0取最小值時(shí)，則下列錯(cuò)誤的是（

）A．SKIPIF1<0圖像的對(duì)稱軸方程為SKIPIF1<0B．SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0C．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D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減52．（2023·山西臨汾·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增B．SKIPIF1<0的最小正周期為SKIPIF1<0C．SKIPIF1<0的值域?yàn)镾KIPIF1<0D．SKIPIF1<0的圖象可以由函數(shù)SKIPIF1<0的圖象，先向左平移SKIPIF1<0個(gè)單位，再向上平移SKIPIF1<0個(gè)單位得到53．（2023·湖南·湖南師大附中校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則（

）A．SKIPIF1<0B．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增C．將函數(shù)SKIPIF1<0圖象上各點(diǎn)橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0（縱坐標(biāo)不變），再將所得圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得函數(shù)SKIPIF1<0的圖象D．函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為754．（2023·遼寧沈陽(yáng)·統(tǒng)考一模）聲音是由物體振動(dòng)產(chǎn)生的聲波，純音的數(shù)學(xué)模型是函數(shù)SKIPIF1<0，我們聽(tīng)到的聲音是由純音合成的，稱之為復(fù)合音．若一個(gè)復(fù)合音的數(shù)學(xué)模型是函數(shù)SKIPIF1<0，則下列結(jié)論中正確的為（

）A．SKIPIF1<0在SKIPIF1<0上是增函數(shù)B．SKIPIF1<0的最小正周期為SKIPIF1<0C．SKIPIF1<0的最大值為SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<0．55．（2023·山東菏澤·統(tǒng)考一模）已知函數(shù)SKIPIF1<0SKIPIF1<0，下列命題正確的有（

）A．SKIPIF1<0