新高考數(shù)學(xué)一輪復(fù)習(xí)提升訓(xùn)練5.3 三角函數(shù)的性質(zhì)（精講）（解析版）

5.3三角函數(shù)的性質(zhì)（精講）（提升版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一值域【例1-1】（2022·湖南·長(zhǎng)郡中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0（SKIPIF1<0）的圖象向左平移SKIPIF1<0個(gè)單位后關(guān)于直線SKIPIF1<0對(duì)稱，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后的圖象表達(dá)式為ySKIPIF1<0，該函數(shù)的圖象關(guān)于直線SKIPIF1<0對(duì)稱，所以SKIPIF1<0，又SKIPIF1<0所以，SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0.故選：A【例1-2】（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的最大值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】C【解析】SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以原函數(shù)可化為SKIPIF1<0，SKIPIF1<0，對(duì)稱軸為SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值，所以函數(shù)的最大值為SKIPIF1<0，即SKIPIF1<0的最大值為SKIPIF1<0，故選：C【例1-3】（2021·河南南陽(yáng)·高三期末）已知SKIPIF1<0，若SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，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】A【解析】SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.SKIPIF1<0SKIPIF1<0，若SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則SKIPIF1<0，即SKIPIF1<0.故選：A.【例1-4】（2022·陜西·武功縣普集高級(jí)中學(xué)高三階段練習(xí)（理））函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)恰有兩個(gè)最小值點(diǎn)，則SKIPIF1<0的范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，函數(shù)有最小值，令SKIPIF1<0時(shí)，有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0內(nèi)恰有兩個(gè)最小值點(diǎn)，SKIPIF1<0，所以有：SKIPIF1<0，故選：B【一隅三反】1．（2021·江蘇泰州·高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0的值域?yàn)镾KIPIF1<0，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】C【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，又SKIPIF1<0的值域?yàn)镾KIPIF1<0，所以SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：C.2．（2022·河南焦作·二模）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0在區(qū)間SKIPIF1<0上恰有5個(gè)實(shí)根，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由方程SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的可能取值為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…，因?yàn)樵匠淘趨^(qū)間SKIPIF1<0上恰有5個(gè)實(shí)根，所以SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D.3．（2022·全國(guó)·高三專題練習(xí)）已知函SKIPIF1<0，對(duì)于任意的x1，x2∈R，都有f(x1)+f(x2)-2SKIPIF1<0≤0，若f(x)在[0，π]上的值域?yàn)镾KIPIF1<0，則實(shí)數(shù)ω的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】f(x)=asinωx+cos(ωx-SKIPIF1<0)=asinωx+cosωxcosSKIPIF1<0+sinωxsinSKIPIF1<0=(SKIPIF1<0+a)sinωx+SKIPIF1<0cosωx=SKIPIF1<0·sin(ωx+φ)，其中tanφ=SKIPIF1<0.對(duì)于任意的x1，x2∈R，都有f(x1)+f(x2)-2SKIPIF1<0≤0，即f(x1)+f(x2)≤2SKIPIF1<0，當(dāng)且僅當(dāng)f(x1)=f(x2)=f(x)max時(shí)取等號(hào)，故2SKIPIF1<0=2SKIPIF1<0，解得a=1或a=-2(舍去)，故f(x)=SKIPIF1<0sinωx+SKIPIF1<0cosωx=SKIPIF1<0sin(ωx+SKIPIF1<0).因?yàn)?≤x≤π，所以SKIPIF1<0≤ωx+SKIPIF1<0≤ωπ+SKIPIF1<0.又f(x)在[0，π]上的值域?yàn)閇SKIPIF1<0]，所以SKIPIF1<0≤ωπ+SKIPIF1<0≤SKIPIF1<0，解得SKIPIF1<0≤ω≤SKIPIF1<0.故選：B.考點(diǎn)二伸縮平移【例2-1】（2022·河南洛陽(yáng)·模擬預(yù)測(cè)（文））已知曲線SKIPIF1<0，SKIPIF1<0，為了得到曲線SKIPIF1<0，則對(duì)曲線SKIPIF1<0的變換正確的是（

）A．先把橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍（縱坐標(biāo)不變），再把得到的曲線向右平移SKIPIF1<0個(gè)單位長(zhǎng)度B．先把橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍（縱坐標(biāo)不變），再把得到的曲線向左平移SKIPIF1<0個(gè)單位長(zhǎng)度C．先把橫坐標(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個(gè)單位長(zhǎng)度D．先把橫坐標(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個(gè)單位長(zhǎng)度【答案】C【解析】A.先把曲線SKIPIF1<0上點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍（縱坐標(biāo)不變），得SKIPIF1<0的圖象，再把得到的曲線向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0的圖象，A錯(cuò)；B.先把曲線SKIPIF1<0上點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍（縱坐標(biāo)不變），得SKIPIF1<0的圖象，再把得到的曲線向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0的圖象，B錯(cuò)；C.先把曲線SKIPIF1<0上點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍（縱坐標(biāo)不變），得SKIPIF1<0的圖象，再把得到的曲線向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0的圖象，C正確；D.先把曲線SKIPIF1<0上點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍（縱坐標(biāo)不變），得SKIPIF1<0的圖象，再把得到的曲線向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0的圖象，D錯(cuò)誤；故選：C．【例2-2】（2022·全國(guó)·高三專題練習(xí)）將函數(shù)SKIPIF1<0的圖象向左平移a(a>0)個(gè)單位長(zhǎng)度得到函數(shù)g(x)=cos2x的圖象，則a的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．π【答案】B【解析】將函數(shù)f(x)=sin(2x-SKIPIF1<0)的圖象向左平移a(a>0)個(gè)單位長(zhǎng)度，可得函數(shù)y=sin[2(x+a)-SKIPIF1<0]=sin[2x+(2a-SKIPIF1<0)]的圖象，所以y=sin[2x+(2a-SKIPIF1<0)]的圖象與g(x)=cos2x的圖象重合.因?yàn)間(x)=cos2x=sin(2x+SKIPIF1<0)，所以2a-SKIPIF1<0=2kπ+SKIPIF1<0，k∈Z，即a=kπ+SKIPIF1<0，k∈Z.當(dāng)k=0時(shí)，可得amin=SKIPIF1<0.故選：B.【一隅三反】1．（2022·陜西）已知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，若將其圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到的圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，則SKIPIF1<0的圖象（

）A．關(guān)于點(diǎn)SKIPIF1<0對(duì)稱 B．關(guān)于SKIPIF1<0對(duì)稱 C．關(guān)于點(diǎn)SKIPIF1<0對(duì)稱 D．關(guān)于SKIPIF1<0對(duì)稱【答案】A【解析】依題意SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，將函數(shù)向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0，因?yàn)镾KIPIF1<0關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，所以SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以函數(shù)關(guān)于SKIPIF1<0對(duì)稱，又SKIPIF1<0，所以函數(shù)關(guān)于SKIPIF1<0對(duì)稱，SKIPIF1<0，所以函數(shù)關(guān)于SKIPIF1<0對(duì)稱；故選：A2．（2022·湖北·一模）函數(shù)SKIPIF1<0，先把函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位，再把圖像上各點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0，得到函數(shù)SKIPIF1<0的圖像，則下列說(shuō)法錯(cuò)誤的是（

）A．函數(shù)SKIPIF1<0是奇函數(shù)，最大值是2B．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增C．函數(shù)SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱D．π是函數(shù)SKIPIF1<0的周期【答案】B【解析】SKIPIF1<0，把函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位，得SKIPIF1<0，再把圖像上各點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0，得SKIPIF1<0，所以可知SKIPIF1<0是奇函數(shù)，最大值是2，最小正周期為SKIPIF1<0，當(dāng)SKIPIF1<0，得SKIPIF1<0，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0也是函數(shù)SKIPIF1<0的對(duì)稱軸，所以錯(cuò)誤的選項(xiàng)為B.故選：B.3．（2022·全國(guó)·模擬預(yù)測(cè)）若將函數(shù)SKIPIF1<0的圖象分別向左平移SKIPIF1<0個(gè)單位長(zhǎng)度與向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得的兩個(gè)函數(shù)圖象恰好重合，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0SKIPIF1<0的圖象,向右平移SKIPIF1<0（SKIPIF1<0）個(gè)單位長(zhǎng)度得SKIPIF1<0SKIPIF1<0的圖象,由題意得SKIPIF1<0（SKIPIF1<0）所以SKIPIF1<0（SKIPIF1<0）又SKIPIF1<0,故SKIPIF1<0的最小值為SKIPIF1<0,故選：A考點(diǎn)三三角函數(shù)的性質(zhì)【例3-1】（2022·全國(guó)·高三專題練習(xí)）（多選）下列函數(shù)中，以SKIPIF1<0為最小正周期的函數(shù)有（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【解析】SKIPIF1<0，其最小正周期為SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0，所以SKIPIF1<0的最小正周期為SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0，所以SKIPIF1<0的最小正周期為SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0故選：BD【例3-2】（2020·河南）已知函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0圖象的對(duì)稱中心完全相同，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】C【解析】由已知SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的對(duì)稱中心為SKIPIF1<0，又SKIPIF1<0的對(duì)稱中心為SKIPIF1<0，所以SKIPIF1<0.故選：C【例3-3】（2022·四川·瀘縣五中二模（文））將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后得到SKIPIF1<0的圖象，則SKIPIF1<0有（

）A．為奇函數(shù)，在SKIPIF1<0上單調(diào)遞減B．為偶函數(shù)，在SKIPIF1<0上單調(diào)遞增C．周期為π，圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱D．最大值為1，圖象關(guān)于直線SKIPIF1<0對(duì)稱【答案】D【解析】將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后，得到函數(shù)SKIPIF1<0的圖象.SKIPIF1<0為偶函數(shù)，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，故A不正確；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0不單調(diào)，故B錯(cuò)誤；SKIPIF1<0的周期為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故C錯(cuò)誤；g（x）最大值為1，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，為最小值，故SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故D正確，故選：D．【例3-4】（2022·山東青島·一模）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【解析】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：A【一隅三反】1．（2022·全國(guó)·高三專題練習(xí)）（多選）下列函數(shù)中，圖象為軸對(duì)稱圖形的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】AC【解析】A.因?yàn)镾KIPIF1<0，所以SKIPIF1<0是偶函數(shù)，函數(shù)圖象關(guān)于y軸對(duì)稱，故正確；B.因?yàn)镾KIPIF1<0的對(duì)稱軸方程為：SKIPIF1<0，SKIPIF1<0的對(duì)稱軸方程為：SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0圖象不是軸對(duì)稱圖形，故錯(cuò)誤；C.將SKIPIF1<0向左平移SKIPIF1<0個(gè)單位可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0是軸對(duì)稱圖形，故正確；D.因?yàn)镾KIPIF1<0的對(duì)稱軸方程為：SKIPIF1<0，SKIPIF1<0的對(duì)稱軸方程為：SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0圖象不是軸對(duì)稱圖形，故錯(cuò)誤；故選：AC2．（2022·北京西城·一模）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位所得函數(shù)圖象關(guān)于原點(diǎn)對(duì)稱，向左平移SKIPIF1<0個(gè)單位所得函數(shù)圖象關(guān)于SKIPIF1<0軸對(duì)稱，其中SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，可得SKIPIF1<0，又由函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，可得SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0又因?yàn)镾KIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，可得SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，可得SKIPIF1<0.故選：D.3．（2022·北京·一模）已知函數(shù)SKIPIF1<0.從下列四個(gè)條件中選擇兩個(gè)作為已知，使函數(shù)SKIPIF1<0存在且唯一確定．(1)求SKIPIF1<0的解析式；(2)設(shè)SKIPIF1<0，求函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)遞增區(qū)間.條件①：SKIPIF1<0；條件②：SKIPIF1<0為偶函數(shù)；條件③：SKIPIF1<0的最大值為1；條件④：SKIPIF1<0圖象的相鄰兩條對(duì)稱軸之間的距離為SKIPIF1<0．注：如果選擇的條件不符合要求，第（1）問(wèn)得0分；如果選擇多個(gè)符合要求的條件分別解答，按第一個(gè)解答計(jì)分.【答案】(1)選擇①④或③④均可得到SKIPIF1<0(2)SKIPIF1<0和SKIPIF1<0【解析】(1)因?yàn)镾KIPIF1<0，所以SKIPIF1<0，顯然當(dāng)SKIPIF1<0時(shí)SKIPIF1<0為奇函數(shù)，故②不能選，若選擇①③，即SKIPIF1<0最大值為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0不能唯一確定，故舍去；若選擇①④，即SKIPIF1<0圖象的相鄰兩條對(duì)稱軸之間的距離為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0；若選擇③④，即SKIPIF1<0圖象的相鄰兩條對(duì)稱軸之間的距離為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0的最大值為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0；(2)由（1）可得SKIPIF1<0SKIPIF1<0令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以函數(shù)的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上的單調(diào)遞增區(qū)間有SKIPIF1<0和SKIPIF1<0；4．（2022·浙江浙江·二模）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間；(2)求函數(shù)SKIPIF1<0的值域．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】(1)由SKIPIF1<0由SKIPIF1<0得SKIPIF1<0所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為：SKIPIF1<0(2)由SKIPIF1<0，則SKIPIF1<0所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0由SKIPIF1<0，則SKIPIF1<0所以函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0考點(diǎn)四三角函數(shù)性質(zhì)與其他知識(shí)的綜合運(yùn)用【例4-1】（2022·江蘇蘇州）若函數(shù)SKIPIF1<0在區(qū)間[0，π)內(nèi)有且只有兩個(gè)極值點(diǎn)，則正數(shù)ω的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因?yàn)镾KIPIF1<0在SKIPIF1<0有2個(gè)極值點(diǎn)，也即SKIPIF1<0在區(qū)間SKIPIF1<0取得一次最大值，一次最小值；又SKIPIF1<0，則當(dāng)SKIPIF1<0，SKIPIF1<0，要使得SKIPIF1<0滿足題意，只需SKIPIF1<0，解得SKIPIF1<0.故選：C.【一隅三反】1．（2022·江蘇南通·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)極值，則SKIPIF1<0的取值范圍是（

）A．（0，5] B．（0，5）C．（0，SKIPIF1<0） D．（0，SKIPIF1<0]【答案】A【解析】由已知條件得SKIPIF1<0，∵函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)極值，∴函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，∴SKIPIF1<0或SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，在此范圍內(nèi)SKIPIF1<0不成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0，故選：SKIPIF1<0.2．（2022·陜西·西安中學(xué)模擬預(yù)測(cè)（文））已知函數(shù)SKIPIF1<0的部分圖像如圖所示，現(xiàn)將SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0的圖像，則方程SKIPIF1<0在SKIPIF1<0上實(shí)數(shù)解的個(gè)數(shù)為（

）A．5 B．6 C．7 D．8【答案】B【解析】根據(jù)函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的部分圖象，可得SKIPIF1<0，SKIPIF1<0．所以SKIPIF1<0，結(jié)合五點(diǎn)法作圖，SKIPIF1<0