高考數(shù)學(xué)三輪沖刺壓軸小題06 與三角函數(shù)相關(guān)的最值問(wèn)題 (解析版)

31/31與三角函數(shù)相關(guān)的最值問(wèn)題與三角函數(shù)相關(guān)的最值問(wèn)題一．方法綜述三角函數(shù)相關(guān)的最值問(wèn)題歷來(lái)是高考的熱點(diǎn)之一，利用三角函數(shù)的性質(zhì)求參數(shù)取值或范圍是往往是解決問(wèn)題的關(guān)鍵，這類(lèi)問(wèn)題一般涉及到值域、單調(diào)性及周期性等性質(zhì)，熟悉三角函數(shù)的圖象和性質(zhì)和掌握轉(zhuǎn)化思想是解題關(guān)鍵．二．解題策略類(lèi)型一與三角函數(shù)的單調(diào)性、奇偶性和對(duì)稱(chēng)性相關(guān)的最值問(wèn)題【例1】（2020·湖北高考模擬（理））已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．1 C．2 D．4【答案】C【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，正弦函數(shù)在SKIPIF1<0上遞增，所以可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0的最大值為2，故選C．2．（2020·山東高考模擬）若函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，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<0SKIPIF1<0而SKIPIF1<0值域?yàn)镾KIPIF1<0，發(fā)現(xiàn)SKIPIF1<0SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0最小值為SKIPIF1<0，選A項(xiàng).3．（2020·河南南陽(yáng)中學(xué)高考模擬）設(shè)>0,函數(shù)y=sin(x+)+2的圖象向右平移個(gè)單位后與原圖象重合，則的最小值是A． B． C． D．3【答案】C【解析】函數(shù)的圖象向右平移個(gè)單位后所以有，故選C【舉一反三】1．已知函數(shù)SKIPIF1<0SKIPIF1<0在區(qū)間SKIPIF1<0上恰有1個(gè)最大值點(diǎn)和1個(gè)最小值點(diǎn)，則ω的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【來(lái)源】安徽省示范高中皖北協(xié)作區(qū)2021屆高三下學(xué)期第23屆聯(lián)考數(shù)學(xué)（文）試題【答案】B【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0上恰有1個(gè)最大值點(diǎn)和1個(gè)最小值點(diǎn)，SKIPIF1<0，解得SKIPIF1<0.故選：B.2.（2020·河南高考模擬）已知函數(shù)SKIPIF1<0，SKIPIF1<0的部分圖象如圖所示，則使SKIPIF1<0成立的SKIPIF1<0的最小正值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】結(jié)合圖象可知，A＝2，f（x）＝2sin（ωx+φ），∵f（0）＝2sinφ＝1，∴sinφSKIPIF1<0，∵|φ|SKIPIF1<0，∴φSKIPIF1<0，f（x）＝2sin（ωxSKIPIF1<0），結(jié)合圖象及五點(diǎn)作圖法可知，ωSKIPIF1<02π，∴ω＝2，f（x）＝2sin（2xSKIPIF1<0），其對(duì)稱(chēng)軸xSKIPIF1<0，k∈Z，∵f（a+x）﹣f（a﹣x）＝0成立，∴f（a+x）＝f（a﹣x）即f（x）的圖象關(guān)于x＝a對(duì)稱(chēng)，結(jié)合函數(shù)的性質(zhì)，滿(mǎn)足條件的最小值aSKIPIF1<03．已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實(shí)數(shù)SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，得SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0是單調(diào)遞減函數(shù)，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0.故選：D.類(lèi)型二轉(zhuǎn)化為SKIPIF1<0型的最值問(wèn)題【例2】（2020·北京人大附中高考模擬）已知函數(shù)SKIPIF1<0的一條對(duì)稱(chēng)軸為SKIPIF1<0，SKIPIF1<0，且函數(shù)SKIPIF1<0在SKIPIF1<0上具有單調(diào)性，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】由題，SKIPIF1<0=SKIPIF1<0，SKIPIF1<0為輔助角，因?yàn)閷?duì)稱(chēng)軸為SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0又因?yàn)镾KIPIF1<0在SKIPIF1<0上具有單調(diào)性，且SKIPIF1<0，所以SKIPIF1<0兩點(diǎn)必須關(guān)于正弦函數(shù)的對(duì)稱(chēng)中心對(duì)稱(chēng)，即SKIPIF1<0所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最小為SKIPIF1<0，故選A【點(diǎn)睛】本題考查了三角函數(shù)綜合知識(shí)，包含圖像與性質(zhì)，輔助角公式化簡(jiǎn)等，熟悉性質(zhì)圖像是解題的關(guān)鍵.【舉一反三】1、（2020·江西高考模擬）已知SKIPIF1<0的最大值為SKIPIF1<0，若存在實(shí)數(shù)SKIPIF1<0、SKIPIF1<0，使得對(duì)任意實(shí)數(shù)SKIPIF1<0總有SKIPIF1<0成立，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】依題意SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，故選C．2．（2020·河北高考模擬）若將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，所得圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，則SKIPIF1<0的最小值是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】∵SKIPIF1<0SKIPIF1<0SKIPIF1<0，函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位可得SKIPIF1<0，所得圖象關(guān)于y軸對(duì)稱(chēng)，根據(jù)三角函數(shù)的對(duì)稱(chēng)性，可得此函數(shù)在y軸處取得函數(shù)的最值，即SKIPIF1<0，解得SKIPIF1<0=SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，令SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0.故選D．3．已知銳角三角形SKIPIF1<0的內(nèi)角SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的對(duì)邊分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.且SKIPIF1<0，則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】依題意SKIPIF1<0，由正弦定理得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由于三角形SKIPIF1<0是銳角三角形，所以SKIPIF1<0.由SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：C4.（2020·河北高考模擬）將函數(shù)的圖像向左平移個(gè)單位長(zhǎng)度后，得到的圖像，若函數(shù)在上單調(diào)遞減，則正數(shù)的最大值為A． B．1 C． D．【答案】A【解析】依題意，，向左平移個(gè)單位長(zhǎng)度得到.故，下面求函數(shù)的減區(qū)間：由，由于故上式可化為，由于函數(shù)在上單調(diào)遞減，故，解得，所以當(dāng)時(shí)，為正數(shù)的最大值.故選A.類(lèi)型三轉(zhuǎn)化為二次函數(shù)型的最值問(wèn)題【例3】函數(shù)SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．3【來(lái)源】安徽省淮北市2020-2021學(xué)年高三上學(xué)期第一次模擬考試?yán)砜茢?shù)學(xué)試題【答案】B【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0則SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)SKIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值，此時(shí)SKIPIF1<0，所以SKIPIF1<0故選：B【舉一反三】1．（2020·湖南高考模擬）已知SKIPIF1<0，則SKIPIF1<0的值域?yàn)椋ǎ〢．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0.故選：B2．（2020·江西高考模擬（理））函數(shù)SKIPIF1<0的值域?yàn)開(kāi)________．【答案】SKIPIF1<0【解析】由題意，可得SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0為增函數(shù)，在SKIPIF1<0為減函數(shù)，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故函數(shù)的值域?yàn)椋篠KIPIF1<0．3、函數(shù)SKIPIF1<0，關(guān)于SKIPIF1<0的為等式SKIPIF1<0對(duì)所有SKIPIF1<0都成立，則實(shí)數(shù)SKIPIF1<0的范圍為_(kāi)_________．【答案】SKIPIF1<0令SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0∴SKIPIF1<0（舍），當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0∴SKIPIF1<0，當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0，綜上所述，SKIPIF1<0，故答案為SKIPIF1<04、求函數(shù)SKIPIF1<0的值域.【解析】SKIPIF1<0[令sinx+cosx=t,則SKIPIF1<0,其中SKIPIF1<0所以SKIPIF1<0,故值域?yàn)镾KIPIF1<0.類(lèi)型四轉(zhuǎn)化為三角函數(shù)函數(shù)型的最值問(wèn)題【例4】（2020·黑龍江高考模擬）已知SKIPIF1<0，在這兩個(gè)實(shí)數(shù)SKIPIF1<0之間插入三個(gè)實(shí)數(shù)，使這五個(gè)數(shù)構(gòu)成等差數(shù)列，那么這個(gè)等差數(shù)列后三項(xiàng)和的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)中間三項(xiàng)為SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以后三項(xiàng)的和為SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以可令SKIPIF1<0，所以SKIPIF1<0，故選SKIPIF1<0【舉一反三】設(shè)點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，則SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【來(lái)源】浙江省寧波市2020-2021學(xué)年高三上學(xué)期期末數(shù)學(xué)試題【答案】D【解析】設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0則有SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取最小值，即SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小值是SKIPIF1<0，故選：D.三．強(qiáng)化訓(xùn)練1．（2020·四川高考模擬）若函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】A【解析】函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后，圖象所對(duì)應(yīng)解析式為：SKIPIF1<0，由SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，則SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，所以SKIPIF1<0，SKIPIF1<0，故選A．2．（2020·陜西高考模擬）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，在向上平移一個(gè)單位，得到g（x）的圖象．若g（x1）g（x2）＝4，且x1，x2∈[﹣2π，2π]，則x1﹣2x2的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，再向上平移一個(gè)單位，得到g（x）＝sin（2x﹣SKIPIF1<0+SKIPIF1<0）+1＝﹣cos2x+1的圖象，故g（x）的最大值為2，最小值為0，若g（SKIPIF1<0）g（SKIPIF1<0）＝4，則g（SKIPIF1<0）＝g（SKIPIF1<0）＝2，或g（SKIPIF1<0）＝g（SKIPIF1<0）＝﹣2（舍去）．故有g(shù)（SKIPIF1<0）＝g（SKIPIF1<0）＝2，即cos2SKIPIF1<0＝cos2SKIPIF1<0＝﹣1，又SKIPIF1<0，x2∈[﹣2π，2π]，∴2SKIPIF1<0，2SKIPIF1<0∈[﹣4π，4π]，要使SKIPIF1<0﹣2SKIPIF1<0取得最大值，則應(yīng)有2SKIPIF1<0＝3π，2SKIPIF1<0＝﹣3π，故SKIPIF1<0﹣2SKIPIF1<0取得最大值為SKIPIF1<0+3π＝SKIPIF1<0．故選A．3．（2020·甘肅高考模擬）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)長(zhǎng)度單位后，所得到的圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，則SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】【詳解】由題意得，SKIPIF1<0，令SKIPIF1<0，可得函數(shù)的圖象對(duì)稱(chēng)軸方程為SKIPIF1<0，取SKIPIF1<0是SKIPIF1<0軸右側(cè)且距離SKIPIF1<0軸最近的對(duì)稱(chēng)軸，因?yàn)閷⒑瘮?shù)的圖象向左平移SKIPIF1<0個(gè)長(zhǎng)度單位后得到的圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，SKIPIF1<0的最小值為SKIPIF1<0，故選B．4．（2020·山東高考模擬）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0SKIPIF1<0個(gè)單位長(zhǎng)度,再將圖象上各點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍(縱坐標(biāo)不變),得到函數(shù)SKIPIF1<0的圖象,若對(duì)任意的SKIPIF1<0均有SKIPIF1<0成立,則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)楹瘮?shù)SKIPIF1<0的圖象向左平移SKIPIF1<0SKIPIF1<0個(gè)單位長(zhǎng)度,所以得到函數(shù)SKIPIF1<0，再將圖象上各點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍(縱坐標(biāo)不變),得到函數(shù)SKIPIF1<0的圖象,所以SKIPIF1<0，對(duì)任意的SKIPIF1<0均有SKIPIF1<0成立，所以SKIPIF1<0在SKIPIF1<0時(shí)，取得最大值，所以有SKIPIF1<0而SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.5．（2020·云南高考模擬）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，所得圖象對(duì)應(yīng)的函數(shù)在區(qū)間SKIPIF1<0上無(wú)極值點(diǎn)，則SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，可得函數(shù)SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0即函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，令SKIPIF1<0，可得函數(shù)的單調(diào)遞增區(qū)間為SKIPIF1<0，又由函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)極值點(diǎn)，則SKIPIF1<0的最大值為SKIPIF1<0，故選A.6．（2020·四川華鎣一中高考模擬）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的圖像，若SKIPIF1<0在SKIPIF1<0上為增函數(shù)，則SKIPIF1<0的最大值為（）A．1 B．2 C．3 D．4【答案】B【解析】由三角函數(shù)的性質(zhì)可得：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，其圖象向左平移SKIPIF1<0個(gè)單位所得函數(shù)的解析式為：SKIPIF1<0，函數(shù)的單調(diào)遞增區(qū)間滿(mǎn)足：SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0可得函數(shù)的一個(gè)單調(diào)遞增區(qū)間為：SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上為增函數(shù)，則：SKIPIF1<0，據(jù)此可得：SKIPIF1<0，則SKIPIF1<0的最大值為2.本題選擇B選項(xiàng).7．（2020·天津高考模擬）已知SKIPIF1<0同時(shí)滿(mǎn)足下列三個(gè)條件：①SKIPIF1<0；②SKIPIF1<0是奇函數(shù)；③SKIPIF1<0.若SKIPIF1<0在SKIPIF1<0上沒(méi)有最小值，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】D【解析】由SKIPIF1<0，可得SKIPIF1<0因?yàn)镾KIPIF1<0是奇函數(shù)，所以SKIPIF1<0是奇函數(shù)，即SKIPIF1<0又因?yàn)镾KIPIF1<0，即SKIPIF1<0所以SKIPIF1<0是奇數(shù)，取k=1，此時(shí)SKIPIF1<0所以函數(shù)SKIPIF1<0因?yàn)镾KIPIF1<0在SKIPIF1<0上沒(méi)有最小值，此時(shí)SKIPIF1<0所以此時(shí)SKIPIF1<0，解得SKIPIF1<0.故選D.8．已知函數(shù)SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，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【來(lái)源】江西省上饒市（天佑中學(xué)、余干中學(xué)等）六校2021屆高三下學(xué)期第一次聯(lián)考數(shù)學(xué)（理）試題【答案】C【解析】SKIPIF1<0且SKIPIF1<0，由題意可知，對(duì)任意的SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，此時(shí)SKIPIF1<0.綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C.9．（2020·廣東高考模擬）已知函數(shù)SKIPIF1<0.若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)沒(méi)有零點(diǎn)，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】SKIPIF1<0SKIPIF1<0，SKIPIF1<0,函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)沒(méi)有零點(diǎn)(1)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，取SKIPIF1<0，SKIPIF1<0SKIPIF1<0；（2）SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，取SKIPIF1<0，SKIPIF1<0；綜上可知：SKIPIF1<0的取值范圍是SKIPIF1<0，選SKIPIF1<0.10．在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0邊上的高為1，則SKIPIF1<0面積的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)BC邊上的高為AD，則AD=1，SKIPIF1<0，如圖所示：所以SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0，設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0=SKIPIF1<0=SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0面積的最小值為SKIPIF1<0.故選：B11．已知實(shí)數(shù)SKIPIF1<0，不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則SKIPIF1<0的最大值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【解析】令SKIPIF1<0，原不等式整理得：SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，兩邊除以SKIPIF1<0得：SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0，故SKIPIF1<0，故SKIPIF1<0為增函數(shù).又SKIPIF1<0，因此SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增，又SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，故SKIPIF1<0.則SKIPIF1<0.故選：B.12．設(shè)函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，其圖象相鄰兩條對(duì)稱(chēng)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(chēng)，則下列判斷正確的是（）A．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增B．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)C．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0D．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【答案】D【解析】由題意可得SKIPIF1<0，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，由于函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.對(duì)于A選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，A選項(xiàng)錯(cuò)誤；對(duì)于B選項(xiàng)，SKIPIF1<0，所以，函數(shù)SKIPIF1<0的圖象不關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，B選項(xiàng)錯(cuò)誤；對(duì)于C選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，C選項(xiàng)錯(cuò)誤；對(duì)于D選項(xiàng)，SKIPIF1<0，所以，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，D選項(xiàng)正確.故選：D.13．設(shè)函數(shù)f（x）＝sin（ωx+φ），SKIPIF1<0，SKIPIF1<0，若存在實(shí)數(shù)φ，使得集合A∩B中恰好有7個(gè)元素，則ω（ω＞0）的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵f′（x0）＝0，∴f（x0）是f（x）的最大值或最小值，又f（x）＝sin（ωx+φ）的最大值或最小值在直線y＝±1上，∴y＝±1代入SKIPIF1<0得，SKIPIF1<0，解得﹣4≤x≤4，又存在實(shí)數(shù)φ，使得集合A∩B中恰好有7個(gè)元素，∴SKIPIF1<0，且ω＞0，解得SKIPIF1<0，∴ω的取值范圍是SKIPIF1<0．故選：B．14．在平面直角坐標(biāo)系SKIPIF1<0中，已知向量SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在圓SKIPIF1<0上，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，若存在正實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．4【來(lái)源】2021年全國(guó)高中名校名師原創(chuàng)預(yù)測(cè)卷理科數(shù)學(xué)全國(guó)卷Ⅰ（第三模擬）【答案】A【解析】解法一設(shè)SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0．即SKIPIF1<0由SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0，又SKIPIF1<0為正實(shí)數(shù)，解得SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0．解法二設(shè)點(diǎn)SKIPIF1<0，又點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0為正實(shí)數(shù)，解得SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0．15．水車(chē)在古代是進(jìn)行灌溉引水的工具，亦稱(chēng)“水轉(zhuǎn)筒車(chē)”，是一種以水流作動(dòng)力，取水灌田的工具.據(jù)史料記載，水車(chē)發(fā)明于隨而盛于唐，距今已有1000多年的歷史是人類(lèi)的一項(xiàng)古老的發(fā)明，也是人類(lèi)利用自然和改造自然的象征.如圖是一個(gè)半徑為SKIPIF1<0的水車(chē)，一個(gè)水斗從點(diǎn)SKIPIF1<0出發(fā)，沿圓周按逆時(shí)針?lè)较騽蛩傩D(zhuǎn)，且旋轉(zhuǎn)一周用時(shí)120秒.經(jīng)過(guò)SKIPIF1<0秒后，水斗旋轉(zhuǎn)到SKIPIF1<0點(diǎn)，設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，其縱坐標(biāo)滿(mǎn)足SKIPIF1<0，則下列敘述正確的是（）A．SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增C．當(dāng)SKIPIF1<0，SKIPIF1<0的最大值為SKIPIF1<0D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0【答案】D【解析】由題意，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0；又點(diǎn)SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0；又SKIPIF1<0，所以SKIPIF1<0．故SKIPIF1<0不正確；所以SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0先增后減，SKIPIF1<0錯(cuò)誤；SKIPIF1<0，SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離的最大值為6，SKIPIF1<0錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的縱坐標(biāo)為SKIPIF1<0，橫坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0正確．故選：SKIPIF1<0．16．已知銳角SKIPIF1<0中，角SKIPIF1<0，SKIPIF1<0，SKIPIF1<0所對(duì)的邊分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0的面積SKIPIF1<0，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】利用正弦定理可知SKIPIF1<0SKIPIF1<0，又SKIPIF1<0為銳角三角形，SKIPIF1<0由銳角SKIPIF1<0可知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，利用正弦函數(shù)性質(zhì)知SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0故選：B17．函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為（）A．-1 B．SKIPIF1<0 C．SKIPIF1<0 D．1【來(lái)源】天一大聯(lián)考2020-2021學(xué)年高三上學(xué)期高中畢業(yè)班階段性測(cè)試（三）理科數(shù)學(xué)【答案】C【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0,因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0.故選：C18．在SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】有正弦定理得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.其中SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最大值為SKIPIF1<0.故選：B19．向量SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最大值為（）A．3 B．4 C．5 D．6【答案】B【解析】由向量的坐標(biāo)運(yùn)算得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，由三角函數(shù)的性質(zhì)得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立.所以SKIPIF1<0.故選：B.20．已知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上的最大值為M，則M的最小值為_(kāi)_______.【來(lái)源】湖南省長(zhǎng)沙市長(zhǎng)郡中學(xué)2021屆高三下學(xué)期月考（七）數(shù)學(xué)試題【答案】SKIPIF1<0【解析】由于函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.不妨取SKIPIF1<0，則SKIPIF1<0.若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，則SKIPIF1<0SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上先增后減，則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0；若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上先減后增，同理可知SKIPIF1<0的最小值為SKIPIF1<0.SKIPIF1<0，綜上可知，SKIPIF1<0的最小值為SKIPIF1<0.21．已知函數(shù)SKIPIF1<0，若對(duì)于任意SKIPIF1<0，均有SKIPIF1<0，則SKIPIF1<0的最大值是___________.【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0

∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有SKIPIF1<0；SKIPIF1<0有SKIPIF1<0；由SKIPIF1<0有SKIPIF1<0，有SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有SKIPIF1<0；SKIPIF1<0有SKIPIF1<0；由SKIPIF1<0有SKIPIF1<0，有SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0：SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞減，SKIPIF1<0上遞增；SKIPIF1<0：SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增；SKIPIF1<0：SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增，SKIPIF1<0上遞增；∴綜上，SKIPIF1<0在SKIPIF1<0上先減后增，則SKIPIF1<0，可得SKIPIF1<0∴SKIPIF1<0恒成立，即SKIPIF1<0的最大值是-1.故答案為：SKIPIF1<0.22．在SKIPIF1<0中，角SKIPIF1<0的對(duì)邊分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0有最大值，則實(shí)數(shù)SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【解析】由于SKIPIF1<0，所以SKIPIF1<0，由正弦定理得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，沒(méi)有最大值，所以SKIPIF1<0，則SKIPIF1<0，其中SKIPIF1<0，要使SKIPIF1<0有最大值，則SKIPIF1<0要能取SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.所以SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<023．若函數(shù)SKIPIF1<0的定義域存在SKIPIF1<0，使SKIPIF1<0成立，則稱(chēng)該函數(shù)為“互補(bǔ)函數(shù)”.若函數(shù)SKIPIF1<0在SKIPIF1<0上為“互補(bǔ)函數(shù)”，則SKIPIF1<0的取值范圍為_(kāi)__________.【來(lái)源】重組卷01-沖刺2021年高考數(shù)學(xué)（理）之精選真題模擬重組卷（新課標(biāo)卷）【答案】SKIPIF1<0【解析】SKIPIF1<0，由“互補(bǔ)函數(shù)”的定義得：存在SKIPIF1<0，SKIPIF1<0，所以令SKIPIF1<0，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在至少兩個(gè)極大值點(diǎn)，則SKIPIF1<0，得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，顯然符合題意；當(dāng)SKIPIF1<0時(shí)，分以下兩種情況討論，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.綜上，SKIPIF1<0的取值范圍為SKIPIF1<0.24．（2020·陜西高考模擬）若向量SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最大值為.【答案】3【解析】根據(jù)題意，由于向量SKIPIF1<0,SKIPIF1<0,則可知SKIPIF1<0=SKIPIF1<0，那么化為單一函數(shù)可知SKIPIF1<0，可知最大值為3，故填寫(xiě)3.24．（2020·浙江高考模擬）定義式子運(yùn)算為將函數(shù)的圖像向左平移個(gè)單位，所得圖像對(duì)應(yīng)的函數(shù)為偶函數(shù)，則