新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練3.4.2 三角函數(shù)的性質(zhì)（2）（基礎(chǔ)版）（解析版）

3.4.2三角函數(shù)的性質(zhì)（2）（精講）（基礎(chǔ)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)一解析式【例1-1】（2022·山東·煙臺(tái)二中）若函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0和SKIPIF1<0的值是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】C【解析】由圖象可知SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0.故選：C【例1-2】（2022·全國(guó)·高三專題練習(xí)）如圖所示，某地一天6~14時(shí)的溫度變化曲線近似滿足函數(shù)SKIPIF1<0，則這段曲線的函數(shù)解析式可以為（）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】A【解析】由于SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，又過(guò)點(diǎn)SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，取SKIPIF1<0，得SKIPIF1<0，符合題意選：A．【例1-3】（2021·貴州·高三階段練習(xí)）函數(shù)f（x）=SKIPIF1<0sin（ωx+φ）+cos（ωx+φ）（ω>0，|φ|<SKIPIF1<0）的部分圖象如圖所示，則φ=（

）A．SKIPIF1<0 B．-SKIPIF1<0 C．-SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：A【一隅三反】1．（2022·甘肅武威）函數(shù)SKIPIF1<0（A，ω，φ為常數(shù)，A>0，ω>0，SKIPIF1<0）的部分圖象如圖所示，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由圖可知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，將SKIPIF1<0代入得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故選：B.2．（2021·陜西省洛南中學(xué)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0的解析式是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖象可得SKIPIF1<0，解得A=2，k=1，由正弦型圖象性質(zhì)可得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A3（2022·廣東·佛山市順德區(qū)容山中學(xué)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則函數(shù)SKIPIF1<0的解析式可能為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)SKIPIF1<0，由圖可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：A.4．（2022·四川南充·二模）函數(shù)SKIPIF1<0的部分圖像如圖所示，SKIPIF1<0，則（

）A．SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱B．SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減D．SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增【答案】C【解析】由圖可知SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，故A錯(cuò)誤；SKIPIF1<0，所以函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，故B錯(cuò)誤；對(duì)于C：由SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故C正確；對(duì)于D：由SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上不單調(diào)，所以SKIPIF1<0在SKIPIF1<0上不單調(diào)，故D錯(cuò)誤；故選：C考點(diǎn)二定義域【例2】（2022·陜西·西安市臨潼區(qū)鐵路中學(xué)）求下列函數(shù)的定義域.(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0.【解析】(1)要使得函數(shù)有意義，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)定義域?yàn)镾KIPIF1<0.(2)要使得函數(shù)有意義，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)定義域?yàn)镾KIPIF1<0.(3)要使得函數(shù)有意義，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)定義域?yàn)镾KIPIF1<0.(1)整式函數(shù)的定義域?yàn)?1)整式函數(shù)的定義域?yàn)镽；(2)分式的分母不為零；(3)偶次根式的被開(kāi)方數(shù)不小于零；(4)對(duì)數(shù)函數(shù)的真數(shù)必須大于零；(5)正切函數(shù)y＝tanx的定義域?yàn)镾KIPIF1<0；(6)x0中x≠0；(7)實(shí)際問(wèn)題中除要考慮函數(shù)解析式有意義外，還應(yīng)考慮實(shí)際問(wèn)題本身的要求溫馨提示【一隅三反】1．（2022·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0的定義域?yàn)椋ǎ〢．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意，得SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．故選：B．2．（2022·江蘇）函數(shù)SKIPIF1<0的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題知，SKIPIF1<0由SKIPIF1<0，解得SKIPIF1<0由SKIPIF1<0解得，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，解得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，區(qū)間SKIPIF1<0和SKIPIF1<0無(wú)交集；當(dāng)SKIPIF1<0時(shí)，區(qū)間SKIPIF1<0和SKIPIF1<0無(wú)交集；所以函數(shù)的定義域SKIPIF1<0.故選：A.3．（2022·四川綿陽(yáng)）函數(shù)SKIPIF1<0的定義域?yàn)锳．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由函數(shù)SKIPIF1<0，則滿足SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0即函數(shù)的定義域?yàn)镾KIPIF1<0，故選C.4．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0(SKIPIF1<0)的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意，得SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0.故選：A.考點(diǎn)三值域【例3-1】（2022·吉林）已知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值的和是___________.【答案】1或SKIPIF1<0【解析】由題設(shè)，SKIPIF1<0，則SKIPIF1<0，在SKIPIF1<0上，當(dāng)SKIPIF1<0則SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0則SKIPIF1<0，故SKIPIF1<0；綜上，最大值與最小值的和為1或SKIPIF1<0.故答案為：1或SKIPIF1<0【例3-2】（2021·全國(guó)·課時(shí)練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最大值和最小值分別為_(kāi)_____．【答案】SKIPIF1<0，6【解析】因SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，于是得SKIPIF1<0，而SKIPIF1<0，因此當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的最大值和最小值分別為SKIPIF1<0，6.故答案為：SKIPIF1<0，6【例3-3】（2021·寧夏·吳忠中學(xué)高三階段練習(xí)（理））當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，則實(shí)數(shù)m的取值范圍為_(kāi)___．【答案】SKIPIF1<0【解析】設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．由題意知m<f(x)<m+2在SKIPIF1<0上恒成立，即SKIPIF1<0∴SKIPIF1<0∴實(shí)數(shù)m的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0【一隅三反】1（2021·天津·高三期中）SKIPIF1<0在區(qū)間SKIPIF1<0的值域是_________.【答案】SKIPIF1<0【解析】SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0，SKIPIF1<0．2．（2022·北京二中）函數(shù)SKIPIF1<0的值域?yàn)開(kāi)_____.【答案】SKIPIF1<0【解析】依題意，原函數(shù)定義域?yàn)镽，SKIPIF1<0，而SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以所求值域是SKIPIF1<0.故答案為：SKIPIF1<03．（2021·全國(guó)·專題練習(xí)）已知函數(shù)SKIPIF1<0．若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有解，則實(shí)數(shù)SKIPIF1<0的取值范圍是________．【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0的值域?yàn)镾KIPIF1<0，關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有解，則關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有解，所以SKIPIF1<0，所以SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0故答案為：SKIPIF1<04．（2022·四川·高三學(xué)業(yè)考試）已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的最小正周期；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的最值.【答案】(1)SKIPIF1<0(2)最大值為SKIPIF1<0，最小值為SKIPIF1<0【解析】(1)∵SKIPIF1<0，∴SKIPIF1<0，即函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0．(2)在區(qū)間SKIPIF1<0上，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0的最大值為SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0．考點(diǎn)四伸縮平移【例4-1】（2022·重慶市育才中學(xué)高三階段練習(xí)）為了得到SKIPIF1<0的圖象，可將函數(shù)SKIPIF1<0的圖象（

）A．向左平移SKIPIF1<0個(gè)單位 B．向右平移SKIPIF1<0個(gè)單位C．向左平移SKIPIF1<0個(gè)單位 D．向右平移SKIPIF1<0個(gè)單位【答案】C【解析】依題意，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0可由SKIPIF1<0向左平移SKIPIF1<0個(gè)單位得到.故選：C【例4-2】（2022·河南省杞縣高中模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0的圖象為C，為了得到函數(shù)SKIPIF1<0的圖象，只要把C上所有點(diǎn)（

）A．橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍，縱坐標(biāo)不變B．橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變C．縱坐標(biāo)伸長(zhǎng)到原來(lái)的2倍，橫坐標(biāo)不變D．縱坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，橫坐標(biāo)不變【答案】B【解析】根據(jù)三角函數(shù)的圖象變換，將SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)縮短為原來(lái)SKIPIF1<0倍，即可得到函數(shù)SKIPIF1<0.故選：B．【例4-3】（2022·陜西·二模）要得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0的圖象（

）A．向左平移是SKIPIF1<0個(gè)單位長(zhǎng)度 B．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向右平移登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】B【解析】因?yàn)楹瘮?shù)SKIPIF1<0，SKIPIF1<0，所以要得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度.故選：B.【例4-4】（2022·山西·懷仁市第一中學(xué)校二模（理））將函數(shù)SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的一半、縱坐標(biāo)不變，然后向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】將函數(shù)SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的一半、縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0，再將函數(shù)SKIPIF1<0向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0，所以SKIPIF1<0.故選：B.【例4-5】（2022·四川達(dá)州·二模（理））將函數(shù)SKIPIF1<0圖象上所有點(diǎn)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是奇函數(shù)，則a的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，則SKIPIF1<0圖象上所有點(diǎn)向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得SKIPIF1<0，因?yàn)镾KIPIF1<0是奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0，故選：D【一隅三反】1．（2022·四川師范大學(xué)附屬中學(xué)二模（文））函數(shù)SKIPIF1<0其中SKIPIF1<0，SKIPIF1<0的圖象如圖所示，為了得到SKIPIF1<0的圖象只要將SKIPIF1<0的圖象（

）A．向右平移SKIPIF1<0個(gè)單位 B．向右平移SKIPIF1<0個(gè)單位C．向左平移SKIPIF1<0個(gè)單位 D．向左平移SKIPIF1<0個(gè)單位【答案】A【解析】由圖可知，SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0，將SKIPIF1<0代入可得，SKIPIF1<0，因此SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，為了得到SKIPIF1<0，所以將函數(shù)SKIPIF1<0向右平移SKIPIF1<0個(gè)單位即可.故選：A.2．（2022·內(nèi)蒙古包頭·一模）把函數(shù)SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)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SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意可知，將函數(shù)SKIPIF1<0的圖象先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，再將所得圖象上所有點(diǎn)的橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0，縱坐標(biāo)不變，可得到函數(shù)SKIPIF1<0的圖象.故選：C.3．（2022·江西·南昌十中高三階段練習(xí)）將函數(shù)SKIPIF1<0的圖象沿SKIPIF1<0軸向左平移SKIPIF1<0個(gè)單位后，得到關(guān)于SKIPIF1<0軸對(duì)稱的圖象，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函數(shù)SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象沿SKIPIF1<0軸向左平移SKIPIF1<0個(gè)單位后，得到函數(shù)SKIPIF1<0，因?yàn)楹瘮?shù)是偶函數(shù)，SKIPIF1<0SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．則SKIPIF1<0的最小值為SKIPIF1<0故選：A4．（2022·陜西·模擬預(yù)測(cè)）把函數(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SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則實(shí)數(shù)a的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題設(shè)，SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0上SKIPIF1<0遞減，即SKIPIF1<0上SKIPIF1<0遞減，由SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0，故a的最大值為SKIPIF1<0故選：A3.4.2三角函數(shù)的性質(zhì)（2）（精練）（基礎(chǔ)版）題組一題組一解析式1．（2022·湖北省廣水市實(shí)驗(yàn)高級(jí)中學(xué)）若函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的部分圖象如圖所示，則SKIPIF1<0的值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖知，A=2，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，又圖象過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0Z，所以SKIPIF1<0Z，因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故選：A2（2022·安徽省宣城中學(xué)高三開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，其中SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意得，SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.∵SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0.故選：C.3．（2022·河南·南陽(yáng)中學(xué)）函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖象可知：SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，不妨設(shè)：SKIPIF1<0，將SKIPIF1<0代入得：SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故A正確，其他選項(xiàng)均不合要求.故選：A4．（2022·福建福州·高三期末）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】D【解析】由圖象可知，函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，因此，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，故選：D.5．（2022·山西呂梁·一模（文））設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0的圖象大致如圖所示，則SKIPIF1<0的最小正周期為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖知SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，此時(shí)SKIPIF1<0，所以SKIPIF1<0，故選：A6．（2022·山西太原）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則函數(shù)的表達(dá)式是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】結(jié)合圖像和選項(xiàng)可知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0.SKIPIF1<0.故選：A.7．（2022·四川宜賓）函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】根據(jù)圖象可得：SKIPIF1<0，所以可得SKIPIF1<0的周期為SKIPIF1<0，根據(jù)SKIPIF1<0，則有：SKIPIF1<0，又SKIPIF1<0解得：SKIPIF1<0，根據(jù)SKIPIF1<0，SKIPIF1<0可得：SKIPIF1<0故選：A8．（2022·四川內(nèi)江）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】D【解析】由圖可知，SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，再由SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以函數(shù)解析式為SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0，所以函數(shù)的單調(diào)遞減區(qū)間為SKIPIF1<0.故選：D9．（2022·廣東深圳）如圖是函數(shù)SKIPIF1<0的部分圖象，則下列說(shuō)法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖像得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0得SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0.故選：A.10．（2022·江蘇·徐州市第七中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖象可知，SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A.11．（2022·四川瀘州）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】A【解析】根據(jù)函數(shù)的圖象，A=2，SKIPIF1<0，所以SKIPIF1<0，根據(jù)函數(shù)在SKIPIF1<0處取得最大值可知，SKIPIF1<0.故選：A.12．（2022·北京東城）某同學(xué)用“五點(diǎn)法”畫函數(shù)SKIPIF1<0在一個(gè)周期內(nèi)的簡(jiǎn)圖時(shí)，列表如下：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0y020SKIPIF1<00則SKIPIF1<0的解析式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由表中數(shù)據(jù)知：SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，可得SKIPIF1<0.∴SKIPIF1<0.故選：D.13．（2022·河南鄭州·高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則下列說(shuō)法錯(cuò)誤的是（

）A．函數(shù)SKIPIF1<0B．函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0中心對(duì)稱C．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)單位得到D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減【答案】D【解析】將點(diǎn)SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0，又SKIPIF1<0為對(duì)稱軸，所以SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，此時(shí)SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0，函數(shù)SKIPIF1<0，A說(shuō)法正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0中心對(duì)稱，B說(shuō)法正確；函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到SKIPIF1<0，C說(shuō)法正確；SKIPIF1<0時(shí)，2x+π3∈2π3,5π3，SKIPIF1<0在2π314．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如下圖所示，若SKIPIF1<0，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，則函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】依題意SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，將點(diǎn)SKIPIF1<0代入可得SKIPIF1<0，因?yàn)镾KIPIF1<0，解得SKIPIF1<0；故SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0.故選：C題組二題組二定義域1．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由函數(shù)式知：SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0.故選：B.2．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】函數(shù)SKIPIF1<0有意義，則SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)的定義域?yàn)镾KIPIF1<0.故選：A3．（2022·湖南·長(zhǎng)沙市明德中學(xué)）函數(shù)SKIPIF1<0的定義域?yàn)锳．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題,SKIPIF1<0,故SKIPIF1<0即SKIPIF1<0解得SKIPIF1<0.即定義域?yàn)镾KIPIF1<0.故選：A4．（2022·全國(guó)·課時(shí)練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0.故選：C.5．（2021·北京市朝陽(yáng)區(qū)人大附中朝陽(yáng)分校）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)_________.【答案】SKIPIF1<0【解析】由題意SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0．6．（2022·甘肅張掖）函數(shù)SKIPIF1<0定義域?yàn)開(kāi)___.【答案】SKIPIF1<0∪SKIPIF1<0【解析】由題意得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，從而函數(shù)的定義域?yàn)镾KIPIF1<0∪SKIPIF1<0.故答案為：SKIPIF1<0∪SKIPIF1<0.7．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)_______________．【答案】SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，對(duì)于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，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴不等式組的解為：SKIPIF1<0或SKIPIF1<0SKIPIF1<0的定義域?yàn)镾KIPIF1<0故答案為：SKIPIF1<08．（2021·陜西·長(zhǎng)安一中高三階段練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)__________.【答案】SKIPIF1<0【解析】由已知可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0.因此，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.9．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的定義域是____________．【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)的定義域?yàn)镾KIPIF1<0故答案為：SKIPIF1<010．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的定義域是________.【答案】SKIPIF1<0【解析】由已知，得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0.因此，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0．11．（2022·甘肅）設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0的定義域?yàn)開(kāi)_________．【答案】SKIPIF1<0或SKIPIF1<0【解析】由題意可知：SKIPIF1<0，故答案為：SKIPIF1<0或SKIPIF1<0.題組三題組三值域1．（2022·陜西咸陽(yáng)·二模（理））函數(shù)SKIPIF1<0的最小值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0.故選：D2．（2022·寧夏吳忠·模擬預(yù)測(cè)（文））函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值是（

）A．1 B．2 C．SKIPIF1<0 D．3【答案】C【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0．故選：C．3．（2022·北京·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，則（

）A．最大值為2，最小值為1B．最大值為SKIPIF1<0，最小值為1C．最大值為SKIPIF1<0，最小值為1D．最大值為SKIPIF1<0，最小值為SKIPIF1<0【答案】B【解析】SKIPIF1<0，SKIPIF1<0時(shí)，sinx∈[SKIPIF1<0，1]，∴當(dāng)sinx＝SKIPIF1<0時(shí)，f(x)最大值為SKIPIF1<0；當(dāng)sinx＝1時(shí)，f(x)最小值為1.故選：B.4．（2022·安徽滁州·二模（理））已知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0.故選：C5．（2022·山西·懷仁市第一中學(xué)校二模）若將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，則SKIPIF1<0在SKIPIF1<0上的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．2【答案】C【解析】因?yàn)镾KIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．故選：C6．（2022·全國(guó)·高三開(kāi)學(xué)考試）函數(shù)SKIPIF1<0的最大值為(

)A．2 B．3 C．4 D．5【答案】D【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0∴f(x)最大值為5，故選：D.7．（2022·安徽·合肥一中高三階段練習(xí)）將函數(shù)f(x)=sinx的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)g(x)的圖象.則函數(shù)y=f(x)·g(x)的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由已知可得SKIPI