新高考數(shù)學(xué)二輪復(fù)習(xí)講義第十三講三角函數(shù)圖象及性質(zhì)（含解析）

第十三講：三角函數(shù)圖象及性質(zhì)【考點(diǎn)梳理】1、用五點(diǎn)法作正弦函數(shù)和余弦函數(shù)的簡(jiǎn)圖(1)在正弦函數(shù)SKIPIF1<0，SKIPIF1<0的圖象中，五個(gè)關(guān)鍵點(diǎn)是：SKIPIF1<0．(2)在余弦函數(shù)SKIPIF1<0，SKIPIF1<0的圖象中，五個(gè)關(guān)鍵點(diǎn)是：SKIPIF1<0．2、SKIPIF1<0的圖象與性質(zhì)（1）最小正周期：SKIPIF1<0.（2）定義域與值域：SKIPIF1<0的定義域?yàn)镽，值域?yàn)閇-A,A].（3）最值（SKIPIF1<0）.對(duì)于SKIPIF1<0，SKIPIF1<0（4）對(duì)稱(chēng)軸與對(duì)稱(chēng)中心.（SKIPIF1<0）對(duì)于SKIPIF1<0，SKIPIF1<0（5）單調(diào)性.（SKIPIF1<0）對(duì)于SKIPIF1<0，SKIPIF1<0（6）平移與伸縮由SKIPIF1<0的圖象變換得到SKIPIF1<0(SKIPIF1<0，SKIPIF1<0)的圖象的兩種方法(1)先平移后伸縮(2)先伸縮后平移【典型題型講解】考點(diǎn)一：三角函數(shù)的性質(zhì)【典例例題】例1．（多選）（2022·廣東汕頭·高三期末）對(duì)于函數(shù)SKIPIF1<0，x∈R，則（

）A．f(x)的最大值為1 B．直線SKIPIF1<0為其對(duì)稱(chēng)軸C．f(x)在SKIPIF1<0上單調(diào)遞增 D．點(diǎn)SKIPIF1<0為其對(duì)稱(chēng)中心【答案】BD【詳解】依題意，SKIPIF1<0，SKIPIF1<0的最大值為SKIPIF1<0，A錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則直線SKIPIF1<0為SKIPIF1<0圖象的對(duì)稱(chēng)軸，B正確；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由SKIPIF1<0得SKIPIF1<0，即在SKIPIF1<0上單調(diào)遞減，C錯(cuò)誤；因SKIPIF1<0，則點(diǎn)SKIPIF1<0為其對(duì)稱(chēng)中心，D正確．故選：BD例2．（2022·廣東珠?！じ呷谀╆P(guān)于函數(shù)SKIPIF1<0，下列說(shuō)法正確的是（

）A．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B．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)C．SKIPIF1<0的表達(dá)式可以改寫(xiě)為SKIPIF1<0D．若函數(shù)SKIPIF1<0在SKIPIF1<0的值域?yàn)镾KIPIF1<0，則m的取值范圍是SKIPIF1<0【答案】BD【詳解】對(duì)于A，由函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位可得到函數(shù)SKIPIF1<0的圖象，所以A選項(xiàng)錯(cuò)誤；對(duì)于B，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以B選項(xiàng)正確；對(duì)于C，SKIPIF1<0，所以C選項(xiàng)錯(cuò)誤；對(duì)于D，由SKIPIF1<0得SKIPIF1<0，又函數(shù)SKIPIF1<0在的值域?yàn)镾KIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以D選項(xiàng)正確．故選：BD【方法技巧與總結(jié)】研究三角函數(shù)的性質(zhì)，關(guān)鍵式將函數(shù)化為SKIPIF1<0與SKIPIF1<0的形式利用正余弦函數(shù)與復(fù)合函數(shù)的性質(zhì)求解.【變式訓(xùn)練】1．（2022·廣東揭陽(yáng)·高三期末）已知函數(shù)SKIPIF1<0，則該函數(shù)的增區(qū)間為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】令SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)的增區(qū)間是SKIPIF1<0.故選：C.2.（2022·廣東茂名·一模）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為_(kāi)_____【答案】3【詳解】由題意，SKIPIF1<0,而SKIPIF1<0，則SKIPIF1<0，所以函數(shù)的最大值為SKIPIF1<0.故答案為：3.3.已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0為奇函數(shù) B．SKIPIF1<0為偶函數(shù)C．SKIPIF1<0為奇函數(shù) D．SKIPIF1<0為偶函數(shù)【答案】B【詳解】因?yàn)镾KIPIF1<0SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)，故A錯(cuò)誤，B正確；又SKIPIF1<0，所以函數(shù)SKIPIF1<0為非奇非偶函數(shù)函數(shù)，故C、D錯(cuò)誤.故選：B.4.設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0，則（

）A．函數(shù)SKIPIF1<0的周期為SKIPIF1<0B．將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，得到的函數(shù)為奇函數(shù)C．當(dāng)SKIPIF1<0，SKIPIF1<0的值域?yàn)镾KIPIF1<0D．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個(gè)數(shù)共有6個(gè)【答案】D【詳解】由題意，得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0選項(xiàng)A不正確；對(duì)于選項(xiàng)B：將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，得到的函數(shù)是SKIPIF1<0為偶函數(shù)，所以選項(xiàng)B錯(cuò)誤；對(duì)于選項(xiàng)C：當(dāng)時(shí)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的值域?yàn)镾KIPIF1<0，選項(xiàng)C不正確；對(duì)于選項(xiàng)D：令SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個(gè)數(shù)共有6個(gè)，D正確，故選：D.5．設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0.若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0的最小正周期大于SKIPIF1<0，則（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】D【詳解】由SKIPIF1<0的最小正周期大于SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0,SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0,SKIPIF1<0，SKIPIF1<0,取SKIPIF1<0，得SKIPIF1<0,SKIPIF1<0，SKIPIF1<0,故選：SKIPIF1<0．6.若函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0的最小值為SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．4 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0的最小值為函數(shù)SKIPIF1<0的最小正周期的SKIPIF1<0，所以，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，因此，SKIPIF1<0.故選：A7.（2022·廣東湛江·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)極大值點(diǎn)，則SKIPIF1<0的最大值為_(kāi)__________.【答案】SKIPIF1<0【詳解】由題意知，SKIPIF1<0,SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0,SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.又SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)極大值點(diǎn)，所以SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，此時(shí)有2個(gè)極大值點(diǎn)，舍去；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，此時(shí)有1個(gè)極大值點(diǎn)，成立，所以SKIPIF1<0的最大值為SKIPIF1<0，故答案為：SKIPIF1<08．（2021·廣東佛山·一模）已知函數(shù)SKIPIF1<0.從下面的兩個(gè)條件中任選其中一個(gè)：①SKIPIF1<0；②若SKIPIF1<0，且SKIPIF1<0的最小值為SKIPIF1<0，SKIPIF1<0，求解下列問(wèn)題：(1)化簡(jiǎn)SKIPIF1<0的表達(dá)式并求SKIPIF1<0的單調(diào)遞增區(qū)間；(2)已知SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0(2)SKIPIF1<0(1)解：若選擇條件①SKIPIF1<0；SKIPIF1<0SKIPIF1<0SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0若選擇條件②，若SKIPIF1<0，即SKIPIF1<0是SKIPIF1<0的最大值點(diǎn)，SKIPIF1<0是SKIPIF1<0的零點(diǎn)且SKIPIF1<0的最小值為SKIPIF1<0，設(shè)SKIPIF1<0的周期為T(mén)，由此可得SKIPIF1<0，即有SKIPIF1<0，∴SKIPIF1<0由SKIPIF1<0，可得SKIPIF1<0，即有SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，再結(jié)合SKIPIF1<0，可得SKIPIF1<0，綜上可得：SKIPIF1<0，(2)解：SKIPIF1<0，可得SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，從而可得SKIPIF1<0，即有SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0.考點(diǎn)二：三角函數(shù)的圖象【典例例題】例1．（2022·廣東·金山中學(xué)高三期末）為了得到函數(shù)SKIPIF1<0的圖象，可以將函數(shù)SKIPIF1<0的圖象（

）A．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 D．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度【答案】A【詳解】SKIPIF1<0．故選：A.例2．（多選）（2022·廣東中山·高三期末）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)B．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)C．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減D．函數(shù)SKIPIF1<0圖象向右平移SKIPIF1<0個(gè)單位可得函數(shù)SKIPIF1<0的圖象【答案】AB【詳解】解：由圖可知SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，將點(diǎn)SKIPIF1<0代入得：SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，對(duì)于A，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，故A正確；對(duì)于B，因?yàn)镾KIPIF1<0，為最小值，所以函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，故B正確；對(duì)于C，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)遞減，故C錯(cuò)誤；對(duì)于D，將函數(shù)SKIPIF1<0圖象向右平移SKIPIF1<0個(gè)單位，可得函數(shù)SKIPIF1<0，故D錯(cuò)誤.故選：AB.【方法技巧與總結(jié)】1.圖象變換過(guò)程中務(wù)必分清式相位變換，還是周期變換.變換只是相對(duì)于其中的自變量SKIPIF1<0而言的，如果SKIPIF1<0的系數(shù)不是1，就要把這個(gè)系數(shù)提取后再確定變換的單位長(zhǎng)度和方向.2.已知函數(shù)SKIPIF1<0的圖象求解析式時(shí)，常采用待定系數(shù)法，由圖中的最高點(diǎn)、最低點(diǎn)或特殊點(diǎn)求A；由函數(shù)的周期確定SKIPIF1<0；確定SKIPIF1<0常根據(jù)“五點(diǎn)法”中的五個(gè)點(diǎn)求解，其中一般把第一個(gè)零點(diǎn)作為突破口，可以從圖象的升降找準(zhǔn)第一個(gè)零點(diǎn)的位置.【變式訓(xùn)練】1．（2022·廣東廣東·一模）將正弦函數(shù)圖象上各點(diǎn)的橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象．則SKIPIF1<0圖象的一個(gè)對(duì)稱(chēng)中心為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】正弦函數(shù)SKIPIF1<0的對(duì)稱(chēng)中心是SKIPIF1<0，若圖象上各點(diǎn)的橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0，縱坐標(biāo)不變，那么對(duì)稱(chēng)中心是SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，對(duì)稱(chēng)中心是SKIPIF1<0，A符合，其他選項(xiàng)不成立.故選：A2．（2022·廣東韶關(guān)·一模）若將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，則平移后的函數(shù)圖象的一條對(duì)稱(chēng)軸為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】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的圖象，則平移后函數(shù)的對(duì)稱(chēng)軸方程為SKIPIF1<0，取SKIPIF1<0可得，SKIPIF1<0，所以直線SKIPIF1<0為平移后的函數(shù)圖象的一條對(duì)稱(chēng)軸，故選：B.3．（2022·廣東廣州·一模）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，得到函數(shù)SKIPIF1<0的圖象，則下列說(shuō)法正確的是（

）A．若SKIPIF1<0，則SKIPIF1<0是偶函數(shù)B．若SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減C．若SKIPIF1<0，則SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)D．若SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增【答案】AC【詳解】由題設(shè)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0為偶函數(shù)，在SKIPIF1<0上有SKIPIF1<0，SKIPIF1<0遞增，故A正確，B錯(cuò)誤；SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)，SKIPIF1<0，即SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，在SKIPIF1<0上有SKIPIF1<0，SKIPIF1<0不單調(diào)，故C正確，D錯(cuò)誤.故選：AC4．(多選)（2022·廣東·鐵一中學(xué)高三期末）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，且SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0為奇函數(shù)B．SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上有4個(gè)極值點(diǎn)D．若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的最大值為5【答案】BCD【詳解】∵SKIPIF1<0∴SKIPIF1<0，且SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0為奇數(shù)，∴SKIPIF1<0為偶函數(shù)，故A錯(cuò).由上得：SKIPIF1<0為奇數(shù)，∴SKIPIF1<0，故B對(duì).由上得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0,由圖象可知SKIPIF1<0在SKIPIF1<0上有4個(gè)極值點(diǎn),故C對(duì)，∵SKIPIF1<0在SKIPIF1<0上單調(diào)，所以SKIPIF1<0,解得：SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0的最大值為5，故D對(duì)故選：BCD.5．(多選)（2022·廣東東莞·高三期末）已知函數(shù)SKIPIF1<0，若SKIPIF1<0且對(duì)任意SKIPIF1<0都有SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后，圖象關(guān)于原點(diǎn)對(duì)稱(chēng)D．SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位后，圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)【答案】BD【詳解】SKIPIF1<0，SKIPIF1<0，又對(duì)任意SKIPIF1<0都有SKIPIF1<0，則SKIPIF1<0為SKIPIF1<0的最大值，SKIPIF1<0，整理得：SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，因此A選項(xiàng)錯(cuò)誤，B正確；SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后得到的圖象對(duì)應(yīng)的函數(shù)解析式為：SKIPIF1<0，該函數(shù)圖象不關(guān)于原點(diǎn)對(duì)稱(chēng)，故C錯(cuò)誤；SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位后，得到函數(shù)SKIPIF1<0的圖象，該圖象關(guān)于y軸對(duì)稱(chēng)，故D正確，故選：BD6．（多選）（2022·廣東清遠(yuǎn)·高三期末）將函數(shù)SKIPIF1<0圖象上所有的點(diǎn)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，則（

）A．SKIPIF1<0的最小值是SKIPIF1<0B．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)C．SKIPIF1<0的最小正周期是SKIPIF1<0D．SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0【答案】ACD【詳解】由題意知，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，SKIPIF1<0的最小值是SKIPIF1<0，最小正周期是SKIPIF1<0，故A，C正確；令SKIPIF1<0，得SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，故B錯(cuò)誤；令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0，故D正確．故選：ACD.7．（多選）（2022·廣東惠州·一模）已知函數(shù)SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的部分圖象如圖所示，則下列結(jié)論正確的是（

）A．函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0直線對(duì)稱(chēng)B．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D．SKIPIF1<0與圖象SKIPIF1<0的所有交點(diǎn)的橫坐標(biāo)之和為SKIPIF1<0【答案】BCD【解析】根據(jù)圖象求出函數(shù)解析式，再判斷各選項(xiàng)．【詳解】由題意SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0不是對(duì)稱(chēng)軸，A錯(cuò)；SKIPIF1<0，∴SKIPIF1<0是對(duì)稱(chēng)中心，B正確；SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，C正確；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，和為SKIPIF1<0，D正確．故選：BCD．9．（2022·廣東茂名·二模）已知函數(shù)SKIPIF1<0的部分圖象如圖所示．將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到SKIPIF1<0的圖象，則（）A．SKIPIF1<0） B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由圖象知，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到SKIPIF1<0的圖象，∴SKIPIF1<0，故選：D．10．（2022·廣東惠州·二模）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則下列結(jié)論中正確的是（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．SKIPIF1<0為奇函數(shù)【答案】ABD【詳解】由圖知SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故SKIPIF1<0，選項(xiàng)A正確；又SKIPIF1<0，所以SKIPIF1<0為函數(shù)的一條對(duì)稱(chēng)軸，故選項(xiàng)B正確；由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故C錯(cuò)誤；SKIPIF1<0為奇函數(shù)，故D正確.故選：ABD.【鞏固練習(xí)】一、單選題1．已知直線SKIPIF1<0是函數(shù)SKIPIF1<0的圖象的一條對(duì)稱(chēng)軸，為了得到函數(shù)SKIPIF1<0的圖象，可把函數(shù)SKIPIF1<0的圖象（

）A．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向左平移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【詳解】依題意，直線SKIPIF1<0是函數(shù)SKIPIF1<0的圖象的一條對(duì)稱(chēng)軸，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0．將SKIPIF1<0的圖象，向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得SKIPIF1<0．故選：B．2．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．2 C．5 D．7【答案】C【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，即函數(shù)SKIPIF1<0是奇函數(shù)，SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0所以SKIPIF1<0.故選：C3．已知函數(shù)SKIPIF1<0，則下列結(jié)論錯(cuò)誤的是（

）A．函數(shù)SKIPIF1<0的最小正周期是SKIPIF1<0B．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減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)單位長(zhǎng)度，再向下平移1個(gè)單位長(zhǎng)度得到D．函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)【答案】C【詳解】SKIPIF1<0，所以函數(shù)SKIPIF1<0的最小正周期是SKIPIF1<0，A正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0單調(diào)遞減，故B正確；函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向下平移1個(gè)單位長(zhǎng)度得到SKIPIF1<0，故C錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于SKIPIF1<0中心對(duì)稱(chēng)，D正確.故選：C4．如圖是函數(shù)SKIPIF1<0的圖象的一部分，則要得到該函數(shù)的圖象，只需要將函數(shù)SKIPIF1<0的圖象（

）A．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向左平移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】A【詳解】由題圖知：SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0SKIPIF1<0，將SKIPIF1<0向左平移SKIPIF1<0得SKIPIF1<0.故選：A.二、多選題5．已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．直線SKIPIF1<0為函數(shù)f(x)圖象的一條對(duì)稱(chēng)軸B．函數(shù)f(x)圖象橫坐標(biāo)縮短為原來(lái)的一半，再向左平移SKIPIF1<0后得到SKIPIF1<0C．函數(shù)f(x)在[－SKIPIF1<0，SKIPIF1<0]上單調(diào)遞增D．函數(shù)SKIPIF1<0的值域?yàn)閇－2，SKIPIF1<0]【答案】AD【詳解】解：對(duì)于A：SKIPIF1<0，選項(xiàng)A正確；對(duì)于B：函數(shù)f(x)圖象橫坐標(biāo)縮短為原來(lái)的一半，得到SKIPIF1<0，再向左平移SKIPIF1<0后得到SKIPIF1<0，選項(xiàng)B錯(cuò)誤；對(duì)于C：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其中SKIPIF1<0，不妨令SKIPIF1<0為銳角，SKIPIF1<0當(dāng)SKIPIF1<0即，SKIPIF1<0時(shí)，f(x)單調(diào)遞增，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，f(x)單調(diào)遞減，選項(xiàng)C錯(cuò)誤；對(duì)于D：2π是函數(shù)的周期，可取一個(gè)周期[－SKIPIF1<0，SKIPIF1<0]探究f(x)值域．而函數(shù)f(x)的對(duì)稱(chēng)軸為：SKIPIF1<0．因此：可取區(qū)間[－SKIPIF1<0，SKIPIF1<0]探究f(x)值域，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0即：SKIPIF1<0，選項(xiàng)D正確．故選：AD.6．設(shè)函數(shù)SKIPIF1<0，則下列結(jié)論中正確的是（

）A．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng) B．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D．SKIPIF1<0在SKIPIF1<0上的最小值為0【答案】ABC【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，A正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，B正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故C正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上的最小值為SKIPIF1<0，D錯(cuò)誤.故選：ABC7．已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱(chēng)C．若SKIPIF1<0，則SKIPIF1<0D．對(duì)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立【答案】ACD【詳解】∵函數(shù)SKIPIF1<0的周期SKIPIF1<0，所以SKIPIF1<0恒成立，故A正確；又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的圖象不關(guān)于原點(diǎn)對(duì)稱(chēng)，故B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C正確；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，即SKIPIF1<0，所以對(duì)SKIPIF1<0有SKIPIF1<0成立，故D正確.故選：ACD.三、填空題8．寫(xiě)出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)SKIPIF1<0___________；已知函數(shù)滿(mǎn)足：①SKIPIF1<0；②SKIPIF1<0；③函數(shù)在SKIPIF1<0上單調(diào)遞減；【答案】SKIPIF1<0（答案不唯一）【詳解】對(duì)于①，若SKIPIF1<0，則SKIPIF1<0的圖象關(guān)于SKIPIF1<0中心對(duì)稱(chēng)，對(duì)于②，若SKIPIF1<0，則SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)，且函數(shù)在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0，得SKIPIF1<0故答案為：SKIPIF1<0（答案不唯一）9．已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0________．【答案】SKIPIF1<0【詳解】解：由SKIPIF1<0知，SKIPIF1<0，由五點(diǎn)法可知，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0故答案為：SKIPIF1<010．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上有最大值，沒(méi)有最小值，則SKIPIF1<0的最大值為_(kāi)_____．【答案】17【詳解】由SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上有最大值，沒(méi)有最小值，可得SKIPIF1<0，所以SKIPIF1<0.由SKIPIF1<0在SKIPIF1<0上有最大值，沒(méi)有最小值，可得SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的最大值為17，,故答案為：17四、解答題11．已知函數(shù)SKIPIF1<0(1)求SKIPIF1<0的值；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的增區(qū)間和值域．【答案】(1)SKIPIF1<0(2)單調(diào)遞增區(qū)間為SKIPIF1<0，值域?yàn)镾KIPIF1<0【解析】(1)解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0(2)解：由（1）可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，即函數(shù)在SKIPIF1<0上的單調(diào)遞增區(qū)間為SKIPIF1<0；12．設(shè)SKIPIF1<0內(nèi)角A，B，C的對(duì)邊分別為a，b，c，函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0的面積；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值，求SKIPIF1<0在SKIPIF1<0上的值域.【答案】(1)若SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0；(2)SKIPIF1<0【解析】(1)因?yàn)镾KIPIF1<0，SKIPIF1<0所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，由正弦定理可得SKIPIF1<0,又SKIPIF1<0，所以SKIPIF1<0，若SKIPIF1<0則SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0則SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0，(2)SKIPIF1<0SKIPIF1<0SKIPIF1<0．因?yàn)镾KIPIF1<0在SKIPIF1<0處取得最大值，所以SKIPIF1<0，即SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0.14．已知函數(shù)SKIPIF1<0，SKIPIF1<0從下面兩個(gè)條件：條件①SKIPIF1<0、條件②SKIPIF1<0中選擇一個(gè)作為已知．(1)求SKIPIF1<0時(shí)函數(shù)SKIPIF1<0的值域；(2)若函數(shù)SKIPIF1<0圖象向右平移m個(gè)單位長(zhǎng)度后與函數(shù)SKIPIF1<0的圖象重合，求正數(shù)m的最小值．【答案】(1)答案見(jiàn)解析；(2)答案見(jiàn)解析.【解析】(1)若選擇條件①作為已知：SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0；若選擇條件②作為已知：SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0；(2)若選擇條件①作為已知：函數(shù)SKIPIF1<0圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后，得到函數(shù)SKIPIF1<0，即SKIPIF1<0的圖象，∵SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象重合．∴SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0