試卷第=page11頁(yè)，共=sectionpages33頁(yè)試卷第=page11頁(yè)，共=sectionpages33頁(yè)第27課函數(shù)y＝Asin(ωx＋φ)的圖象及三角函數(shù)模型的簡(jiǎn)單應(yīng)用學(xué)校:___________姓名：___________班級(jí)：___________考號(hào)：___________【基礎(chǔ)鞏固】1．（2022·廣東·大埔縣虎山中學(xué)高三階段練習(xí)）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0后，所得圖象對(duì)應(yīng)的函數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0后，所得圖象對(duì)應(yīng)的函數(shù)為SKIPIF1<0SKIPIF1<0.故選：B2．（2022·浙江·高考真題）為了得到函數(shù)SKIPIF1<0的圖象，只要把函數(shù)SKIPIF1<0圖象上所有的點(diǎn)（

）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】D【解析】因?yàn)镾KIPIF1<0，所以把函數(shù)SKIPIF1<0圖象上的所有點(diǎn)向右平移SKIPIF1<0個(gè)單位長(zhǎng)度即可得到函數(shù)SKIPIF1<0的圖象．故選：D.3．（2022·湖南師大附中三模）某智能主動(dòng)降噪耳機(jī)工作的原理是利用芯片生成與噪音的相位相反的聲波，通過(guò)兩者疊加完全抵消掉噪音（如圖），已知噪音的聲波曲線SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的振幅為1，周期為2，初相位為SKIPIF1<0，則用來(lái)降噪的聲波曲線的解析式是（

）

A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意，SKIPIF1<0且SKIPIF1<0則SKIPIF1<0，所以SKIPIF1<0，則降噪的聲波曲線為SKIPIF1<0.故選：D.4．（2022·江蘇鹽城·三模）把函數(shù)SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的2倍，縱坐標(biāo)不變，再把所得曲線向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，則函數(shù)SKIPIF1<0的解析式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】根據(jù)題意，先將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到函數(shù)SKIPIF1<0的圖象，再將該函數(shù)圖象上各點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象.故選：A.5．（2022·全國(guó)·模擬預(yù)測(cè)）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，若SKIPIF1<0滿(mǎn)足SKIPIF1<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】解法一：SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0滿(mǎn)足SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0．故選:A．解法二SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0滿(mǎn)足SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0．故選：A．6．（2022·天津南開(kāi)·二模）函數(shù)SKIPIF1<0SKIPIF1<0，其圖象的一個(gè)最低點(diǎn)是SKIPIF1<0，距離SKIPIF1<0點(diǎn)最近的對(duì)稱(chēng)中心為SKIPIF1<0，則（

）A．SKIPIF1<0B．SKIPIF1<0是函數(shù)SKIPIF1<0圖象的一條對(duì)稱(chēng)軸C．SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增D．SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位后得到SKIPIF1<0的圖象，若SKIPIF1<0是奇函數(shù)，則SKIPIF1<0的最小值是SKIPIF1<0【答案】C【解析】解：SKIPIF1<0函數(shù)SKIPIF1<0，SKIPIF1<0的圖象的一個(gè)最低點(diǎn)是SKIPIF1<0，距離SKIPIF1<0點(diǎn)最近的對(duì)稱(chēng)中心為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0，故A錯(cuò)誤；SKIPIF1<0，故函數(shù)關(guān)于SKIPIF1<0對(duì)稱(chēng)，故B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，故C正確；把SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位后得到SKIPIF1<0的圖象，若SKIPIF1<0是奇函數(shù)，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0的最小值是SKIPIF1<0，故D錯(cuò)誤，故選：C7．（2022·山東·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的部分圖象如圖所示，則下列敘述正確的是（

）A．函數(shù)SKIPIF1<0的圖象可由SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到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在區(qū)間SKIPIF1<0上單調(diào)遞增D．函數(shù)SKIPIF1<0圖象的對(duì)稱(chēng)中心為SKIPIF1<0【答案】C【解析】由圖象可知SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，由圖可知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0.對(duì)于A選項(xiàng)，因?yàn)镾KIPIF1<0，所以，函數(shù)SKIPIF1<0的圖象可由SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到，A錯(cuò)；對(duì)于B選項(xiàng)，因?yàn)镾KIPIF1<0，所以，函數(shù)SKIPIF1<0的圖象不關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，B錯(cuò)；對(duì)于C選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，C對(duì)；對(duì)于D選項(xiàng)，令SKIPIF1<0，則SKIPIF1<0，D錯(cuò).故選：C.8．（2022·浙江·海寧中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，下列說(shuō)法正確的是（

）A．若SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上存在零點(diǎn)B．若SKIPIF1<0，則將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得圖象關(guān)于原點(diǎn)對(duì)稱(chēng)C．若函數(shù)SKIPIF1<0在SKIPIF1<0上取到最大值，則SKIPIF1<0的最小值為SKIPIF1<0D．若函數(shù)SKIPIF1<0在SKIPIF1<0上存在兩個(gè)最值，則SKIPIF1<0的取值范圍是SKIPIF1<0【答案】C【解析】解：對(duì)于選項(xiàng)A，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上不存在零點(diǎn)，所以選項(xiàng)A錯(cuò)誤；對(duì)于選項(xiàng)B，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得函數(shù)為SKIPIF1<0是偶函數(shù)，其圖象關(guān)于y軸對(duì)稱(chēng)，所以選項(xiàng)B錯(cuò)誤；對(duì)于選項(xiàng)C，因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上取到最大值，所以SKIPIF1<0，即有SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0，所以選項(xiàng)C正確；對(duì)于選項(xiàng)D，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，要使函數(shù)SKIPIF1<0在SKIPIF1<0上存在兩個(gè)最值，則SKIPIF1<0，解得SKIPIF1<0，所以選項(xiàng)D不正確.故選：C.9．（多選）（2022·山東濟(jì)南·三模）將函數(shù)SKIPIF1<0圖像上所有的點(diǎn)向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖像，則下列說(shuō)法正確的是（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．SKIPIF1<0圖像的一個(gè)對(duì)稱(chēng)中心為SKIPIF1<0C．SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0D．SKIPIF1<0的圖像與函數(shù)SKIPIF1<0的圖像重合【答案】ABC【解析】根據(jù)題意，SKIPIF1<0，則周期SKIPIF1<0，A正確；對(duì)B，令SKIPIF1<0，B正確；對(duì)C，令SKIPIF1<0，即函數(shù)的減區(qū)間為SKIPIF1<0，C正確；對(duì)D，因?yàn)镾KIPIF1<0，D錯(cuò)誤.故選：ABC.10．（多選）（2022·河北衡水·高三階段練習(xí)）已知函數(shù)SKIPIF1<0，則（

）A．若SKIPIF1<0，SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后，所得函數(shù)圖象關(guān)于y軸對(duì)稱(chēng)，則SKIPIF1<0的最小值為SKIPIF1<0B．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，所得函數(shù)圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱(chēng)，則SKIPIF1<0的最小值為0.C．若SKIPIF1<0，對(duì)SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上至少有2個(gè)零點(diǎn)，至多有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是SKIPIF1<0D．若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)，函數(shù)SKIPIF1<0的圖象向右平移π個(gè)單位長(zhǎng)度后，所得圖象與原來(lái)的圖象重合，則SKIPIF1<0【答案】BC【解析】對(duì)于A選項(xiàng)，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，其圖象關(guān)于y軸對(duì)稱(chēng)，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值，為SKIPIF1<0，所以選項(xiàng)A不正確；對(duì)于B選項(xiàng)，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，其圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱(chēng)，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值，為0，所以選項(xiàng)B正確；對(duì)于C選項(xiàng)，函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，即方程SKIPIF1<0的根的個(gè)數(shù).因?yàn)镾KIPIF1<0，所以SKIPIF1<0.又SKIPIF1<0，方程SKIPIF1<0在SKIPIF1<0上至少有2個(gè)不同實(shí)根，至多有3個(gè)不同實(shí)根，則必須滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0，所以選項(xiàng)C正確；對(duì)于D選項(xiàng)，函數(shù)SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，于是函數(shù)SKIPIF1<0.又SKIPIF1<0在SKIPIF1<0上單調(diào)，所以SKIPIF1<0，解得SKIPIF1<0，函數(shù)SKIPIF1<0的圖象向右平移π個(gè)單位長(zhǎng)度后，得到函數(shù)SKIPIF1<0的圖象與原來(lái)的圖象重合，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0.由SKIPIF1<0，因此SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上不單調(diào)，不合題意，舍去，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0符合題意，所以選項(xiàng)D不正確故選：BC.11．（多選）（2022·湖北·黃岡中學(xué)三模）已知函數(shù)SKIPIF1<0SKIPIF1<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的圖象可由SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到B．直線SKIPIF1<0是SKIPIF1<0圖象的一條對(duì)稱(chēng)軸C．若SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0D．直線SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象有SKIPIF1<0個(gè)交點(diǎn)【答案】BCD【解析】對(duì)于A選項(xiàng)，由圖可知，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0，又因?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，故函數(shù)SKIPIF1<0的圖象可由SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到，A錯(cuò)；對(duì)于B選項(xiàng)，SKIPIF1<0，所以，直線SKIPIF1<0是SKIPIF1<0圖象的一條對(duì)稱(chēng)軸，B對(duì)；對(duì)于C選項(xiàng)，因?yàn)镾KIPIF1<0，所以，SKIPIF1<0的最小值為SKIPIF1<0，C對(duì)；對(duì)于D選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0可知SKIPIF1<0的可能取值集合為SKIPIF1<0，所以，直線SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象有SKIPIF1<0個(gè)交點(diǎn)，D對(duì).故選：BCD.12．（多選）（2022·山東·勝利一中模擬預(yù)測(cè)）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【解析】∵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,由圖像可知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.13．（2022·遼寧·沈陽(yáng)二中模擬預(yù)測(cè)）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SKIPIF1<0的圖象，則SKIPIF1<0__________.【答案】0【解析】解：由題意可知，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SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故答案為：0.14．（2022·浙江溫州·三模）已知函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0___________，SKIPIF1<0的圖象至少向左平移___________個(gè)單位長(zhǎng)度得到SKIPIF1<0的圖象.【答案】

SKIPIF1<0

SKIPIF1<0【解析】由題有：SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0.∴SKIPIF1<0設(shè)SKIPIF1<0向左平移m個(gè)單位長(zhǎng)度，則SKIPIF1<0∴SKIPIF1<0，即SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0.15．（2022·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的部分圖象如圖所示，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0________．【答案】SKIPIF1<0﹔【解析】由題意知，函數(shù)SKIPIF1<0中，周期SKIPIF1<0，所以SKIPIF1<0，又函數(shù)圖象過(guò)點(diǎn)SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0；由SKIPIF1<0，得圖象的最高點(diǎn)坐標(biāo)為SKIPIF1<0，因?yàn)镾KIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<0.16．（2022·湖北·襄陽(yáng)五中模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0的圖象，若SKIPIF1<0為SKIPIF1<0的一條對(duì)稱(chēng)軸，則SKIPIF1<0__________．【答案】SKIPIF1<0【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0是SKIPIF1<0的一條對(duì)稱(chēng)軸，∴SKIPIF1<0SKIPIF1<0，即SKIPIF1<0.故答案為SKIPIF1<017．（2022·山東臨沂·二模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0．(1)求SKIPIF1<0的解析式；(2)將函數(shù)SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)縮小為原來(lái)的SKIPIF1<0，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0，求SKIPIF1<0的值．【解】(1)因?yàn)镾KIPIF1<0，所以周期SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0；(2)SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.18．（2022·浙江省浦江中學(xué)高三期末）設(shè)SKIPIF1<0，將奇函數(shù)SKIPIF1<0圖象向左平移SKIPIF1<0個(gè)單位，再將圖象上各點(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的圖像．(1)求a的值及函數(shù)SKIPIF1<0的解析式；(2)設(shè)SKIPIF1<0，SKIPIF1<0，求函數(shù)SKIPIF1<0的值域．【解】(1)因?yàn)镾KIPIF1<0是奇函數(shù)，且在SKIPIF1<0處有定義，可知SKIPIF1<0，得到SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0圖象向左平移SKIPIF1<0個(gè)單位得到SKIPIF1<0，再將圖象上各點(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的圖像，可得SKIPIF1<0.(2)由（1）可得：SKIPIF1<0SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.【素養(yǎng)提升】1．（2022·天津·耀華中學(xué)一模）已知函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上有且僅有2個(gè)最大值點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0第1次取到最大值，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0第2次取到最大值，由SKIPIF1<0知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0第3次取到最大值．∴SKIPIF1<0．故選：C2．（2022·北京·101中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，將SKIPIF1<0的圖象上的所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0，縱坐標(biāo)保持不變，得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0，則SKIPIF1<0的值不可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小正周期SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，不妨設(shè)SKIPIF1<0SKIPIF1<0與SKIPIF1<0分別對(duì)應(yīng)SKIPIF1<0的最大值點(diǎn)和最小值點(diǎn)，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故選：C3．（多選）（2022·山東濰坊·高三期末）已知函數(shù)SKIPIF1<0，現(xiàn)有如下四個(gè)命題：甲：該函數(shù)的最小值為SKIPIF1<0；乙：該函數(shù)圖像的相鄰兩條對(duì)稱(chēng)軸之間的距離為π；丙：該函數(shù)的一個(gè)零點(diǎn)為SKIPIF1<0；?。涸摵瘮?shù)圖像可以由SKIPIF1<0的圖像平移得到．如果有且只有一個(gè)假命題，那么下列說(shuō)法正確的是（

）A．乙一定是假命題．B．φ的值可唯一確定C．函數(shù)f（x）的極大值點(diǎn)為SKIPIF1<0D．函數(shù)f（x）圖像可以由SKIPIF1<0圖像伸縮變換得到【答案】BD【解析】若甲命題正確：該函數(shù)的最小值為SKIPIF1<0，則SKIPIF1<0；若乙命題正確：該函數(shù)圖像的相鄰兩條對(duì)稱(chēng)軸之間的距離為π，則SKIPIF1<0；若丙命題正確：該函數(shù)的一個(gè)零點(diǎn)為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0；若丁命題正確：該函數(shù)圖像可以由SKIPIF1<0的圖像平移得到.由SKIPIF1<0，可知SKIPIF1<0，SKIPIF1<0故命題乙與命題丁矛盾.由甲乙丙丁有且只有一個(gè)假命題可知，二者必一真一假，則命題甲與命題丙均為真命題.由命題甲為真命題，可知SKIPIF1<0，由命題丙為真命題可知SKIPIF1<0若命題乙為真命題，則SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0此時(shí)SKIPIF1<0.若命題丁為真命題，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0又SKIPIF1<0，則不存在符合條件的SKIPIF1<0.不合題意.綜上，命題丁為假命題，命題甲、乙、丙均為真命題.選項(xiàng)A：乙一定是假命題．判斷錯(cuò)誤；選項(xiàng)B：φ的值可唯一確定.判斷正確；選項(xiàng)C：函數(shù)f（x）的極大值點(diǎn)為SKIPIF1<0.由SKIPIF1<0，可得SKIPIF1<0，即函數(shù)f（x）的極大值點(diǎn)為SKIPIF1<0.判斷錯(cuò)誤；選項(xiàng)D：函數(shù)f（x）圖像可以由SKIPIF1<0圖像伸縮變換得到.由SKIPIF1<0可知，把SKIPIF1<0的圖像上每一點(diǎn)的橫坐標(biāo)保持不變，縱坐標(biāo)伸長(zhǎng)為原來(lái)的SKIPIF1<0倍，可以得到SKIPIF1<0的圖像.判斷正確.故選：BD4．（2022·山東濰坊·三模）已知函數(shù)SKIPIF1<0向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到SKIPIF1<0．若對(duì)于任意的SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_____．【答案】SKIPIF1<0【解析】函數(shù)SKIPIF1<0向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)閷?duì)于任意的SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0，所以SKIPIF1<0的取值范圍應(yīng)包含SKIPIF1<0，根據(jù)余弦函數(shù)的性質(zhì)，為使SKIPIF1<0取最小值，只需函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)且值域?yàn)镾KIPIF1<0即可.由SKIPIF1<0可得SKIPIF1<0，因此SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.5．（2022·重慶八中模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示．(1)求函數(shù)SKIPIF1<0的解析式；(2)將函數(shù)SKIPIF1<0的圖象上所有的點(diǎn)向右平移SKIPIF1<0個(gè)單位，再將所得圖象上每一個(gè)點(diǎn)的橫坐標(biāo)變?yōu)樵瓉?lái)的2倍（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象．當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0恰有三個(gè)不相等的實(shí)數(shù)根SKIPIF1<0，求實(shí)數(shù)a的取值范圍和SKIPIF1<0的值．【解】(1)解：由圖示得：SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0；(2)解：由已知得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0