新高考數(shù)學(xué)一輪復(fù)習(xí) 專項(xiàng)分層精練第20課 函數(shù)y＝Asin(ωx＋φ)的圖象（解析版）

第20課函數(shù)y＝Asin(ωx＋φ)的圖象（分層專項(xiàng)精練）【一層練基礎(chǔ)】一、單選題1．（2023·全國(guó)·高三專題練習(xí)）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【答案】C【分析】A選項(xiàng)，利用三角恒等變換得到SKIPIF1<0，從而求出最小正周期；B選項(xiàng)，整體代入檢驗(yàn)是否是單調(diào)遞減區(qū)間；C選項(xiàng)，利用函數(shù)平移左加右減，上加下減進(jìn)行平移，求出平移后的解析式；D選項(xiàng)，代入檢驗(yàn)是否是對(duì)稱中心.【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，D正確.故選：C2．（2023·江西南昌·南昌市八一中學(xué)?？既＃┖瘮?shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位得到函數(shù)SKIPIF1<0的圖像，若函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)圖像平移得函數(shù)SKIPIF1<0的解析式，由函數(shù)SKIPIF1<0是偶函數(shù)，解出SKIPIF1<0，可得SKIPIF1<0.【詳解】函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位，得SKIPIF1<0的圖像，又函數(shù)SKIPIF1<0是偶函數(shù)，則有SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0；所以SKIPIF1<0.故選：C．3．（2023·全國(guó)·高一專題練習(xí)）已知函數(shù)SKIPIF1<0，將函數(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【分析】首先根據(jù)平移后得到函數(shù)SKIPIF1<0的解析式，再根據(jù)圖象求函數(shù)的解析式，即可求值.【詳解】平移不改變振幅和周期，所以由圖象可知SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0,函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，得SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0.故選：A4．（2022秋·全國(guó)·高一期末）已知函數(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的圖象關(guān)于y軸對(duì)稱，則SKIPIF1<0的最小值為（

）A．1 B．2 C．SKIPIF1<0 D．5【答案】D【分析】根據(jù)輔助角公式，結(jié)合正弦型函數(shù)的奇偶性進(jìn)行求解即可.【詳解】SKIPIF1<0，因?yàn)樵摵瘮?shù)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，所以SKIPIF1<0，因?yàn)镾KIPIF1<0的圖象關(guān)于y軸對(duì)稱，所以SKIPIF1<0是偶函數(shù)，因此有SKIPIF1<0，因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值，最小值為5，故選：D二、多選題5．（2023秋·廣西貴港·高三平南縣中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值域?yàn)镾KIPIF1<0C．將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度可得函數(shù)SKIPIF1<0的圖象D．將函數(shù)SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來的2倍，縱坐標(biāo)不變，得到的函數(shù)圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】ACD【分析】先根據(jù)SKIPIF1<0中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的幾何意義，求得SKIPIF1<0的解析式，再結(jié)合正弦函數(shù)的圖象與性質(zhì)，函數(shù)圖象的變換，逐一分析選項(xiàng)即可．【詳解】由圖可知，SKIPIF1<0，函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0，故A正確；由SKIPIF1<0，知SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，對(duì)于B，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的值域?yàn)镾KIPIF1<0，故B錯(cuò)誤；對(duì)于C，將函數(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，故C正確；對(duì)于D，將函數(shù)SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來的2倍，縱坐標(biāo)不變，得到SKIPIF1<0的圖象，因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以得到的函數(shù)圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，故D正確．故選：ACD．6．（2023春·浙江金華·高一浙江省東陽中學(xué)校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，那么（

）A．函數(shù)SKIPIF1<0為奇函數(shù)B．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．若SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0D．函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象【答案】AC【分析】利用SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，即可求出SKIPIF1<0的值，從而得出SKIPIF1<0的解析式，再利用三角函數(shù)的性質(zhì)逐一判斷四個(gè)選項(xiàng)即可．【詳解】因?yàn)镾KIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，所以SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，對(duì)于A：SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)成立，故選項(xiàng)A正確；對(duì)于B：SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上不是單調(diào)函數(shù)；故選項(xiàng)B不正確；對(duì)于C：因?yàn)镾KIPIF1<0，SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0的最小值為半個(gè)周期，即SKIPIF1<0，故選項(xiàng)C正確；對(duì)于D：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，故選項(xiàng)D不正確；故選：AC7．（2023·全國(guó)·高一期中）已知函數(shù)SKIPIF1<0，則下列說法正確的有（

）A．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱B．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減D．將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，可以得到SKIPIF1<0的圖象【答案】AC【分析】用余弦函數(shù)的圖像與性質(zhì)，采用整體代入的思想對(duì)選項(xiàng)逐一判斷即可.【詳解】由SKIPIF1<0可知，SKIPIF1<0解得SKIPIF1<0，所以函數(shù)的對(duì)稱中心為SKIPIF1<0，SKIPIF1<0故A選項(xiàng)正確；令SKIPIF1<0解得SKIPIF1<0，所以函數(shù)的對(duì)稱軸為SKIPIF1<0，SKIPIF1<0，故B選項(xiàng)錯(cuò)誤；令SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)的單調(diào)遞減區(qū)間為SKIPIF1<0，故C選項(xiàng)正確；將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得SKIPIF1<0，故D選項(xiàng)錯(cuò)誤；故選：AC三、填空題8．（2023春·福建福州·高三?？茧A段練習(xí)）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度．得到函數(shù)g(x)的圖象，若g(x)是奇函數(shù)，則φ＝．【答案】SKIPIF1<0【分析】首先根據(jù)平移規(guī)律求函數(shù)SKIPIF1<0的解析式，再根據(jù)函數(shù)是奇函數(shù)，求SKIPIF1<0的值.【詳解】函數(shù)SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0是奇函數(shù)，所以SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<09．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，現(xiàn)將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向下平移兩個(gè)單位長(zhǎng)度，得到SKIPIF1<0的圖象，則滿足SKIPIF1<0的SKIPIF1<0的取值集合為．【答案】SKIPIF1<0【分析】先利用三角函數(shù)圖象變換規(guī)律求出SKIPIF1<0的解析式，再由SKIPIF1<0求解即可.【詳解】解：由題意可知，SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，故取值集合為SKIPIF1<0．故答案為：SKIPIF1<0四、解答題10．（2023秋·天津薊州·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示.(1)求SKIPIF1<0的最小正周期及解析式；(2)將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象，求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值和最小值.【答案】(1)SKIPIF1<0，SKIPIF1<0(2)SKIPIF1<0.【分析】（1）由圖象可知SKIPIF1<0，相鄰的對(duì)稱中心和對(duì)稱軸距離相差SKIPIF1<0，再代入關(guān)鍵點(diǎn)可得解析式；（2）根據(jù)圖象的變換得到SKIPIF1<0解析式，再根據(jù)正弦函數(shù)的圖象與性質(zhì)可得其在區(qū)間上SKIPIF1<0最值.【詳解】（1）由圖象可知SKIPIF1<0的最大值為1，最小值-1，故SKIPIF1<0；又SKIPIF1<0∴SKIPIF1<0，將點(diǎn)SKIPIF1<0代入SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0故答案為：SKIPIF1<0，SKIPIF1<0.（2）由SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0∵SKIPIF1<0∴SKIPIF1<0∴當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0故答案為：SKIPIF1<0【二層練綜合】一、單選題1．（2023·全國(guó)·統(tǒng)考高考真題）函數(shù)SKIPIF1<0的圖象由函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到，則SKIPIF1<0的圖象與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為（

）A．1 B．2 C．3 D．4【答案】C【分析】先利用三角函數(shù)平移的性質(zhì)求得SKIPIF1<0，再作出SKIPIF1<0與SKIPIF1<0的部分大致圖像，考慮特殊點(diǎn)處SKIPIF1<0與SKIPIF1<0的大小關(guān)系，從而精確圖像，由此得解.【詳解】因?yàn)镾KIPIF1<0向左平移SKIPIF1<0個(gè)單位所得函數(shù)為SKIPIF1<0，所以SKIPIF1<0，而SKIPIF1<0顯然過SKIPIF1<0與SKIPIF1<0兩點(diǎn)，作出SKIPIF1<0與SKIPIF1<0的部分大致圖像如下，

考慮SKIPIF1<0，即SKIPIF1<0處SKIPIF1<0與SKIPIF1<0的大小關(guān)系，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；所以由圖可知，SKIPIF1<0與SKIPIF1<0的交點(diǎn)個(gè)數(shù)為SKIPIF1<0.故選：C.二、多選題2．（2023春·河南南陽·高一河南省桐柏縣第一高級(jí)中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0（其中A，SKIPIF1<0，SKIPIF1<0是常數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的部分圖象如圖所示，則下列說法正確的是（

）A．SKIPIF1<0的值域?yàn)镾KIPIF1<0B．SKIPIF1<0的最小正周期為πC．SKIPIF1<0D．將函數(shù)f（x）的圖象向左平移SKIPIF1<0個(gè)單位，得到函數(shù)SKIPIF1<0的圖象【答案】AB【分析】對(duì)A、B、C：根據(jù)函數(shù)圖象求SKIPIF1<0，即可分析判斷；對(duì)D：根據(jù)圖象變換結(jié)合誘導(dǎo)公式求解析式，即可得結(jié)果.【詳解】對(duì)A：由圖可知：SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0的值域?yàn)镾KIPIF1<0，A正確；對(duì)B：由圖可得：SKIPIF1<0，則SKIPIF1<0，B正確；對(duì)C：∵SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0，由圖可得：SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，C錯(cuò)誤；對(duì)D：可得：SKIPIF1<0，將函數(shù)f（x）的圖象向左平移SKIPIF1<0個(gè)單位，得到SKIPIF1<0，D錯(cuò)誤；故選：AB.三、填空題3．（2023春·江西宜春·高三江西省宜豐中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位得到SKIPIF1<0的圖象，若不等式SKIPIF1<0在SKIPIF1<0，上恒成立，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【分析】先根據(jù)圖象的變換規(guī)律求出SKIPIF1<0的解析式，進(jìn)而求出SKIPIF1<0在SKIPIF1<0上的值域SKIPIF1<0，再利用換元法，結(jié)合函數(shù)性質(zhì)，求出最值解決問題．【詳解】解：依題意有SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，由圖知，函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0滿足：SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，原不等式即化為SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上恒成立，令SKIPIF1<0，該二次函數(shù)開口向上，要使上式恒成立，只需：SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0的范圍是SKIPIF1<0．故答案為：SKIPIF1<0．四、解答題4．（2023春·四川眉山·高一統(tǒng)考期中）已知數(shù)SKIPIF1<0的相鄰兩對(duì)稱軸間的距離為SKIPIF1<0.(1)求SKIPIF1<0的解析式；(2)將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再把各點(diǎn)的橫坐標(biāo)縮小為原來的SKIPIF1<0（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象，當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的值域；(3)對(duì)于第（2）問中的函數(shù)SKIPIF1<0，記方程SKIPIF1<0在SKIPIF1<0上的根從小到大依次為SKIPIF1<0，若SKIPIF1<0SKIPIF1<0，試求SKIPIF1<0與SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【分析】（1）先整理化簡(jiǎn)得SKIPIF1<0，利用周期求得SKIPIF1<0，即可得到SKIPIF1<0；（2）利用圖像變換得到SKIPIF1<0，用換元法求出函數(shù)SKIPIF1<0的值域；（3）由方程SKIPIF1<0，得到SKIPIF1<0，借助于正弦函數(shù)SKIPIF1<0的圖象，求出SKIPIF1<0與SKIPIF1<0的值.【詳解】（1）由題意，函數(shù)SKIPIF1<0SKIPIF1<0因?yàn)楹瘮?shù)SKIPIF1<0圖象的相鄰兩對(duì)稱軸間的距離為SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0.故SKIPIF1<0（2）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得SKIPIF1<0的圖象.再把橫坐標(biāo)縮小為原來的SKIPIF1<0，得到函數(shù)SKIPIF1<0的圖象.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最小值，最小值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最大值，最大值為SKIPIF1<0，故函數(shù)SKIPIF1<0的值域SKIPIF1<0.（3）由方程SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，其中SKIPIF1<0，即SKIPIF1<0，結(jié)合正弦函數(shù)SKIPIF1<0的圖象，

可得方程SKIPIF1<0在區(qū)間SKIPIF1<0有5個(gè)解，即SKIPIF1<0，

其中SKIPIF1<0，即SKIPIF1<0解得SKIPIF1<0所以SKIPIF1<0SKIPIF1<0.綜上，SKIPIF1<0【點(diǎn)睛】（1）三角函數(shù)問題通常需要把它化為“一角一名一次”的結(jié)構(gòu)，借助于SKIPIF1<0或SKIPIF1<0的性質(zhì)解題；（2）求y=Asin(ωx+φ)+B的值域通常用換元法；【三層練能力】一、多選題1．（2023春·河南焦作·高二博愛縣第一中學(xué)?？计谀┮阎瘮?shù)SKIPIF1<0（SKIPIF1<0為正整數(shù)，SKIPIF1<0）的最小正周期SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后所得圖象關(guān)于原點(diǎn)對(duì)稱，則下列關(guān)于函數(shù)SKIPIF1<0的說法正確的是（

）A．SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)零點(diǎn) B．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱C．方程SKIPIF1<0在SKIPIF1<0上有三個(gè)解 D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減【答案】ABD【分析】先由周期范圍及SKIPIF1<0為正整數(shù)求得SKIPIF1<0，再由SKIPIF1<0平移后關(guān)于原點(diǎn)對(duì)稱求得SKIPIF1<0，從而得到SKIPIF1<0，對(duì)于AB，將SKIPIF1<0與SKIPIF1<0代入檢驗(yàn)即可；對(duì)于C，利用換元法得到SKIPIF1<0在SKIPIF1<0內(nèi)只有兩個(gè)解，從而可以判斷；對(duì)于D，利用整體法及SKIPIF1<0的單調(diào)性即可判斷.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0為正整數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后所得圖象對(duì)應(yīng)的函數(shù)SKIPIF1<0，（點(diǎn)撥：函數(shù)SKIPIF1<0的圖象經(jīng)過平移變換得到SKIPIF1<0的圖象時(shí)，不是平移SKIPIF1<0個(gè)單位長(zhǎng)度，而是平移SKIPIF1<0個(gè)單位長(zhǎng)度），由題意知，函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，故SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，對(duì)于A，SKIPIF1<0，故A正確；對(duì)于B，SKIPIF1<0，故B正確；對(duì)于A，令SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0內(nèi)只有SKIPIF1<0，SKIPIF1<0兩個(gè)解，即方程SKIPIF1<0在SKIPIF1<0上只有兩個(gè)解，故C錯(cuò)誤；對(duì)于A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D正確.故選：ABD.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：求解此類問題的關(guān)鍵是會(huì)根據(jù)三角函數(shù)的圖象變換法則求出變換后所得圖象對(duì)應(yīng)的函數(shù)解析式，注意口訣“左加右減，上加下減，橫變SKIPIF1<0，縱變A”在解題中的應(yīng)用.二、填空題2．（2022·四川廣安·廣安二中校考模擬預(yù)測(cè)）已知曲線SKIPIF1<0相鄰對(duì)稱軸之間的距離為SKIPIF1<0，且函數(shù)SKIPIF1<0在SKIPIF1<0處取得最大值，則下列結(jié)論正確的序號(hào)是.①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的取值范圍是SKIPIF1<0；②將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后所對(duì)應(yīng)的函數(shù)為偶函數(shù)；③函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0；④函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn).【答案】①③【分析】根據(jù)題意確定函數(shù)周期，求得SKIPIF1<0，先討論SKIPIF1<0時(shí)情況，對(duì)于①，由函數(shù)SKIPIF1<0在SKIPIF1<0處取得最大值，可得SKIPIF1<0