新高考數(shù)學(xué)一輪復(fù)習(xí)分層提升練習(xí)第11練 對(duì)數(shù)與對(duì)數(shù)函數(shù)（含解析）

第11練對(duì)數(shù)與對(duì)數(shù)函數(shù)（精練）【A組

在基礎(chǔ)中考查功底】一、單選題1．（2023·天津·統(tǒng)考二模）已知SKIPIF1<0，則SKIPIF1<0（

）A．3 B．5 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)指對(duì)運(yùn)算化簡(jiǎn)SKIPIF1<0，再根據(jù)對(duì)數(shù)運(yùn)算法則計(jì)算SKIPIF1<0的值.【詳解】SKIPIF1<0，SKIPIF1<0SKIPIF1<0.故選：A.2．（2023·山西陽(yáng)泉·統(tǒng)考三模）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0存在零點(diǎn)．則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用函數(shù)的單調(diào)性的性質(zhì)及函數(shù)零點(diǎn)的存在性定理即可求解.【詳解】由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，得函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0存在零點(diǎn)，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)m的取值范圍是SKIPIF1<0.故選：B.3．（2023·浙江紹興·統(tǒng)考模擬預(yù)測(cè)）基本再生數(shù)SKIPIF1<0與世代間隔SKIPIF1<0是新冠肺炎的流行病學(xué)基本參數(shù)，基本再生數(shù)指一個(gè)感染者傳染的平均人數(shù)，世代間隔指相鄰兩代間傳染所需的平均時(shí)間.在新冠肺炎疫情初始階段，可以用指數(shù)模型：SKIPIF1<0（其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)）描述累計(jì)感染病例數(shù)SKIPIF1<0隨時(shí)間SKIPIF1<0（單位：天）的變化規(guī)律，指數(shù)增長(zhǎng)率SKIPIF1<0與SKIPIF1<0，SKIPIF1<0近似滿足SKIPIF1<0.有學(xué)者基于已有數(shù)據(jù)估計(jì)出SKIPIF1<0，SKIPIF1<0，據(jù)此，在新冠肺炎疫情初始階段，累計(jì)感染病例數(shù)增加SKIPIF1<0倍需要的時(shí)間約為（

）（參考數(shù)據(jù)：SKIPIF1<0，SKIPIF1<0）A．SKIPIF1<0天 B．SKIPIF1<0天 C．SKIPIF1<0天 D．SKIPIF1<0天【答案】B【分析】根據(jù)所給模型求得SKIPIF1<0，令SKIPIF1<0，求得SKIPIF1<0，根據(jù)條件可得方程SKIPIF1<0，然后解出SKIPIF1<0即可．【詳解】把SKIPIF1<0，SKIPIF1<0代入SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，兩邊取對(duì)數(shù)得SKIPIF1<0，解得SKIPIF1<0．故選：B.4．（2023春·貴州·高三校聯(lián)考期中）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】SKIPIF1<0用對(duì)數(shù)函數(shù)的單調(diào)性和SKIPIF1<0比較，SKIPIF1<0用指數(shù)函數(shù)的單調(diào)性和SKIPIF1<0比較，SKIPIF1<0用對(duì)數(shù)函數(shù)的單調(diào)性和SKIPIF1<0比較，即可判斷大小關(guān)系.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0為減函數(shù)，所以SKIPIF1<0，即SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0為增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0為增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．故選：D5．（2023·云南·校聯(lián)考二模）函數(shù)SKIPIF1<0的圖象大致形如（

）A． B． C． D．【答案】A【分析】根據(jù)函數(shù)的奇偶性和函數(shù)值等知識(shí)確定正確答案.【詳解】依題意SKIPIF1<0，SKIPIF1<0為偶函數(shù)，則SKIPIF1<0為偶函數(shù)，又SKIPIF1<0，則SKIPIF1<0.故選A．6．（2023春·黑龍江哈爾濱·高三哈爾濱市第十三中學(xué)校?？奸_(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0．若SKIPIF1<0，且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】C【分析】根據(jù)函數(shù)圖象得SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，利用對(duì)勾函數(shù)的圖象與性質(zhì)即可求出其范圍.【詳解】由SKIPIF1<0得SKIPIF1<0．根據(jù)函數(shù)SKIPIF1<0的圖象及SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．令SKIPIF1<0，根據(jù)對(duì)勾函數(shù)的圖像與性質(zhì)易得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0．故SKIPIF1<0，故選：C.7．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0，過(guò)定點(diǎn)SKIPIF1<0的直線SKIPIF1<0與坐標(biāo)軸的正半軸相交，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】求出SKIPIF1<0，代入直線方程，再根據(jù)基本不等式可求出結(jié)果.【詳解】令SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0．當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，等號(hào)成立，故選：C【點(diǎn)睛】易錯(cuò)點(diǎn)睛：利用基本不等式求最值時(shí)，要注意其必須滿足的三個(gè)條件：（1）“一正二定三相等”“一正”就是各項(xiàng)必須為正數(shù)；（2）“二定”就是要求和的最小值，必須把構(gòu)成和的二項(xiàng)之積轉(zhuǎn)化成定值；要求積的最大值，則必須把構(gòu)成積的因式的和轉(zhuǎn)化成定值；（3）“三相等”是利用基本不等式求最值時(shí)，必須驗(yàn)證等號(hào)成立的條件，若不能取等號(hào)則這個(gè)定值就不是所求的最值，這也是最容易發(fā)生錯(cuò)誤的地方.8．（2023秋·江蘇無(wú)錫·高三統(tǒng)考期末）函數(shù)SKIPIF1<0的部分圖象大致為（

）．A． B．C． D．【答案】A【分析】先求出定義域，由SKIPIF1<0得到SKIPIF1<0為偶函數(shù)，結(jié)合函數(shù)在SKIPIF1<0上函數(shù)值的正負(fù)，排除BC，結(jié)合函數(shù)圖象的走勢(shì)，排除D，得到正確答案.【詳解】SKIPIF1<0變形為SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，故SKIPIF1<0為偶函數(shù)，關(guān)于y軸對(duì)稱．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，排除BC，又SKIPIF1<0時(shí)，SKIPIF1<0，故排除D，A正確.故選：A．9．（2023·河南周口·統(tǒng)考模擬預(yù)測(cè)）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】運(yùn)用對(duì)數(shù)的運(yùn)算法則和指數(shù)函數(shù)的性質(zhì)求解.【詳解】SKIPIF1<0，對(duì)于指數(shù)函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；故選：A.10．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，若命題“SKIPIF1<0，SKIPIF1<0”為假命題，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】A【分析】根據(jù)題意，轉(zhuǎn)化為命題“SKIPIF1<0，SKIPIF1<0”為真命題．利用不等式恒成立得出關(guān)于SKIPIF1<0的不等式求解.【詳解】由題意知SKIPIF1<0且SKIPIF1<0，命題“SKIPIF1<0，SKIPIF1<0”為真命題，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，其最小值為SKIPIF1<0，則由SKIPIF1<0恒成立得SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0為增函數(shù)，故SKIPIF1<0，得SKIPIF1<0．綜上，SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選：A11．（2023·河北·高三學(xué)業(yè)考試）若函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）在區(qū)間SKIPIF1<0上的最大值比最小值多2，則SKIPIF1<0（

）A．2或SKIPIF1<0 B．3或SKIPIF1<0 C．4或SKIPIF1<0 D．2或SKIPIF1<0【答案】A【分析】分別討論SKIPIF1<0和SKIPIF1<0，然后利用對(duì)數(shù)函數(shù)的單調(diào)性列方程即可得解.【詳解】由題意SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0（舍去），①當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在定義域內(nèi)為增函數(shù)，則由題意得SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去）；②當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在定義域內(nèi)為減函數(shù)，則由題意得SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，解得SKIPIF1<0；綜上可得：SKIPIF1<0或SKIPIF1<0.故選：A.【點(diǎn)睛】本題考查了分類討論思想的應(yīng)用，考查了對(duì)數(shù)函數(shù)單調(diào)性的應(yīng)用，屬于基礎(chǔ)題.12．（2023·全國(guó)·高三專題練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)題意，化為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即可比較SKIPIF1<0的大小關(guān)系，然后SKIPIF1<0作商即可比較SKIPIF1<0的大小，從而得到結(jié)果.【詳解】由題設(shè)知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，于是SKIPIF1<0.故選：C.二、多選題13．（2023·湖南·鉛山縣第一中學(xué)校聯(lián)考二模）下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【分析】根據(jù)指數(shù)以及對(duì)數(shù)的運(yùn)算性質(zhì)即可結(jié)合選項(xiàng)逐一求解.【詳解】對(duì)于A,SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，故A錯(cuò)誤，對(duì)于B,由于SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故B正確，對(duì)于C,SKIPIF1<0，所以C錯(cuò)誤，對(duì)于D，由于SKIPIF1<0，所以SKIPIF1<0，故D正確，故選：BD14．（2023·全國(guó)·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，若SKIPIF1<0，則滿足條件的實(shí)數(shù)SKIPIF1<0可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【分析】根據(jù)對(duì)數(shù)函數(shù)和正弦函數(shù)的圖象，對(duì)a分類討論，結(jié)合對(duì)數(shù)函數(shù)、正弦函數(shù)的單調(diào)性求解即可.【詳解】函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，如圖，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，由圖可知，函數(shù)SKIPIF1<0在SKIPIF1<0上，有SKIPIF1<0，得SKIPIF1<0所以SKIPIF1<0，解得SKIPIF1<0，結(jié)合選項(xiàng)，實(shí)數(shù)a可以是SKIPIF1<0和SKIPIF1<0.故選：BD.三、填空題15．（2023·上?！じ呷龑ｎ}練習(xí)）若實(shí)數(shù)SKIPIF1<0、SKIPIF1<0滿足SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0______________．【答案】SKIPIF1<0【分析】根據(jù)指數(shù)式與對(duì)數(shù)式的關(guān)系，將SKIPIF1<0轉(zhuǎn)化為指數(shù)式，再根據(jù)指數(shù)運(yùn)算公式求值.【詳解】由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.16．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0且SKIPIF1<0的圖像恒過(guò)定點(diǎn)SKIPIF1<0，且點(diǎn)SKIPIF1<0在圓SKIPIF1<0外，則符合條件的整數(shù)SKIPIF1<0的取值可以為_(kāi)_________.（寫(xiě)出一個(gè)值即可）【答案】SKIPIF1<0（不唯一，取SKIPIF1<0的整數(shù)即可）【分析】先求定點(diǎn)SKIPIF1<0的坐標(biāo)，結(jié)合點(diǎn)在圓外以及圓的限制條件可得SKIPIF1<0的取值.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的圖像恒過(guò)定點(diǎn)SKIPIF1<0，所以SKIPIF1<0；因?yàn)辄c(diǎn)SKIPIF1<0在圓SKIPIF1<0外，所以SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；又SKIPIF1<0為整數(shù)，所以SKIPIF1<0的取值可以為SKIPIF1<0.故答案為：SKIPIF1<0（不唯一，取SKIPIF1<0的整數(shù)即可）.17．（2023·全國(guó)·高三專題練習(xí)）一種藥在病人血液中的量保持1000mg以上才有療效，而低于500mg病人就有危險(xiǎn)．現(xiàn)給某病人靜脈注射了這種藥2000mg，如果藥在血液中以每小時(shí)10%的比例衰減，為了充分發(fā)揮藥物的利用價(jià)值，那么從現(xiàn)在起經(jīng)過(guò)______小時(shí)內(nèi)向病人的血液補(bǔ)充這種藥，才能保持療效．（附：SKIPIF1<0，SKIPIF1<0，精確到0.1h）【答案】6.6【分析】寫(xiě)出血液中藥物含量關(guān)于時(shí)間的關(guān)系式，解不等式求出答案.【詳解】設(shè)SKIPIF1<0h后血液中的藥物量為SKIPIF1<0mg，則有SKIPIF1<0，令SKIPIF1<0得：SKIPIF1<0故從現(xiàn)在起經(jīng)過(guò)6.6h內(nèi)向病人的血液補(bǔ)充這種藥，才能保持療效.故答案為：6.618．（2023·河南平頂山·葉縣高級(jí)中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的取值范圍為_(kāi)_____．【答案】SKIPIF1<0【分析】令SKIPIF1<0，即可判斷SKIPIF1<0在SKIPIF1<0上的單調(diào)性，依題意可得SKIPIF1<0在SKIPIF1<0上為減函數(shù)，即可得到不等式組，解得即可.【詳解】令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0為減函數(shù)，所以由復(fù)合函數(shù)的單調(diào)性可知SKIPIF1<0在SKIPIF1<0上為減函數(shù)，則SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0的取值范圍為SKIPIF1<0．故答案為：SKIPIF1<0【B組

在綜合中考查能力】一、單選題1．（2023·天津河西·統(tǒng)考一模）已知SKIPIF1<0SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】?jī)蛇吶?duì)數(shù)，根據(jù)對(duì)數(shù)的運(yùn)算性質(zhì)、法則化簡(jiǎn)即可得解.【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0（舍去）故選：C2．（2023·江西南昌·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0，則SKIPIF1<0（

）A．11或SKIPIF1<0 B．11或SKIPIF1<0 C．12或SKIPIF1<0 D．10或SKIPIF1<0【答案】A【分析】對(duì)SKIPIF1<0兩邊同時(shí)取對(duì)數(shù)，可解得SKIPIF1<0或SKIPIF1<0，討論SKIPIF1<0或SKIPIF1<0時(shí)SKIPIF1<0的值，即可得出答案.【詳解】由SKIPIF1<0，兩邊取對(duì)數(shù)得SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<08，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，綜上，SKIPIF1<0SKIPIF1<0或SKIPIF1<0，故選：A.3．（2023·全國(guó)·高三專題練習(xí)）如圖為一臺(tái)冷軋機(jī)的示意圖.冷軋機(jī)由若干對(duì)軋輥組成，厚度為SKIPIF1<0（單位：SKIPIF1<0）的帶鋼從一端輸入，經(jīng)過(guò)各對(duì)車輥逐步減薄后輸出，厚度變?yōu)镾KIPIF1<0（單位：SKIPIF1<0）.若SKIPIF1<0，每對(duì)軋輥的減薄率SKIPIF1<0不超過(guò)4%，則冷軋機(jī)至少需要安裝軋輥的對(duì)數(shù)為（

）（一對(duì)軋輥減薄率SKIPIF1<0）A．14 B．15 C．16 D．17【答案】D【分析】根據(jù)題意可得SKIPIF1<0，兩邊取對(duì)數(shù)能求出冷軋機(jī)至少需要安裝軋輥的對(duì)數(shù).【詳解】厚度為SKIPIF1<0SKIPIF1<0的帶鋼從一端輸入經(jīng)過(guò)減薄率為4%的SKIPIF1<0對(duì)軋輥后厚度為SKIPIF1<0，過(guò)各對(duì)車輥逐步減薄后輸出，厚度變?yōu)镾KIPIF1<0，則SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故選：D.4．（2023秋·安徽宣城·高三統(tǒng)考期末）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】先構(gòu)造函數(shù)SKIPIF1<0，對(duì)函數(shù)求導(dǎo)，利用導(dǎo)函數(shù)的單調(diào)性可得到SKIPIF1<0，且SKIPIF1<0，再結(jié)合SKIPIF1<0，即可得到SKIPIF1<0，進(jìn)而即可得到答案．【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0，且SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，故選：D．5．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，若對(duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】求出函數(shù)SKIPIF1<0的最大值，結(jié)合已知條件可得出SKIPIF1<0，進(jìn)而可求得實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.所以，SKIPIF1<0.若對(duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，則SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0.因此，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B.【點(diǎn)睛】結(jié)論點(diǎn)睛：利用參變量分離法求解函數(shù)不等式恒（能）成立，可根據(jù)以下原則進(jìn)行求解：（1）SKIPIF1<0，SKIPIF1<0；（2）SKIPIF1<0，SKIPIF1<0；（3）SKIPIF1<0，SKIPIF1<0；（4）SKIPIF1<0，SKIPIF1<0.6．（2023·安徽銅陵·統(tǒng)考三模）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用換底公式得到SKIPIF1<0，再利用基本不等式比較即可；同理得到SKIPIF1<0的大小.【詳解】解：因?yàn)镾KIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0；因?yàn)镾KIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故選：A7．（2023·全國(guó)·高三專題練習(xí)）若SKIPIF1<0且SKIPIF1<0在SKIPIF1<0上恒正，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)題意，結(jié)合對(duì)數(shù)型復(fù)合函數(shù)的性質(zhì)，列出不等式，即可容易列出不等式，即可容易求得參數(shù)范圍.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0,SKIPIF1<0且SKIPIF1<0，在SKIPIF1<0上恒正,令SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的對(duì)稱軸方程為SKIPIF1<0，知SKIPIF1<0，即SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，滿足SKIPIF1<0或SKIPIF1<0或SKIPIF1<0解不等式得：SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：SKIPIF1<0.【點(diǎn)睛】本題考查對(duì)數(shù)型復(fù)合函數(shù)的性質(zhì)，注意函數(shù)定義域即可，屬中檔題.二、多選題8．（2023春·廣東·高三校聯(lián)考階段練習(xí)）若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【分析】構(gòu)造函數(shù)SKIPIF1<0，通過(guò)函數(shù)單調(diào)性及SKIPIF1<0，比較出各式的大小關(guān)系.【詳解】設(shè)函數(shù)SKIPIF1<0，易得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：ABD9．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0單調(diào)遞增B．SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減C．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱D．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】BC【分析】由題可得函數(shù)的定義域，化簡(jiǎn)函數(shù)SKIPIF1<0，分析函數(shù)的單調(diào)性和對(duì)稱性，從而判斷選項(xiàng).【詳解】函數(shù)的定義域滿足SKIPIF1<0，即SKIPIF1<0，即函數(shù)的定義域是SKIPIF1<0，∵SKIPIF1<0，設(shè)SKIPIF1<0，則函數(shù)在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，又函數(shù)SKIPIF1<0單調(diào)遞增，由復(fù)合函數(shù)單調(diào)性可知函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，故A錯(cuò)誤，B正確；因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0，即函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0對(duì)稱，故C正確；又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以D錯(cuò)誤．故選：BC．三、填空題10．（2023·全國(guó)·高三專題練習(xí)）若SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)__________.【答案】SKIPIF1<0【分析】分離參數(shù)，將恒成立問(wèn)題轉(zhuǎn)化為函數(shù)最值問(wèn)題，根據(jù)單調(diào)性可得.【詳解】因?yàn)镾KIPIF1<0，不等式SKIPIF1<0恒成立，所以SKIPIF1<0對(duì)SKIPIF1<0恒成立.記SKIPIF1<0，SKIPIF1<0，只需SKIPIF1<0.因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<011．（2023·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0(SKIPIF1<0且SKIPIF1<0)的圖像經(jīng)過(guò)定點(diǎn)SKIPIF1<0，則函數(shù)SKIPIF1<0的最大值為_(kāi)__________.【答案】SKIPIF1<0【分析】由題知SKIPIF1<0，進(jìn)而得SKIPIF1<0，進(jìn)而結(jié)合復(fù)合函數(shù)求值域即可.【詳解】由于函數(shù)SKIPIF1<0是由函數(shù)SKIPIF1<0(SKIPIF1<0且SKIPIF1<0)向左平移SKIPIF1<0個(gè)單位,再向下平移SKIPIF1<0個(gè)單位得到,所以函數(shù)SKIPIF1<0(SKIPIF1<0且SKIPIF1<0)的圖像經(jīng)過(guò)定點(diǎn)SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立.故答案為:SKIPIF1<0.【點(diǎn)睛】本題考查指數(shù)型函數(shù)過(guò)定點(diǎn)問(wèn)題，對(duì)數(shù)型復(fù)合函數(shù)值域求解問(wèn)題，考查運(yùn)算求解能力，是中檔題.本題解題的關(guān)鍵在于根據(jù)已知條件得SKIPIF1<0，進(jìn)而利用配方法得SKIPIF1<0，再結(jié)合二次函數(shù)值域求解即可.12．（2023·海南?？凇ば？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0和SKIPIF1<0的圖象分別交于點(diǎn)SKIPIF1<0，則SKIPIF1<0________．【答案】SKIPIF1<0【分析】確定SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，根據(jù)函數(shù)單調(diào)遞增得到SKIPIF1<0，得到答案.【詳解】SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0，即SKIPIF1<0，設(shè)SKIPIF1<0，函數(shù)在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.四、解答題13．（2023春·全國(guó)·高三校聯(lián)考開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0的最小值為0．(1)求實(shí)數(shù)a的值；(2)設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，判斷SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大?。敬鸢浮?1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）求出函數(shù)的導(dǎo)函數(shù)，分SKIPIF1<0、SKIPIF1<0兩種情況討論，分別求出函數(shù)的單調(diào)區(qū)間，即可得到函數(shù)的最小值為SKIPIF1<0，從而得到SKIPIF1<0，再令SKIPIF1<0，利用導(dǎo)數(shù)說(shuō)明函數(shù)的單調(diào)性，即可得到SKIPIF1<0值，從而得解；（2）由（1）可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)兩邊取對(duì)數(shù)得到SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，根據(jù)函數(shù)值的情況判斷SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，即可判斷SKIPIF1<0，從而得解.【詳解】（1）解：由題意得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，無(wú)最小值，不滿足題意．當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．所以SKIPIF1<0的最小值為SKIPIF1<0，即SKIPIF1<0．設(shè)SKIPIF1<0，則SKIPIF1<0．令SKIPIF1<0，得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0．故SKIPIF1<0的解只有SKIPIF1<0，綜上所述，SKIPIF1<0．（2）解：由（1）可得SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立．當(dāng)SKIPIF1<0時(shí)，不等式兩邊取對(duì)數(shù)，得SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立．當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．故SKIPIF1<0，所以SKIPIF1<0．綜上所述，SKIPIF1<0．【C組

在創(chuàng)新中考查思維】一、解答題1．（2023·河北邢臺(tái)·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0，證明：(1)SKIPIF1<0；(2)SKIPIF1<0．【答案】(1)證明見(jiàn)解析(2)證明見(jiàn)解析【分析】（1）利用導(dǎo)數(shù)研究函數(shù)SKIPIF1<0的單調(diào)性可得SKIPIF1<0，即證SKIPIF1<0，進(jìn)而SKIPIF1<0，即證SKIPIF1<0，原不等式即可證明；（2）易知SKIPIF1<0時(shí)不等式成立；當(dāng)SKIPIF1<0時(shí)，利用二階導(dǎo)數(shù)研究函數(shù)SKIPIF1<0的單調(diào)性可得SKIPIF1<0，即SKIPIF1<0(SKIPIF1<0)，變形即可證明.【詳解】（1）令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，等號(hào)僅當(dāng)SKIPIF1<0時(shí)成立，即SKIPIF1<0，從而SKIPIF1<0，所以SKIPIF1<0．綜上，SKIPIF1<0．（2）顯然SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0成立．令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，等號(hào)僅當(dāng)SKIPIF1<0時(shí)成立，從而可得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減．由（1）知，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．又當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0．故SKIPIF1<0時(shí)，SKIPIF1<0．【點(diǎn)睛】在解決類似的問(wèn)題時(shí)，要熟練應(yīng)用導(dǎo)數(shù)研究函數(shù)的單調(diào)性與最值，善于培養(yǎng)轉(zhuǎn)化的數(shù)學(xué)思想，學(xué)會(huì)構(gòu)造新函數(shù)，利用導(dǎo)數(shù)或二階求導(dǎo)研究新函數(shù)的性質(zhì)即可解決問(wèn)題.二、單選題2．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)對(duì)數(shù)的運(yùn)算，計(jì)算可得SKIPIF1<0，則SKIPIF1<0.構(gòu)造函數(shù)SKIPIF1<0，根據(jù)導(dǎo)函數(shù)得到函數(shù)的單調(diào)性，即可得出SKIPIF1<0，根據(jù)對(duì)數(shù)函數(shù)的單調(diào)性即可得出SKIPIF1<0；先證明當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.然后根據(jù)二倍角公式以及不等式的性質(zhì)，推得SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0，所以，SKIPIF1<0.令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0恒成立，所以，SKIPIF1<0在R上單調(diào)遞減，所以，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0.故選：B．【點(diǎn)睛】方法點(diǎn)睛：對(duì)SKIPIF1<0變形后，作差構(gòu)造函數(shù)，根據(jù)導(dǎo)函數(shù)得到函數(shù)的單調(diào)性，即可得出值的大小關(guān)系.3．（2023·全國(guó)·高三專題練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用正切函數(shù)單調(diào)性借助1比較b，c大小；根據(jù)對(duì)數(shù)結(jié)構(gòu)構(gòu)造函數(shù)SKIPIF1<0比較a，b大小，即可解答.【詳解】因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，于是SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：A4．（2023·陜西寶雞·?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有6個(gè)不同的零點(diǎn)，且最小的零點(diǎn)為SKIPIF1<0，則SKIPIF1<0（

）．A．6 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】B【分析】根據(jù)函數(shù)圖象變換，畫(huà)出圖像，找到對(duì)稱軸，進(jìn)而數(shù)形結(jié)合求解即可.【詳解】由函數(shù)SKIPIF1<0的圖象，經(jīng)過(guò)翻折變換，可得函數(shù)SKIPIF1<0的圖象，再經(jīng)過(guò)向右平移1個(gè)單位，可得SKIPIF1<0的圖象，最終經(jīng)過(guò)翻折變換，可得SKIPIF1<0的圖象，如下圖：則函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，令SKIPIF1<0因?yàn)楹瘮?shù)SKIPIF1<0最小的零點(diǎn)為SKIPIF1<0，且SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有4個(gè)零點(diǎn)，所以，要使函數(shù)SKIPIF1<0有6個(gè)不同的零點(diǎn)，且最小的零點(diǎn)為SKIPIF1<0,則SKIPIF1<0，或SKIPIF1<