新高考數(shù)學(xué)一輪復(fù)習(xí)分層提升練習(xí)第10練 指數(shù)與指數(shù)函數(shù)（含解析）

第10練指數(shù)與指數(shù)函數(shù)（精練）【A組

在基礎(chǔ)中考查功底】一、單選題1．（2023·全國(guó)·學(xué)軍中學(xué)校聯(lián)考二模）已知集合SKIPIF1<0或x≤?2}，則SKIPIF1<0（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或x≤?2}C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】解法一：根據(jù)題意求集合M，進(jìn)而根據(jù)交集運(yùn)算求解；解法二：取特值檢驗(yàn)排除.【詳解】解法一：由題可得SKIPIF1<0或SKIPIF1<0或x≤?2}，所以SKIPIF1<0或x≤?2}.故選：B.解法二：由題可得SKIPIF1<0，所以SKIPIF1<0，故排除A、D；又SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，故排除C.故選：B.2．（2023·北京朝陽(yáng)·高三專題練習(xí)）“SKIPIF1<0”是“SKIPIF1<0”的（

）A．充分而不必要條件 B．必要而不充分條件C．充分必要條件 D．既不充分也不必要條件【答案】A【分析】根據(jù)充分條件和必要條件的定義，結(jié)合指數(shù)函數(shù)的單調(diào)性即可得出答案.【詳解】因?yàn)橹笖?shù)函數(shù)SKIPIF1<0單調(diào)遞增，由SKIPIF1<0可得：SKIPIF1<0，充分性成立，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，但不一定SKIPIF1<0，必要性不成立，故選：A3．（2023·江西南昌·南昌縣蓮塘第一中學(xué)校聯(lián)考二模）為了預(yù)防某種病毒，某學(xué)校需要通過(guò)噴灑藥物對(duì)教室進(jìn)行全面消毒．出于對(duì)學(xué)生身體健康的考慮，相關(guān)部門(mén)規(guī)定空氣中這種藥物的濃度不超過(guò)0.25毫克/立方米時(shí)，學(xué)生方可進(jìn)入教室．已知從噴灑藥物開(kāi)始，教室內(nèi)部的藥物濃度y（毫克/立方米）與時(shí)間t（分鐘）之間的函數(shù)關(guān)系為SKIPIF1<0，函數(shù)的圖像如圖所示．如果早上7：30就有學(xué)生進(jìn)入教室，那么開(kāi)始噴灑藥物的時(shí)間最遲是（

）A．7：00 B．6：40 C．6：30 D．6：00【答案】A【分析】函數(shù)的圖像過(guò)點(diǎn)SKIPIF1<0，代入函數(shù)的解析式求得未知系數(shù)a，解函數(shù)不等式即可.【詳解】根據(jù)函數(shù)的圖像，可得函數(shù)的圖像過(guò)點(diǎn)SKIPIF1<0，由函數(shù)圖像連續(xù)，代入函數(shù)的解析式，可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．所以如果7：30學(xué)生進(jìn)入教室，那么開(kāi)始噴灑藥物的時(shí)間最遲是7：00．故選：A．4．（2023·陜西商洛·統(tǒng)考二模）函數(shù)SKIPIF1<0的部分圖象大致是（

）A． B．C． D．【答案】C【分析】奇偶性定義判斷函數(shù)奇偶性，結(jié)合SKIPIF1<0上函數(shù)符號(hào)，應(yīng)用排除法即可得答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0且定義域?yàn)镽，所以SKIPIF1<0是奇函數(shù)，則SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，排除A、B.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除D.故選：C5．（2023秋·吉林遼源·高三校聯(lián)考期末）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】構(gòu)造函數(shù)SKIPIF1<0，研究函數(shù)SKIPIF1<0的單調(diào)性與奇偶性，利用函數(shù)性質(zhì)解不等式.【詳解】令SKIPIF1<0，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，所以函數(shù)SKIPIF1<0為定義域內(nèi)的奇函數(shù)，且在SKIPIF1<0上單調(diào)遞增；則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0為定義域內(nèi)的奇函數(shù)，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故實(shí)數(shù)a的取值范圍是SKIPIF1<0.故選：C6．（2023秋·湖南長(zhǎng)沙·高三?？茧A段練習(xí)）設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過(guò)SKIPIF1<0的最大整數(shù)，則SKIPIF1<0稱取整函數(shù)，例如：SKIPIF1<0，SKIPIF1<0已知SKIPIF1<0則函數(shù)SKIPIF1<0的值域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】分離常數(shù)得SKIPIF1<0，進(jìn)而求得SKIPIF1<0，從而可得答案.【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0當(dāng)SKIPIF1<0.故選:D二、多選題7．（2023春·重慶渝中·高三重慶巴蜀中學(xué)?？茧A段練習(xí)）若SKIPIF1<0，其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，則下列命題正確的是（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 B．SKIPIF1<0在SKIPIF1<0上單調(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中心對(duì)稱【答案】BC【分析】根據(jù)復(fù)合函數(shù)的單調(diào)性判斷A、B，根據(jù)奇偶性的定義判斷函數(shù)為偶函數(shù)，即可判斷C、D.【詳解】因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故A錯(cuò)誤，B正確；又SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)，函數(shù)圖象關(guān)于SKIPIF1<0軸對(duì)稱，即關(guān)于直線SKIPIF1<0對(duì)稱，故C正確，D錯(cuò)誤；故選：BC8．（2023春·云南昭通·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，下列說(shuō)法中正確的是（

）A．SKIPIF1<0不是周期函數(shù) B．SKIPIF1<0在(0，SKIPIF1<0)上是單調(diào)遞增函數(shù)C．SKIPIF1<0在(0，SKIPIF1<0)內(nèi)有且只有一個(gè)零點(diǎn) D．SKIPIF1<0關(guān)于點(diǎn)(SKIPIF1<0，0)對(duì)稱【答案】BCD【分析】根據(jù)周期函數(shù)的定義、指數(shù)函數(shù)、正弦函數(shù)、余弦函數(shù)的單調(diào)性，結(jié)合零點(diǎn)定義和點(diǎn)對(duì)稱的性質(zhì)逐一判斷即可.【詳解】∵SKIPIF1<0，∴SKIPIF1<0是周期函數(shù)，A錯(cuò)誤；當(dāng)x∈(0，SKIPIF1<0)時(shí)，sinx是增函數(shù)，cosx是減函數(shù)，∴SKIPIF1<0是增函數(shù)，SKIPIF1<0是減函數(shù)，SKIPIF1<0是增函數(shù)，∴SKIPIF1<0是增函數(shù)，B對(duì)；由SKIPIF1<0得sinx=cosx，因?yàn)镾KIPIF1<0，所以有SKIPIF1<0，C對(duì)；∵SKIPIF1<0，∴SKIPIF1<0關(guān)于點(diǎn)(SKIPIF1<0，0)對(duì)稱，D對(duì)，故選：BCD.三、填空題9．（2023·全國(guó)·高三專題練習(xí)）SKIPIF1<0________.【答案】19【分析】根據(jù)指數(shù)冪的運(yùn)算性質(zhì)即可求解.【詳解】SKIPIF1<0SKIPIF1<0.故答案為：1910．（2023·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0______.【答案】4【分析】根據(jù)函數(shù)的奇偶性，結(jié)合代入法進(jìn)行求解即可.【詳解】設(shè)SKIPIF1<0，SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0.故答案為：SKIPIF1<011．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的值域?yàn)開(kāi)_______.【答案】SKIPIF1<0【分析】根據(jù)二次函數(shù)的圖像性質(zhì)和指數(shù)函數(shù)的性質(zhì)求解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故答案為:SKIPIF1<0.12．（2023·全國(guó)·高三專題練習(xí)）寫(xiě)出一個(gè)同時(shí)滿足下列三個(gè)性質(zhì)的函數(shù)SKIPIF1<0__________．①若SKIPIF1<0，則SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)函數(shù)的三個(gè)性質(zhì)，列出符合條件的函數(shù)即可.【詳解】比如SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，又SKIPIF1<0，也即SKIPIF1<0成立，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．故答案為：SKIPIF1<0.13．（2023春·河南鄭州·高三校考階段練習(xí)）已知函數(shù)SKIPIF1<0，若實(shí)數(shù)a，SKIPIF1<0滿足SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0___________.【答案】SKIPIF1<0【分析】根據(jù)指數(shù)式的運(yùn)算和指數(shù)函數(shù)函數(shù)值的運(yùn)算求解.【詳解】SKIPIF1<0,由SKIPIF1<0，可得SKIPIF1<0所以SKIPIF1<0①，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0②，聯(lián)立①②解得SKIPIF1<0SKIPIF1<0，故答案為:SKIPIF1<0.14．（2023·全國(guó)·高三練習(xí)）若關(guān)于SKIPIF1<0的方程SKIPIF1<0有實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)_____．【答案】SKIPIF1<0【分析】方程SKIPIF1<0有實(shí)根可轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上有解，根據(jù)SKIPIF1<0的范圍求解SKIPIF1<0的取值范圍即可.【詳解】方程SKIPIF1<0有實(shí)根，所以SKIPIF1<0有實(shí)根，令SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有解，又因?yàn)楫?dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0四、解答題15．（2023·全國(guó)·高三專題練習(xí)）已知向量SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，求：（1）實(shí)數(shù)m的取值范圍；（2）函數(shù)SKIPIF1<0定義域.【答案】（1）SKIPIF1<0；（2）SKIPIF1<0.【分析】（1）根據(jù)數(shù)量積的坐標(biāo)表示，求解不等式即可得出答案；（2）根據(jù)（1）中m的取值范圍，再運(yùn)用指數(shù)函數(shù)的單調(diào)性求解定義域即可.【詳解】（1）由題意得，SKIPIF1<0，SKIPIF1<0，即m的取值范圍為SKIPIF1<0；（2）由題意知SKIPIF1<0，即SKIPIF1<0，由（1）知SKIPIF1<0，根據(jù)指數(shù)函數(shù)的單調(diào)性得：SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.16．（2023·全國(guó)·高三專題練習(xí)）已知a>0，且a≠1，若函數(shù)y＝|ax－2|與y＝3a的圖象有兩個(gè)交點(diǎn)，求實(shí)數(shù)a的取值范圍．【答案】SKIPIF1<0【分析】討論0<a<1或a>1，作出函數(shù)y＝|a－2|與y＝3a的圖象，由數(shù)形結(jié)合即可求解.【詳解】①當(dāng)0<a<1時(shí)，在同一平面直角坐標(biāo)系中作出函數(shù)y＝|a－2|與y＝3a的圖象如圖1．若直線y＝3a與函數(shù)y＝|a－2|(0<a<1)的圖象有兩個(gè)交點(diǎn)，則由圖象可知0<3a<2，所以0<a<SKIPIF1<0．②當(dāng)a>1時(shí)，在同一平面直角坐標(biāo)系中作出函數(shù)y＝|a－2|與y＝3a的圖象如圖2．若直線y＝3a與函數(shù)y＝|a－2|(a>1)的圖象有兩個(gè)交點(diǎn)，則由圖象可知0<3a<2，此時(shí)無(wú)解．所以實(shí)數(shù)a的取值范圍是SKIPIF1<0．【B組

在綜合中考查能力】一、單選題1．（2023·甘肅武威·統(tǒng)考三模）函數(shù)SKIPIF1<0的圖象大致是（

）A． B． C． D．【答案】D【分析】SKIPIF1<0，排除BC，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，A不滿足，排除，得到答案.【詳解】SKIPIF1<0，排除BC；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，A不滿足，排除.故選：D2．（2023·四川綿陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0在定義域SKIPIF1<0上滿足SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上是減函數(shù)，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)題意可得函數(shù)SKIPIF1<0在定義域SKIPIF1<0上奇函數(shù)，進(jìn)而可得SKIPIF1<0在SKIPIF1<0上是減函數(shù)，根據(jù)題意結(jié)合單調(diào)性解不等式即可.【詳解】∵SKIPIF1<0，即SKIPIF1<0，故函數(shù)SKIPIF1<0在定義域SKIPIF1<0上奇函數(shù)，若SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0在SKIPIF1<0上是減函數(shù)，∵SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0.故選：A.3．（2023·湖南益陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】作商與1比較進(jìn)而比較指數(shù)冪的大小,再構(gòu)造函數(shù)SKIPIF1<0，根據(jù)單調(diào)性比較函數(shù)值大小即可.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0,令設(shè)SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，且SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0故選:SKIPIF1<04．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0是偶函數(shù)，SKIPIF1<0是奇函數(shù)，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用函數(shù)奇偶性的定義可求得函數(shù)SKIPIF1<0的解析式，再利用基本不等式可求得SKIPIF1<0的最小值.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0，即SKIPIF1<0，①又因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，即SKIPIF1<0，②聯(lián)立①②可得SKIPIF1<0，由基本不等式可得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，故函數(shù)SKIPIF1<0的最小值為SKIPIF1<0.故選：B.5．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0則實(shí)數(shù)a，b，c的大小關(guān)系為（

）A．c＞a＞b B．a(chǎn)＞b＞cC．a(chǎn)＞c＞b D．c＞b＞a【答案】A【分析】先利用作商法比較a，b的大小，再借助中間值“0.5”得到SKIPIF1<0，得到a＜c，即可得到結(jié)果．【詳解】易知SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0由SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0，所以a＜c．所以實(shí)數(shù)a，b，c的大小關(guān)系為c＞a＞b．故選：A．6．（2023·貴州·統(tǒng)考模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，則SKIPIF1<0(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)x的范圍分類討論去掉SKIPIF1<0的絕對(duì)值符號(hào)，再根據(jù)二次函數(shù)的性質(zhì)和f(x)的最小值即可求出關(guān)于a的方程SKIPIF1<0，令SKIPIF1<0，根據(jù)g(a)的單調(diào)性即可求出a的范圍．【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0的最小值為SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0是R上的增函數(shù)，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．故選：C．二、多選題7．（2023·山東青島·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0 B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0 D．當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有兩個(gè)解【答案】AC【分析】結(jié)合SKIPIF1<0在定義域內(nèi)單調(diào)性，判斷A；利用導(dǎo)數(shù)判斷函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)性，由此判斷B；判斷函數(shù)SKIPIF1<0，在SKIPIF1<0上的單調(diào)性，由此判斷C；，舉反例判斷D.【詳解】SKIPIF1<0在定義域內(nèi)單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，故A正確；當(dāng)SKIPIF1<0時(shí)，要證明SKIPIF1<0，只需證明SKIPIF1<0，故考慮構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以B錯(cuò)誤；設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，C正確；取SKIPIF1<0可得，方程SKIPIF1<0等價(jià)于SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0時(shí)，方程SKIPIF1<0只有一個(gè)解，D錯(cuò)誤.故選：AC.8．（2023·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0，SKIPIF1<0為SKIPIF1<0導(dǎo)函數(shù)，SKIPIF1<0，SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0為偶函數(shù) B．當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0恒成立C．SKIPIF1<0的值域?yàn)镾KIPIF1<0 D．SKIPIF1<0與曲線SKIPIF1<0無(wú)交點(diǎn)【答案】AD【分析】對(duì)A，由偶函數(shù)定義判斷；對(duì)B，結(jié)合指數(shù)函數(shù)的非負(fù)性，判斷存在SKIPIF1<0即可；對(duì)C，化簡(jiǎn)SKIPIF1<0，求指數(shù)函數(shù)復(fù)合型函數(shù)的值域即可.對(duì)D，聯(lián)立兩曲線，判斷方程的解即可.【詳解】對(duì)A，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0為偶函數(shù)，A對(duì)；對(duì)B，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0，SKIPIF1<0，B錯(cuò)；對(duì)C，由SKIPIF1<0可得SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，C錯(cuò)；對(duì)D，由SKIPIF1<0，方程無(wú)解，∴SKIPIF1<0與曲線SKIPIF1<0無(wú)交點(diǎn)，D對(duì).故選：AD三、填空題9．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0所過(guò)的定點(diǎn)在一次函數(shù)SKIPIF1<0的圖像上，則SKIPIF1<0的最小值為_(kāi)_________.【答案】SKIPIF1<0【分析】由指數(shù)函數(shù)性質(zhì)與基本不等式求解，【詳解】令SKIPIF1<0得SKIPIF1<0，由題意得SKIPIF1<0過(guò)的定點(diǎn)為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0的最小值為SKIPIF1<0，故答案為：SKIPIF1<010．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0(SKIPIF1<0且SKIPIF1<0)在區(qū)間SKIPIF1<0上是減函數(shù)，則實(shí)數(shù)SKIPIF1<0的取值范圍是________．【答案】SKIPIF1<0【分析】令SKIPIF1<0，則SKIPIF1<0，則由題意可得SKIPIF1<0在區(qū)間SKIPIF1<0上為減函數(shù)，SKIPIF1<0為增函數(shù)，則SKIPIF1<0，所以得SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，從而可得SKIPIF1<0，進(jìn)而可求出結(jié)果.【詳解】令SKIPIF1<0，則SKIPIF1<0，由于SKIPIF1<0且SKIPIF1<0，內(nèi)層函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上為減函數(shù)，所以外層函數(shù)SKIPIF1<0為增函數(shù)，所以SKIPIF1<0．由題意可知，不等式SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，所以SKIPIF1<0，解得SKIPIF1<0．綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<011．（2023·北京朝陽(yáng)·高三專題練習(xí)）若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則實(shí)數(shù)SKIPIF1<0的一個(gè)取值可以為_(kāi)__________.【答案】1【分析】考察函數(shù)SKIPIF1<0的圖像，就是先把SKIPIF1<0向上或向下平移SKIPIF1<0個(gè)單位（取決于SKIPIF1<0的符號(hào)），如果圖像存在小于零的部分，則再把小于零的部分以x軸為對(duì)稱軸翻折上去，最后再把整個(gè)圖像向下平移一個(gè)單位.【詳解】如果SKIPIF1<0，SKIPIF1<0SKIPIF1<0，其值域?yàn)镾KIPIF1<0，SKIPIF1<0，不符合題意；如果SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0就是把函數(shù)SKIPIF1<0的部分以x軸為對(duì)稱軸翻折上去，∴此時(shí)SKIPIF1<0的最小值為0，SKIPIF1<0的最小值為-1，值域?yàn)镾KIPIF1<0，所以SKIPIF1<0，不妨取SKIPIF1<0；故答案為：1.12．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0是定義在R上的奇函數(shù)，若不等式SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，則實(shí)數(shù)m的取值范圍是__________．【答案】SKIPIF1<0【分析】利用函數(shù)的奇偶性和單調(diào)性可得不等式SKIPIF1<0在SKIPIF1<0恒成立，換元法討論函數(shù)在給定區(qū)間的單調(diào)性和最值，結(jié)合分類討論即可求SKIPIF1<0的范圍.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0解得SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0函數(shù)為奇函數(shù)，滿足題意，所以SKIPIF1<0，因?yàn)镾KIPIF1<0在R上單調(diào)遞增，所以SKIPIF1<0在R上單調(diào)遞減，所以SKIPIF1<0在R上單調(diào)遞增，所以由SKIPIF1<0可得，SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0在SKIPIF1<0恒成立，令SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式可化為SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式SKIPIF1<0顯然成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式可化為SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0；綜上，SKIPIF1<0,故答案為:SKIPIF1<0.四、解答題13．（2023·全國(guó)·高三專題練習(xí)）設(shè)定義在SKIPIF1<0上的偶函數(shù)SKIPIF1<0和奇函數(shù)SKIPIF1<0滿足SKIPIF1<0（其中SKIPIF1<0），且SKIPIF1<0.(1)求函數(shù)SKIPIF1<0和SKIPIF1<0的解析式；(2)若SKIPIF1<0的最小值為SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的值.【答案】(1)SKIPIF1<0，SKIPIF1<0(2)SKIPIF1<0【分析】(1)由已知可得SKIPIF1<0，結(jié)合奇函數(shù)和偶函數(shù)的性質(zhì)變形求解即可；(2)令SKIPIF1<0，函數(shù)SKIPIF1<0可化為關(guān)于SKIPIF1<0的函數(shù)，結(jié)合二次函數(shù)性質(zhì)求其最小值，列方程求SKIPIF1<0的值.【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為偶函數(shù)，函數(shù)SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0(舍)，從而SKIPIF1<0，SKIPIF1<0.（2）因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0：所以SKIPIF1<0，因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.【C組

在創(chuàng)新中考查思維】一、單選題1．（2023·浙江溫州·統(tǒng)考三模）已知函數(shù)SKIPIF1<0，存在實(shí)數(shù)SKIPIF1<0使得SKIPIF1<0成立，若正整數(shù)SKIPIF1<0的最大值為6，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】分類討論SKIPIF1<0的值域，然后根據(jù)值域端點(diǎn)的倍數(shù)關(guān)系可解.【詳解】記SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0顯然存在任意正整數(shù)SKIPIF1<0，使得SKIPIF1<0成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0要使正整數(shù)SKIPIF1<0的最大值為6，則SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0顯然存在任意正整數(shù)SKIPIF1<0，使得SKIPIF1<0成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0要使正整數(shù)SKIPIF1<0的最大值為6，則SKIPIF1<0，解得SKIPIF1<0綜上，SKIPIF1<0的取值范圍為SKIPIF1<0故選：C2．（2023·北京朝陽(yáng)·二模）已知函數(shù)SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.若關(guān)于x的方程SKIPIF1<0有且僅有兩個(gè)不相等的實(shí)數(shù)解則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】利用奇函數(shù)性質(zhì)求分段函數(shù)解析式，根據(jù)指數(shù)函數(shù)性質(zhì)畫(huà)出SKIPIF1<0函數(shù)圖象，數(shù)形結(jié)合判斷不同值域范圍的函數(shù)值對(duì)應(yīng)自變量的個(gè)數(shù)，再由SKIPIF1<0有兩個(gè)解，對(duì)應(yīng)SKIPIF1<0的解的個(gè)數(shù)確定SKIPIF1<0范圍，進(jìn)而求m的范圍.【詳解】由題設(shè)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，值域?yàn)镽，函數(shù)圖象如下：當(dāng)SKIPIF1<0時(shí)，只有一個(gè)SKIPIF1<0與之對(duì)應(yīng)；當(dāng)SKIPIF1<0時(shí)，有兩個(gè)對(duì)應(yīng)自變量，記為SKIPIF1<0，則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，有三個(gè)對(duì)應(yīng)自變量且SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，有兩個(gè)對(duì)應(yīng)自變量，記為SKIPIF1<0，則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，有一個(gè)SKIPIF1<0與之對(duì)應(yīng)；令SKIPIF1<0，則SKIPIF1<0，要使SKIPIF1<0有且僅有兩個(gè)不相等的實(shí)數(shù)解，若SKIPIF1<0有三個(gè)解，則SKIPIF1<0，此時(shí)SKIPIF1<0有5個(gè)解，不滿足；若SKIPIF1<0有兩個(gè)解SKIPIF1<0且SKIPIF1<0，此時(shí)SKIPIF1<0和SKIPIF1<0各有一個(gè)解，結(jié)合圖象知，不存在這樣的SKIPIF1<0，故不存在對(duì)應(yīng)的m；若SKIPIF1<0有一個(gè)解SKIPIF1<0，則SKIPIF1<0有兩個(gè)解，此時(shí)SKIPIF1<0，所以對(duì)應(yīng)的SKIPIF1<0，綜上，SKIPIF1<0.故選：C.二、多選題3．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0，下列結(jié)論中一定成立的結(jié)論的序號(hào)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【分析】先研究值域?yàn)镾KIPIF1<0時(shí)函數(shù)的定義域，再研究使得值域?yàn)镾KIPIF1<0得函數(shù)的最小值的自變量的取值集合，研究函數(shù)值取1，2時(shí)對(duì)應(yīng)的自變量的取值，由此可判斷各個(gè)選項(xiàng).【詳解】由于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0當(dāng)函數(shù)的最小值為1時(shí)，僅有SKIPIF1<0滿足，所以SKIPIF1<0，故D正確；當(dāng)函數(shù)的最大值為2時(shí)，僅有SKIPIF1<0滿足，所以SKIPIF1<0，故C正確；即當(dāng)SKIPIF1<0時(shí)，函數(shù)的值域?yàn)镾KIPIF1<0，故SKIPIF1<0，故SKIPIF1<0不一定正確，故A正確，B錯(cuò)誤；故選：ACD【點(diǎn)睛】關(guān)鍵點(diǎn)睛：本題考查函數(shù)的定義域及其求法，解題的關(guān)鍵是通過(guò)函數(shù)的值域求出函數(shù)的定義域，再利用元素與集合關(guān)系的判斷，集合的包含關(guān)系判斷，考查了學(xué)生的邏輯推理與轉(zhuǎn)化能力，屬于基礎(chǔ)題.三、填空題4．（2023·北京東城·統(tǒng)考二模）定義在區(qū)間SKIPIF1<0上的函數(shù)SKIPIF1<0的圖象是一條連續(xù)不斷的曲線，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0給出下列四個(gè)結(jié)論：①若SKIPIF1<0為遞增數(shù)列，則SKIPIF1<0存在最大值；②若SKIPIF1<0為遞增數(shù)列，則SKIPIF1<0存在最小值；③若SKIPIF1<0，且SKIPIF1<0存在最小值，則SKIPIF1<0存在最小值；④若SKIPIF1<0，且SKIPIF1<0存在最大值，則SKIPIF1<0存在最大值.其中所有錯(cuò)誤結(jié)論的序號(hào)有_______．【答案】①③④【分析】結(jié)合函數(shù)的單調(diào)性判斷最值，即可判斷①②，利用取反例，判斷③④.【詳解】①由條件可知，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0那么在區(qū)間SKIPIF1<0，函數(shù)的最大值是SKIPIF1<0，若數(shù)列SKIPIF1<0為遞增數(shù)列，則函數(shù)SKIPIF1<0不存在最大值，故①錯(cuò)誤；②由條件可知，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減，若SKIPIF1<0為遞增數(shù)列，那么在區(qū)間SKIPIF1<0的最小值是SKIPIF1<0，且SKIPIF1<0為遞增數(shù)列，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的最小值是SKIPIF1<0，故②正確；③若SKIPIF1<0，取SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0，存在最小值，但此時(shí)SKIPIF1<0的最小值是SKIPIF1<0的最小值，函數(shù)單調(diào)遞減，無(wú)最小值，故③錯(cuò)誤；④若SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0恒成立，則SKIPIF1<0有最大值，但SKIPIF1<0的最大值是SKIPIF1<0的最大值，函數(shù)單調(diào)遞增，無(wú)最大值，故④錯(cuò)誤.故答案為：①③④5．（2023·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的取值范圍為_(kāi)__________.【答案】SKIPIF1<0【分析】先討論SKIPIF1<0、SKIPIF1<0與1的大小關(guān)系確定SKIPIF1<0、SKIPIF1<0，進(jìn)而確定SKIPIF1<0的取值范圍，再結(jié)合函數(shù)的單調(diào)性進(jìn)行求解.【詳解】①當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，SKIPIF1<0，又由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以不存在SKIPIF1<0，使得SKIPIF1<0；③當(dāng)SKIPIF1<0