新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第06講 指數(shù)函數(shù)與對(duì)數(shù)函數(shù)（練）（原卷版）

第03講指數(shù)函數(shù)與對(duì)數(shù)函數(shù)1．SKIPIF1<0（

）A．4 B．3 C．2 D．12、已知函數(shù)SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03、已知SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象是（）A． B．C． D．5．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)__________.6．已知SKIPIF1<0，則SKIPIF1<0的值等于__．7．函數(shù)SKIPIF1<0的大致圖象為（

）A． B．C． D．8．已知函數(shù)SKIPIF1<0，則SKIPIF1<0________，函數(shù)SKIPIF1<0的零點(diǎn)有________個(gè)．9．在下列區(qū)間中，函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．已知函數(shù)SKIPIF1<0是偶函數(shù).(1)當(dāng)SKIPIF1<0，函數(shù)SKIPIF1<0存在零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)設(shè)函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象只有一個(gè)公共點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.11．設(shè)SKIPIF1<0為實(shí)數(shù)，函數(shù)SKIPIF1<0在SKIPIF1<0上有零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)_______．1、設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02、已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0只有兩個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．已知SKIPIF1<0，不等式SKIPIF1<0恒成立，實(shí)數(shù)SKIPIF1<0取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有5個(gè)不同的實(shí)數(shù)解，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05、根據(jù)有關(guān)資料，圍棋狀態(tài)空間復(fù)雜度的上限SKIPIF1<0約為SKIPIF1<0，而可觀測(cè)宇宙中普通物質(zhì)的原子總數(shù)SKIPIF1<0約為SKIPIF1<0已知SKIPIF1<0，則下列各數(shù)中與SKIPIF1<0最接近的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0恰有SKIPIF1<0個(gè)不同的實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值范圍是_________.7．已知SKIPIF1<0（SKIPIF1<0且SKIPIF1<0，SKIPIF1<0且SKIPIF1<0），則函數(shù)SKIPIF1<0與SKIPIF1<0的圖像可能是（

）A． B．C． D．8、高斯是德國(guó)著名的數(shù)學(xué)家，近代數(shù)學(xué)奠基者之一，享有“數(shù)學(xué)王子”的美譽(yù)，用其名字命名的“高斯函數(shù)”：設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過(guò)x的最大整數(shù)，則SKIPIF1<0稱為高斯函數(shù)，也稱取整函數(shù)，例如：SKIPIF1<0，SKIPIF1<0，已知SKIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)開(kāi)_____．（多選）9．已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0為偶函數(shù) B．SKIPIF1<0是增函數(shù)C．SKIPIF1<0不是周期函數(shù) D．SKIPIF1<0的最小值為SKIPIF1<010．已知函數(shù)SKIPIF1<0，SKIPIF1<0有意義時(shí)SKIPIF1<0的取值范圍為SKIPIF1<0，其中SKIPIF1<0為實(shí)數(shù)．(1)求SKIPIF1<0的值；(2)寫(xiě)出函數(shù)SKIPIF1<0的單調(diào)區(qū)間，并求函數(shù)SKIPIF1<0的最大值．11、已知集合SKIPIF1<0，集合SKIPIF1<0．記集合SKIPIF1<0中最小元素為SKIPIF1<0，集合SKIPIF1<0中最大元素為SKIPIF1<0．(1)求SKIPIF1<0及SKIPIF1<0，SKIPIF1<0的值；(2)證明：函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；并用上述結(jié)論比較SKIPIF1<0與SKIPIF1<0的大?。?．（2022·浙江·高考真題）已知SKIPIF1<0，則SKIPIF1<0（

）A．25 B．5 C．SKIPIF1<0 D．SKIPIF1<02．（2020·山東·高考真題）已知函數(shù)SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則該函數(shù)在SKIPIF1<0上的圖像大致是（

）A． B．C． D．3．（2017·全國(guó)·高考真題（理））某城市為了解游客人數(shù)的變化規(guī)律，提高旅游服務(wù)質(zhì)量，收集并整理了2014年1月至2016年12月期間月接待游客量（單位：萬(wàn)人）的數(shù)據(jù)，繪制了如圖所示的折線圖.根據(jù)該折線圖，下列結(jié)論錯(cuò)誤的是（

）A．月接待游客量逐月增加B．年接待游客量逐年增加C．各年的月接待游客量高峰期大致在7，8月D．各年1月至6月的月接待游客量相對(duì)于7月至12月，波動(dòng)性更小，變化比較平穩(wěn)4．（2021·全國(guó)·高考真題（文））青少年視力是社會(huì)普遍關(guān)注的問(wèn)題，視力情況可借助視力表測(cè)量．通常用五分記錄法和小數(shù)記錄法記錄視力數(shù)據(jù)，五分記錄法的數(shù)據(jù)L和小數(shù)記錄表的數(shù)據(jù)V的滿足SKIPIF1<0．已知某同學(xué)視力的五分記錄法的數(shù)據(jù)為4.9，則其視力的小數(shù)記錄法的數(shù)據(jù)為（

）（SKIPIF1<0）A．1.5 B．1.2 C．0.8 D．0.65．（2013·浙江·高考真題（理））已知x，y為正實(shí)數(shù)，則（）A．2lgx+lgy=2lgx+2lgy B．2lg（x+y）=2lgx?2lgyC．2lgx?lgy=2lgx+2lgy D．2lg（xy）=2lgx?2lgy6．（2010·浙江·高考真題（文））已知SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)零點(diǎn)，若SKIPIF1<0，則（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<07．（2020·全國(guó)·高考真題（文））Logistic模型是常用數(shù)學(xué)模型之一，可應(yīng)用于流行病學(xué)領(lǐng)域．有學(xué)者根據(jù)公布數(shù)據(jù)建立了某地區(qū)新冠肺炎累計(jì)確診病例數(shù)I(t)(t的單位：天)的Logistic模型：SKIPIF1<0，其中K為最大確診病例數(shù)．當(dāng)I(SKIPIF1<0)=0.95K時(shí)，標(biāo)志著已初步遏制疫情，則SKIPIF1<0約為（

）（ln19≈3）A．60 B．63 C．66 D．698．（2022·天津·高考真題）已知函數(shù)SKIPIF1<0，若SKIPIF1<0至少有SKIPIF1<0個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是______.9．（2022·天津·高考真題）化簡(jiǎn)SKIPIF1<0____________10．（2014·廣東·高考真題（理））若等比數(shù)列的各項(xiàng)均為正數(shù)，且，則SKIPIF1<0.11．（2019·江蘇·高考真題）設(shè)SKIPIF1<0是定義在SKIPIF1<0上的兩個(gè)周期函數(shù)，SKIPIF1<0的周期為4，SKIPIF1<0的周期為2，且SKIPIF1<0是奇函數(shù).當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0.若在區(qū)間SKIPIF1<0上，關(guān)于SKIPIF1<0的方程