新高考數(shù)學(xué)一輪復(fù)習(xí)講練測專題3.5指數(shù)與指數(shù)函數(shù)（練）解析版

專題3.5指數(shù)與指數(shù)函數(shù)練基礎(chǔ)練基礎(chǔ)1．（2021·山東）設(shè)全集SKIPIF1<0，集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】利用指數(shù)函數(shù)的性質(zhì)求解集合B，再求集合的補(bǔ)集，交集即可.【詳解】由題知SKIPIF1<0，SKIPIF1<0又SKIPIF1<0，則SKIPIF1<0，故選：B.2.（2019·貴州省織金縣第二中學(xué)高一期中）函數(shù)且過定點（）A． B． C． D．【答案】D【解析】令，所以函數(shù)且過定點．3．（2021·江西高三二模（文））下列函數(shù)中，在SKIPIF1<0上單調(diào)遞增的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】利用二次函數(shù)的性質(zhì)判定A；利用分段函數(shù)的圖象可以判定B；根據(jù)冪函數(shù)和對數(shù)函數(shù)的性質(zhì)判定C,D．【詳解】A中，SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對稱，開口向下的拋物線，在SKIPIF1<0上單調(diào)遞減，故Ａ不對；B中，SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對稱，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故排除B；C中，由冪函數(shù)的性質(zhì)可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C正確；D中，根據(jù)指數(shù)函數(shù)的性質(zhì)可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故排除D；故選：C．4．（2020·浙江高三月考）當(dāng)SKIPIF1<0時，“函數(shù)SKIPIF1<0的值恒小于1”的一個充分不必要條件是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由指數(shù)函數(shù)的圖象與性質(zhì)可得原命題等價于SKIPIF1<0，再由充分不必要條件的概念即可得解.【詳解】若當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的值恒小于1，則SKIPIF1<0即SKIPIF1<0，所以當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的值恒小于1的一個充分不必要條件是SKIPIF1<0.故選：D.5．（2019·浙江高三專題練習(xí)）已知函數(shù)SKIPIF1<0（其中SKIPIF1<0的圖象如圖所示，則函數(shù)SKIPIF1<0的圖象是（）A． B．C． D．【答案】C【解析】由二次函數(shù)的圖象確定SKIPIF1<0的取值范圍，然后可確定SKIPIF1<0的圖象．【詳解】由函數(shù)的圖象可知，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0為增函數(shù)，SKIPIF1<0，SKIPIF1<0過定點SKIPIF1<0，故選：SKIPIF1<0.6．（2021·浙江高三專題練習(xí)）不等式SKIPIF1<0的解集是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】根據(jù)題意得SKIPIF1<0，再解絕對值不等式即可得答案.【詳解】解：由指數(shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，所以SKIPIF1<0，進(jìn)而得SKIPIF1<0，即SKIPIF1<0.故選：A.7．（2021·浙江高三專題練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）的圖象恒過定點SKIPIF1<0，若點SKIPIF1<0在冪函數(shù)SKIPIF1<0的圖象上，則冪函數(shù)SKIPIF1<0的圖象大致是（）A． B．C． D．【答案】B【解析】由指數(shù)函數(shù)性質(zhì)求得定點坐標(biāo)，由定點求得冪函數(shù)解析式，確定圖象．【詳解】由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，即定點為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，圖象為B．故選：B．8．（2021·山東高三三模）已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系正確的為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】根據(jù)指數(shù)函數(shù)與冪函數(shù)的單調(diào)性即可求解.【詳解】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0指數(shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，即SKIPIF1<0，又冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故選：B.9．【多選題】（2021·全國高三專題練習(xí)）函數(shù)SKIPIF1<0的圖象可能為（）A．B．

C． D．【答案】ABD【解析】根據(jù)函數(shù)解析式的形式，以及圖象的特征，合理給SKIPIF1<0賦值，判斷選項.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，圖象A滿足；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，此時函數(shù)是偶函數(shù)，關(guān)于SKIPIF1<0軸對稱，圖象B滿足；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，此時函數(shù)是奇函數(shù)，關(guān)于原點對稱，圖象D滿足；圖象C過點SKIPIF1<0，此時SKIPIF1<0，故C不成立.故選：ABD10．【多選題】（2021·全國高三專題練習(xí)）已知SKIPIF1<0（k為常數(shù)），那么函數(shù)SKIPIF1<0的圖象不可能是（）A． B．C． D．【答案】AD【解析】根據(jù)選項，四個圖象可知備選函數(shù)都具有奇偶性．當(dāng)SKIPIF1<0時，SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0為奇函數(shù)，再根據(jù)單調(diào)性進(jìn)行分析得出答案．【詳解】由選項的四個圖象可知，備選函數(shù)都具有奇偶性．當(dāng)SKIPIF1<0時，SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0且單調(diào)遞增，而SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故選項C正確，D錯誤；當(dāng)SKIPIF1<0時，SKIPIF1<0為奇函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0且單調(diào)遞增，而SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故選項B正確，A錯誤.故選:AD．練提升TIDHNEG練提升TIDHNEG1．（2021·浙江金華市·高三其他模擬）已知函數(shù)SKIPIF1<0，若對于任意一個正數(shù)SKIPIF1<0，不等式SKIPIF1<0在SKIPIF1<0上都有解，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由不等式可知，SKIPIF1<0或SKIPIF1<0，結(jié)合圖象，分析可得SKIPIF1<0的取值范圍.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，不能滿足SKIPIF1<0都有解；當(dāng)SKIPIF1<0時，SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，如圖，當(dāng)SKIPIF1<0或SKIPIF1<0時，只需滿足SKIPIF1<0或SKIPIF1<0，滿足條件.所以SKIPIF1<0，SKIPIF1<0時，滿足條件.故選：A2．（2021·安徽蕪湖市·高三二模（理））函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且當(dāng)SKIPIF1<0時，SKIPIF1<0.若對任意的SKIPIF1<0，均有SKIPIF1<0，則實數(shù)SKIPIF1<0的最大值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．SKIPIF1<0【答案】A【解析】首先根據(jù)函數(shù)是偶函數(shù)，求出函數(shù)的解析式，結(jié)合不等式的關(guān)系進(jìn)行轉(zhuǎn)化，利用單調(diào)性轉(zhuǎn)化為不等式恒成立問題即可求解.【詳解】∵SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且當(dāng)SKIPIF1<0時，SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0時為增函數(shù)，∴SKIPIF1<0，則SKIPIF1<0等價于SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0對任意SKIPIF1<0恒成立，設(shè)SKIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0.故選：A.3．（2021·遼寧沈陽市·高三三模）已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】根據(jù)指數(shù)函數(shù)的單調(diào)性，將問題轉(zhuǎn)化為比較當(dāng)SKIPIF1<0時SKIPIF1<0的大小，利用特值法即可求得結(jié)果.【詳解】因為SKIPIF1<0，函數(shù)SKIPIF1<0是單調(diào)增函數(shù)，所以比較a，b，c的大小，只需比較當(dāng)SKIPIF1<0時SKIPIF1<0的大小即可.用特殊值法，取SKIPIF1<0，容易知SKIPIF1<0，再對其均平方得SKIPIF1<0，顯然SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0故選：B.4．（2021·江蘇蘇州市·高三其他模擬）生物體死亡后，它機(jī)體內(nèi)原有的碳14含量P會按確定的比率衰減(稱為衰減率)，P與死亡年數(shù)t之間的函數(shù)關(guān)系式為SKIPIF1<0(其中a為常數(shù))，大約每經(jīng)過5730年衰減為原來的一半，這個時間稱為“半衰期”.若2021年某遺址文物出土?xí)r碳14的殘余量約占原始含量的79%，則可推斷該文物屬于（）參考數(shù)據(jù)：SKIPIF1<0.參考時間軸：A．戰(zhàn)國 B．漢 C．唐 D．宋【答案】B【解析】根據(jù)“半衰期”得SKIPIF1<0，進(jìn)而解方程SKIPIF1<0得SKIPIF1<0，進(jìn)而可推算其所處朝代.【詳解】由題可知，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時，解方程SKIPIF1<0，兩邊取以SKIPIF1<0為底的對數(shù)得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，所以可推斷該文物屬于漢朝.故選：B5．（2021·河南高三月考（理））設(shè)實數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．無法比較【答案】A【解析】從選項A或C出發(fā)，分析其對立面，推理導(dǎo)出矛盾結(jié)果或成立的結(jié)果即可得解.【詳解】假設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，因函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0；由SKIPIF1<0得SKIPIF1<0，因函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0；即有SKIPIF1<0與假設(shè)SKIPIF1<0矛盾，所以SKIPIF1<0，故選：A6．【多選題】（2021·全國高三專題練習(xí)）若函數(shù)SKIPIF1<0，則下述正確的有（）A．SKIPIF1<0在R上單調(diào)遞增 B．SKIPIF1<0的值域為SKIPIF1<0C．SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱 D．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱【答案】AC【解析】A.由SKIPIF1<0和SKIPIF1<0的單調(diào)性判斷；B.取SKIPIF1<0判斷；C.D.判斷SKIPIF1<0是否等于零即可.【詳解】因為SKIPIF1<0是定義在R上的增函數(shù)，SKIPIF1<0是定義在R上的減函數(shù)，所以SKIPIF1<0在R上單調(diào)遞增，故A正確；因為SKIPIF1<0，故B錯誤；因為SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，故C正確，D錯誤．故選：AC．7．【多選題】（2020·山東省青島第十六中學(xué)高三月考）已知函數(shù)SKIPIF1<0，則下列正確的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0的值域為SKIPIF1<0【答案】BD【解析】對選項A，根據(jù)計算SKIPIF1<0，即可判斷A錯誤，對選項B，根據(jù)計算SKIPIF1<0，即可判斷B正確；對選項C，根據(jù)計算SKIPIF1<0，即可判斷C錯誤，對選項D，分別求SKIPIF1<0和SKIPIF1<0的值域即可得到答案.【詳解】對選項A，SKIPIF1<0，SKIPIF1<0，故A錯誤；對選項B，SKIPIF1<0，SKIPIF1<0，故B正確.對選項C，因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故C錯誤；對選項D，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0的值域為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0的值域為SKIPIF1<0，又因為SKIPIF1<0時，SKIPIF1<0，是周期為SKIPIF1<0的函數(shù)，所以當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的值域為SKIPIF1<0，綜上，函數(shù)SKIPIF1<0的值域為SKIPIF1<0，故D正確.故選：BD8．【多選題】（2020·河北冀州中學(xué)（衡水市冀州區(qū)第一中學(xué)）高三月考）高斯是德國著名的數(shù)學(xué)家，近代數(shù)學(xué)奠基者之一，享有“數(shù)學(xué)王子”的稱號，他和阿基米德、牛頓并列為世界三大數(shù)學(xué)家，用其名字命名的“高斯函數(shù)”為：設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)，則SKIPIF1<0稱為高斯函數(shù)，例如：SKIPIF1<0，SKIPIF1<0.已知函數(shù)SKIPIF1<0，則關(guān)于函數(shù)SKIPIF1<0的敘述中正確的是（）A．SKIPIF1<0是偶函數(shù) B．SKIPIF1<0是奇函數(shù)C．SKIPIF1<0在SKIPIF1<0上是增函數(shù) D．SKIPIF1<0的值域是SKIPIF1<0【答案】BC【解析】由SKIPIF1<0判斷A；由奇函數(shù)的定義證明B；把SKIPIF1<0的解析式變形，由SKIPIF1<0的單調(diào)性結(jié)合復(fù)合函數(shù)的單調(diào)性判斷C正確；求出SKIPIF1<0的范圍，進(jìn)一步求得SKIPIF1<0的值域判斷D．【詳解】SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0不是偶函數(shù)，故A錯誤；SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為奇函數(shù)，故B正確；SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上是增函數(shù)，故C正確；SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0．SKIPIF1<0，故D錯誤．故選：BC．9．【多選題】（2020·重慶市第十一中學(xué)校高三月考）已知函數(shù)SKIPIF1<0（SKIPIF1<0為常數(shù)），函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，則實數(shù)SKIPIF1<0的取值可以是（）A．-1 B．2 C．1 D．0【答案】CD【解析】由已知求得當(dāng)SKIPIF1<0時，SKIPIF1<0的最小值為SKIPIF1<0，問題轉(zhuǎn)化為當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，對SKIPIF1<0分類討論求得SKIPIF1<0的范圍，結(jié)合選項得答案．【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，且當(dāng)SKIPIF1<0時，函數(shù)取得最小值為SKIPIF1<0；要使原分段函數(shù)有最小值為SKIPIF1<0，則當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，當(dāng)SKIPIF1<0時，滿足；當(dāng)SKIPIF1<0時，需SKIPIF1<0，即SKIPIF1<0．綜上，實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．結(jié)合選項可得，實數(shù)SKIPIF1<0的取值可以是1，0．故選：CD．10．【多選題】（2021·南京市中華中學(xué)高三期末）“懸鏈線”進(jìn)入公眾視野，源于達(dá)芬奇的畫作《抱銀貂的女人》.這幅畫作中，女士脖頸上懸掛的黑色珍珠鏈與主人相互映襯，顯現(xiàn)出不一樣的美與光澤.而達(dá)芬奇卻心生好奇：“固定項鏈的兩端，使其在重力作用下自然下垂，那么項鏈所形成的曲線是什么？”隨著后人研究的深入，懸鏈線的廬山真面目被揭開.法國著名昆蟲學(xué)家、文學(xué)家法布爾，在《昆蟲記》里有這樣的記載：“每當(dāng)?shù)匦囊蛿_性同時發(fā)生作用時，懸鏈線就在現(xiàn)實中出現(xiàn)了.當(dāng)一條懸鏈彎曲成兩點不在同一垂直線（注：垂直于地面的直線）上的曲線時，人們便把這曲線稱為懸鏈線.這就是一條軟繩子兩端抓住而垂下來的形狀，這就是一張被風(fēng)鼓起來的船帆外形的那條線條．”建立適當(dāng)?shù)钠矫嬷苯亲鴺?biāo)系，可以寫出懸鏈線的函數(shù)解析式：SKIPIF1<0，其中SKIPIF1<0為懸鏈線系數(shù)．當(dāng)SKIPIF1<0時，SKIPIF1<0稱為雙曲余弦函數(shù)，記為SKIPIF1<0．類似的雙曲正弦函數(shù)SKIPIF1<0．直線SKIPIF1<0與SKIPIF1<0和SKIPIF1<0的圖像分別交于點SKIPIF1<0、SKIPIF1<0．下列結(jié)論正確的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0隨SKIPIF1<0的增大而減小 D．SKIPIF1<0與SKIPIF1<0的圖像有完全相同的漸近線【答案】AC【解析】由函數(shù)的定義，代入化簡可得A正確，B不正確；由SKIPIF1<0可得C正確；由函數(shù)的圖象變化可得D不正確.【詳解】SKIPIF1