新高考數(shù)學一輪復(fù)習第2章 第05講 指數(shù)與指數(shù)函數(shù) 精講+精練（教師版）

第05講指數(shù)與指數(shù)函數(shù)(精講+精練）目錄第一部分：知識點精準記憶第二部分：課前自我評估測試第三部分：典型例題剖析高頻考點一：指數(shù)與指數(shù)冪的運算高頻考點二：指數(shù)函數(shù)的概念高頻考點三：指數(shù)函數(shù)的圖象①判斷指數(shù)型函數(shù)的圖象；②根據(jù)指數(shù)型函數(shù)圖象求參數(shù)③指數(shù)型函數(shù)圖象過定點問題；④指數(shù)函數(shù)圖象應(yīng)用高頻考點四：指數(shù)（型）函數(shù)定義域高頻考點五：指數(shù)（型）函數(shù)的值域①指數(shù)函數(shù)在區(qū)間SKIPIF1<0上的值域；②指數(shù)型復(fù)合函數(shù)值域③根據(jù)指數(shù)函數(shù)值域（最值）求參數(shù)高頻考點六：指數(shù)函數(shù)單調(diào)性①判斷指數(shù)函數(shù)單調(diào)性；②由指數(shù)（型）函數(shù)單調(diào)性求參數(shù)③判斷指數(shù)型復(fù)合函數(shù)單調(diào)性；④比較大?、莞鶕?jù)指數(shù)函數(shù)單調(diào)性解不等式高頻考點七：指數(shù)函數(shù)的最值①求已知指數(shù)型函數(shù)的值域②根據(jù)指數(shù)函數(shù)最值求參數(shù)③含參指數(shù)（型）函數(shù)最值第四部分：高考真題感悟第五部分：第05講指數(shù)與指數(shù)函數(shù)（精練）第一部分：知識點精準記憶第一部分：知識點精準記憶1、根式的概念及性質(zhì)（1）概念：式子SKIPIF1<0叫做根式，其中SKIPIF1<0叫做根指數(shù)，SKIPIF1<0叫做被開方數(shù)．（2）性質(zhì)：①SKIPIF1<0(SKIPIF1<0且SKIPIF1<0)；②當SKIPIF1<0為奇數(shù)時，SKIPIF1<0；當SKIPIF1<0為偶數(shù)時，SKIPIF1<02、分數(shù)指數(shù)冪①正數(shù)的正分數(shù)指數(shù)冪的意義是SKIPIF1<0(SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0)；②正數(shù)的負分數(shù)指數(shù)冪的意義是SKIPIF1<0(SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0)；③0的正分數(shù)指數(shù)冪等于0；0的負分數(shù)指數(shù)冪沒有意義．3、指數(shù)冪的運算性質(zhì)①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0.4、指數(shù)函數(shù)及其性質(zhì)（1）指數(shù)函數(shù)的概念函數(shù)SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)叫做指數(shù)函數(shù)，其中指數(shù)SKIPIF1<0是自變量，函數(shù)的定義域是SKIPIF1<0．（2）指數(shù)函數(shù)SKIPIF1<0的圖象和性質(zhì)底數(shù)SKIPIF1<0SKIPIF1<0圖象性質(zhì)定義域為SKIPIF1<0，值域為SKIPIF1<0圖象過定點SKIPIF1<0當SKIPIF1<0時，恒有SKIPIF1<0；當SKIPIF1<0時，恒有SKIPIF1<0當SKIPIF1<0時，恒有SKIPIF1<0；當SKIPIF1<0時，恒有SKIPIF1<0在定義域SKIPIF1<0上為增函數(shù)在定義域SKIPIF1<0上為減函數(shù)注意指數(shù)函數(shù)SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)的圖象和性質(zhì)與SKIPIF1<0的取值有關(guān)，應(yīng)分SKIPIF1<0與SKIPIF1<0來研究第二部分：課前自我評估測試第二部分：課前自我評估測試一、判斷題1．（2021·江西·貴溪市實驗中學高二階段練習）函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象必過定點SKIPIF1<0()【答案】正確解：令SKIPIF1<0得，SKIPIF1<0，此時SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0的圖象必過定點SKIPIF1<0，故答案為：正確2．（2021·江西·貴溪市實驗中學高二階段練習）SKIPIF1<0

()【答案】正確SKIPIF1<0，判斷正確故答案為：正確．二、單選題1．（2022·寧夏·銀川一中高二期末（文））函數(shù)SKIPIF1<0在SKIPIF1<0的最大值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：因為函數(shù)SKIPIF1<0是單調(diào)遞增函數(shù)，所以函數(shù)SKIPIF1<0也是單調(diào)遞增函數(shù)，所以SKIPIF1<0.故選：C2．（2022·江蘇南通·高一期末）已知指數(shù)函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0），且SKIPIF1<0，則SKIPIF1<0的取值范圍（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A解：由指數(shù)函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0），且SKIPIF1<0根據(jù)指數(shù)函數(shù)單調(diào)性可知SKIPIF1<0所以SKIPIF1<0，故選：A3．（2022·北京·高三專題練習）若函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖像經(jīng)過定點P，則點P的坐標是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B因為SKIPIF1<0，所以當SKIPIF1<0，即SKIPIF1<0時，函數(shù)值為定值0，所以點P坐標為SKIPIF1<0.另解：因為SKIPIF1<0可以由SKIPIF1<0向右平移一個單位長度后，再向下平移1個單位長度得到，由SKIPIF1<0過定點SKIPIF1<0，所以SKIPIF1<0過定點SKIPIF1<0.故選：B4．（2022·河北廊坊·高一期末）指數(shù)函數(shù)SKIPIF1<0在R上單調(diào)遞減，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D因為指數(shù)函數(shù)SKIPIF1<0在R上單調(diào)遞減，所以SKIPIF1<0，得SKIPIF1<0，所以實數(shù)a的取值范圍是SKIPIF1<0，故選：D5．（2022·北京·高三專題練習）若函數(shù)SKIPIF1<0是指數(shù)函數(shù)，則SKIPIF1<0等于（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C由題意可得SKIPIF1<0，解得SKIPIF1<0.故選：C.第三部分：典型例題剖析第三部分：典型例題剖析高頻考點一：指數(shù)與指數(shù)冪的運算1．（2022·廣東肇慶·高一期末）設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．2 D．3【答案】B∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故選：B2．（2022·上海楊浦·高一期末）設(shè)SKIPIF1<0，下列計算中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】CSKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0故選：C3．（2022·廣東深圳·高一期末）下列根式與分數(shù)指數(shù)冪的互化正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B解：對A：SKIPIF1<0，故選項A錯誤；對B：SKIPIF1<0，故選項B正確；對C：SKIPIF1<0，SKIPIF1<0不能化簡為SKIPIF1<0，故選項C錯誤；對D：因為SKIPIF1<0，所以SKIPIF1<0，故選項D錯誤.故選：B.4．（2022·全國·高三專題練習）化簡SKIPIF1<0的結(jié)果為（

）A．－SKIPIF1<0 B．－SKIPIF1<0C．－SKIPIF1<0 D．－6ab【答案】C原式＝SKIPIF1<0.故選：C.高頻考點二：指數(shù)函數(shù)的概念1．（2022·浙江·高三專題練習）函數(shù)SKIPIF1<0，且a≠1)的圖象經(jīng)過點SKIPIF1<0，則f(-2)=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．9【答案】D由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0．故選：D.2．（2022·黑龍江·嫩江市第一中學校高一期末）已知指數(shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的值為（

）A．3 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】BSKIPIF1<0解得SKIPIF1<0，又函數(shù)在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，SKIPIF1<0故選：B3．（2022·全國·高一課時練習）函數(shù)SKIPIF1<0是指數(shù)函數(shù)，則（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0且SKIPIF1<0【答案】C因為函數(shù)SKIPIF1<0是指數(shù)函數(shù)所以SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0.故選：C.4．（2022·浙江·高三專題練習）若指數(shù)函數(shù)SKIPIF1<0在[－1,1]上的最大值與最小值的差是1，則底數(shù)a等于A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D指數(shù)函數(shù)SKIPIF1<0在[－1,1]上的最大值與最小值的差是1則SKIPIF1<0

解得a=SKIPIF1<0故選D高頻考點三：指數(shù)函數(shù)的圖象①判斷指數(shù)型函數(shù)的圖象1．（2022·上海市復(fù)興高級中學高一階段練習）函數(shù)SKIPIF1<0的大致圖像是（

）A．B．C．D．【答案】C解：由函數(shù)SKIPIF1<0，得SKIPIF1<0，所以函數(shù)為偶函數(shù)，故排除AB，當SKIPIF1<0時，SKIPIF1<0，所以函數(shù)在SKIPIF1<0上是減函數(shù)，故排除D.故選：C.2．（2022·上海市進才中學高二階段練習）函數(shù)SKIPIF1<0的圖像的大致形狀是（

）A． B．C．D．【答案】D根據(jù)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0是減函數(shù)，SKIPIF1<0是增函數(shù).SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增故選：D.3．（2022·全國·高三專題練習）已知0＜m＜n＜1，則指數(shù)函數(shù)①y＝mx，②y＝nx的圖象為（）．A． B．C．D．【答案】C試題分析：由，在上單調(diào)遞減，所以排除；令，，C正確．4．（2022·全國·高三專題練習（文））函數(shù)SKIPIF1<0的圖象可能是

()A． B．C． D．【答案】C①當SKIPIF1<0時，函數(shù)SKIPIF1<0可以看做函數(shù)SKIPIF1<0的圖象向下平移SKIPIF1<0個單位，由于SKIPIF1<0，則A錯誤；又SKIPIF1<0時，SKIPIF1<0，則函數(shù)SKIPIF1<0過點SKIPIF1<0，故B錯誤；②當SKIPIF1<0時，函數(shù)SKIPIF1<0可以看做函數(shù)SKIPIF1<0的圖象向下平移SKIPIF1<0個單位，由于SKIPIF1<0，則D錯誤；又SKIPIF1<0時，SKIPIF1<0，則函數(shù)SKIPIF1<0過點SKIPIF1<0，故C正確；故選：C②根據(jù)指數(shù)型函數(shù)圖象求參數(shù)1．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0的圖象如圖所示，其中SKIPIF1<0，SKIPIF1<0為常數(shù)，則下列結(jié)論正確的是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】A由SKIPIF1<0，可得SKIPIF1<0，因為由圖像可知函數(shù)是減函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選：A2．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0與SKIPIF1<0的圖象如圖，則下列不等式一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C由圖可知，SKIPIF1<0單調(diào)遞增，則SKIPIF1<0；SKIPIF1<0單調(diào)遞減，則SKIPIF1<0，A：SKIPIF1<00不一定成立，如SKIPIF1<0；B：SKIPIF1<0不一定成立，如SKIPIF1<0；C：SKIPIF1<0，成立；D：SKIPIF1<0不成立，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：C.3．（2021·全國·高一專題練習）函數(shù)SKIPIF1<0的圖像如圖所示，其中SKIPIF1<0SKIPIF1<0為常數(shù)，則下列結(jié)論正確的是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】D由函數(shù)SKIPIF1<0的圖像可知，函數(shù)SKIPIF1<0在定義域上單調(diào)遞減，SKIPIF1<0，排除AB選項；分析可知：函數(shù)SKIPIF1<0圖像是由SKIPIF1<0向左平移所得，SKIPIF1<0，SKIPIF1<0.故D選項正確.故選：D4．（2021·全國·高一專題練習）若函數(shù)SKIPIF1<0的圖象如圖所示，則（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】D根據(jù)圖象，函數(shù)SKIPIF1<0是單調(diào)遞減的所以指數(shù)函數(shù)的底SKIPIF1<0根據(jù)圖象的縱截距，令SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0即SKIPIF1<0，SKIPIF1<0故選：D.③指數(shù)型函數(shù)圖象過定點問題1．（2022·吉林·長春市第二中學高一期末）函數(shù)SKIPIF1<0且SKIPIF1<0的圖象恒過定點（

）A．(－2，0) B．(－1，0)C．(0，－1) D．(－1，－2)【答案】A由題意，函數(shù)SKIPIF1<0且SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖象過定點SKIPIF1<0．故選：A2．（2022·全國·高三專題練習）若函數(shù)SKIPIF1<0過定點SKIPIF1<0，以SKIPIF1<0為頂點且過原點的二次函數(shù)SKIPIF1<0的解析式為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A對于函數(shù)SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0過定點SKIPIF1<0SKIPIF1<0，設(shè)以SKIPIF1<0SKIPIF1<0為頂點且過原點的二次函數(shù)SKIPIF1<0，因為SKIPIF1<0過原點SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0的解析式為：SKIPIF1<0，故選：A.3．（2022·河南焦作·高一期末）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象過定點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】D當SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0，所以不等式為SKIPIF1<0，解得SKIPIF1<0，所以不等式的解集為SKIPIF1<0.故選：D4．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0(SKIPIF1<0，SKIPIF1<0)恒過定點SKIPIF1<0，則函數(shù)SKIPIF1<0的圖像不經(jīng)過（

）A．第一象限 B．第二象限C．第三象限 D．第四象限【答案】BSKIPIF1<0且SKIPIF1<0恒過定點SKIPIF1<0SKIPIF1<0則函數(shù)SKIPIF1<0恒過定點SKIPIF1<0且是單調(diào)遞增函數(shù)，其圖象不經(jīng)過第二象限.故選：B④指數(shù)函數(shù)圖象應(yīng)用1．（2021·重慶市涪陵第二中學校高一階段練習）函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】C當SKIPIF1<0時，SKIPIF1<0是減函數(shù)，SKIPIF1<0，SKIPIF1<0，D選項錯誤，C選項符合.當SKIPIF1<0時，SKIPIF1<0是增函數(shù)，SKIPIF1<0，SKIPIF1<0，AB選項錯誤.故選：C2．（2021·全國·高一課時練習）函數(shù)SKIPIF1<0，且SKIPIF1<0）與SKIPIF1<0的圖像大致是A． B． C． D．【答案】A由題知，直線SKIPIF1<0的斜率為SKIPIF1<0，故排除選項C、D，又由選項A、B中的圖像知SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以A正確，B錯誤．故選A．3．（2021·全國·高一課時練習）若SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的圖像一定經(jīng)過（

）A．第一、二、三象限 B．第一、三、四象限C．第二、三、四象限 D．第一、二、四象限【答案】A由SKIPIF1<0可得函數(shù)SKIPIF1<0的圖像單調(diào)遞增，且過第一、二象限，由SKIPIF1<0可得把SKIPIF1<0的圖像向下平移SKIPIF1<0個單位可得SKIPIF1<0的圖像，結(jié)合SKIPIF1<0可知，圖像過第一、二、三象限．故答案為A高頻考點四：指數(shù)（型）函數(shù)定義域1．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B解：由題意得：SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，解得：SKIPIF1<0，故函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，故選：B．2．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0的定義域為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0.故選：D.3．（2021·江蘇·高一專題練習）函數(shù)y＝SKIPIF1<0的定義域是(－∞，0]，則a的取值范圍為()A．a(chǎn)＞0 B．a(chǎn)＜1C．0＜a＜1 D．a(chǎn)≠1【答案】C要使函數(shù)SKIPIF1<0且SKIPIF1<0有意義,則SKIPIF1<0,即SKIPIF1<0,當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0，因為SKIPIF1<0的定義域為SKIPIF1<0所以可得SKIPIF1<0符合題意，SKIPIF1<0的取值范圍為SKIPIF1<0，故選C.4．（2021·廣西河池·高一階段練習）設(shè)函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A由SKIPIF1<0即SKIPIF1<0可得SKIPIF1<0所以SKIPIF1<0的定義域為SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，故選：A．高頻考點五：指數(shù)（型）函數(shù)的值域①指數(shù)函數(shù)在區(qū)間SKIPIF1<0上的值域1．（2022·全國·高一）當xSKIPIF1<0[-1，1]時，函數(shù)f(x)=3x-2的值域為________【答案】SKIPIF1<0因為xSKIPIF1<0[-1，1]，所以SKIPIF1<0，所以SKIPIF1<0，所以f(x)=3x-2的值域為SKIPIF1<0，故答案為：SKIPIF1<02．（2022·全國·高三專題練習）已知函數(shù)f（x）=9x﹣aSKIPIF1<03x+1+a2（x∈[0，1]，a∈R），記f（x）的最大值為g（a）．（Ⅰ）求g（a）解析式；（Ⅱ）若對于任意t∈[﹣2，2]，任意a∈R，不等式g（a）≥﹣m2+tm恒成立，求實數(shù)m的范圍．【答案】（Ⅰ）g（a）=SKIPIF1<0；（Ⅱ）m≤﹣SKIPIF1<0或m≥SKIPIF1<0．解：（Ⅰ）令u=3x∈[1，3]，則f（x）=h（u）=u2﹣3au+a2．當SKIPIF1<0≤2，即a≤SKIPIF1<0時，g（a）=h（u）max=h（3）=a2﹣9a+9；當SKIPIF1<0，即a＞SKIPIF1<0時，g（a）=h（u）max=h（1）=a2﹣3a+1；故g（a）=SKIPIF1<0；（Ⅱ）當a≤SKIPIF1<0時，g（a）=a2﹣9a+9，g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0；當aSKIPIF1<0時，g（a）=a2﹣3a+1，g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0；因此g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0；對于任意任意a∈R，不等式g（a）≥﹣m2+tm恒成立等價于﹣m2+tm≤﹣SKIPIF1<0．令h（t）=mt﹣m2，由于h（t）是關(guān)于t的一次函數(shù)，故對于任意t∈[﹣2，2]都有h（t）≤﹣SKIPIF1<0等價于SKIPIF1<0，即SKIPIF1<0，解得m≤﹣SKIPIF1<0或m≥SKIPIF1<0．3．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0.當SKIPIF1<0時，求函數(shù)SKIPIF1<0在SKIPIF1<0的值域；【答案】（1）SKIPIF1<0；（1）當SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，故值域為SKIPIF1<0；4．（2022·江西省豐城中學高一開學考試）函數(shù)SKIPIF1<0且SKIPIF1<0，函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的解析式；(2)若關(guān)于SKIPIF1<0的方程SKIPIF1<0在區(qū)間SKIPIF1<0上有實數(shù)根，求實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.(1)由SKIPIF1<0，可得：SKIPIF1<0，解得：SKIPIF1<0，∴SKIPIF1<0；(2)由SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，則原問題等價于y＝m與y＝h(t)＝SKIPIF1<0在SKIPIF1<0上有交點，數(shù)形結(jié)合可知m∈[h(SKIPIF1<0)，h(4)]＝SKIPIF1<0.故實數(shù)SKIPIF1<0的取值范圍為：SKIPIF1<0.②指數(shù)型復(fù)合函數(shù)值域1．（2022·山西·臨汾第一中學校高一期末）函數(shù)SKIPIF1<0的值域為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D令SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴函數(shù)SKIPIF1<0的值域為SKIPIF1<0，故選：D2．（2022·湖南邵陽·高一期末）函數(shù)SKIPIF1<0的值域為______．【答案】SKIPIF1<0由于SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0.故答案為：SKIPIF1<0.3．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0的值域是___________.【答案】SKIPIF1<0因為SKIPIF1<0，設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0故答案為：SKIPIF1<0.4．（2022·河南·洛寧縣第一高級中學高一階段練習）已知函數(shù)SKIPIF1<0.(1)當SKIPIF1<0時，求SKIPIF1<0的值域；(2)若SKIPIF1<0有最大值16，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(1)當SKIPIF1<0時，SKIPIF1<0.因為SKIPIF1<0在R上單調(diào)遞增，且SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0的值域為SKIPIF1<0.(2)令SKIPIF1<0，因為函數(shù)SKIPIF1<0在其定義域內(nèi)單調(diào)遞增，所以要使函數(shù)SKIPIF1<0有最大值16，則SKIPIF1<0的最大值為4，故SKIPIF1<0解得SKIPIF1<0.故SKIPIF1<0的值為SKIPIF1<0.5．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0.（1）求SKIPIF1<0在SKIPIF1<0上的值域；【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【詳解】（1）令SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，則可將原函數(shù)轉(zhuǎn)化為SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0；③根據(jù)指數(shù)函數(shù)值域（最值）求參數(shù)1．（2022·廣東湛江·高一期末）已知函數(shù)SKIPIF1<0的定義域和值域都是SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】A當SKIPIF1<0時，SKIPIF1<0，方程組無解當SKIPIF1<0時，SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0故選：A.2．（2022·遼寧鞍山·高一期末）若函數(shù)SKIPIF1<0的值域為SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C因為SKIPIF1<0，且SKIPIF1<0的值域為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故選：C.3．（2022·全國·高一）已知函數(shù)SKIPIF1<0且SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值比最小值大SKIPIF1<0,求SKIPIF1<0的值.【答案】SKIPIF1<0或SKIPIF1<0SKIPIF1<0時，SKIPIF1<0是減函數(shù)，SKIPIF1<0，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0是增函數(shù)，SKIPIF1<0，SKIPIF1<0．綜上，SKIPIF1<0或SKIPIF1<0．4．（2022·湖南·高一期末）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的值域；(2)當SKIPIF1<0時，SKIPIF1<0的最大值為7，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0(1)設(shè)SKIPIF1<0，則SKIPIF1<0.因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0的值域為SKIPIF1<0.(2)函數(shù)SKIPIF1<0圖象的對稱軸為直線SKIPIF1<0.當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去）所以SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），因為SKIPIF1<0，所以SKIPIF1<0.綜上，SKIPIF1<0或SKIPIF1<0.5．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0）是SKIPIF1<0上的奇函數(shù)．(1)求實數(shù)SKIPIF1<0的值；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的值域為SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(1)由SKIPIF1<0是奇函數(shù)得SKIPIF1<0，SKIPIF1<0，此時SKIPIF1<0是奇函數(shù)；(2)由復(fù)合函數(shù)的性質(zhì)得SKIPIF1<0在定義域內(nèi)是增函數(shù)，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0（舍去），SKIPIF1<0，所以SKIPIF1<0．高頻考點六：指數(shù)函數(shù)單調(diào)性①判斷指數(shù)函數(shù)單調(diào)性1．（2022·廣西南寧·高一期末）設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．是偶函數(shù)，且在SKIPIF1<0單調(diào)遞增 B．是偶函數(shù)，且在SKIPIF1<0單調(diào)遞減C．是奇函數(shù)，且在SKIPIF1<0單調(diào)遞增 D．是奇函數(shù)，且在SKIPIF1<0單調(diào)遞減【答案】D函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0為奇函數(shù).而SKIPIF1<0，可知函數(shù)SKIPIF1<0為定義域SKIPIF1<0上的減函數(shù)，因此，函數(shù)SKIPIF1<0為奇函數(shù)，且是SKIPIF1<0上的減函數(shù).故選：D.2．（2022·福建寧德·高一期末）已知SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0．(1)求SKIPIF1<0的解析式；(2)判斷SKIPIF1<0的單調(diào)性，并根據(jù)定義證明．【答案】(1)SKIPIF1<0(2)見解析(1)已知SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，(2)根據(jù)指數(shù)函數(shù)的單調(diào)性可判斷得SKIPIF1<0為增函數(shù).下證明：設(shè)SKIPIF1<0是SKIPIF1<0上任意給定的兩個實數(shù)，且SKIPIF1<0，

則SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0

SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增函數(shù)3．（2021·貴州·六盤水紅橋?qū)W校高一階段練習）若函數(shù)SKIPIF1<0是指數(shù)函數(shù)(1)求SKIPIF1<0，SKIPIF1<0的值；(2)求解不等式SKIPIF1<0【答案】(1)SKIPIF1<0且SKIPIF1<0；(2)SKIPIF1<0；(1)因函數(shù)SKIPIF1<0是指數(shù)函數(shù)，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0的值是：SKIPIF1<0且SKIPIF1<0.(2)由(1)知，SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0在R上單調(diào)遞增，由SKIPIF1<0得：SKIPIF1<0，解得SKIPIF1<0，所以不等式SKIPIF1<0的解集是：SKIPIF1<0.4．（2021·全國·高一期末）設(shè)函數(shù)SKIPIF1<0，(1)判斷SKIPIF1<0的單調(diào)性，并證明你的結(jié)論；【答案】(1)增函數(shù)，證明見解析；(1)依題意，函數(shù)SKIPIF1<0定義域為R，SKIPIF1<0是R上的增函數(shù)，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，因SKIPIF1<0為R上的增函數(shù)，則由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，于是得SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù).②由指數(shù)（型）函數(shù)單調(diào)性求參數(shù)1．（2022·遼寧朝陽·高一開學考試）若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A∵函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，∴SKIPIF1<0，解得SKIPIF1<0，實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：A.2．（2022·內(nèi)蒙古·赤峰二中高一期末（文））若函數(shù)SKIPIF1<0是R上的減函數(shù)，則實數(shù)a的取值范圍是___．【答案】SKIPIF1<0由題知SKIPIF1<0．故答案為：SKIPIF1<0.3．（2022·河北張家口·高一期末）已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實數(shù)SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0因為分段函數(shù)在SKIPIF1<0上單調(diào)遞減，所以每段都單調(diào)遞減，即SKIPIF1<0，并且在分界點處需滿足SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<04．（2022·湖南·高一課時練習）若函數(shù)SKIPIF1<0是指數(shù)函數(shù)，且為指數(shù)增長型函數(shù)模型，則實數(shù)SKIPIF1<0________.【答案】1依題意知SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0（舍SKIPIF1<0）故答案為：15．（2022·安徽·歙縣教研室高一期末）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上為增函數(shù)，則實數(shù)SKIPIF1<0的取值范圍為______.【答案】SKIPIF1<0由復(fù)合函數(shù)的同增異減性質(zhì)可得，SKIPIF1<0在SKIPIF1<0上嚴格單調(diào)遞減，二次函數(shù)開口向上，對稱軸為SKIPIF1<0所以SKIPIF1<0，即SKIPIF1<0故答案為：SKIPIF1<06．（2022·湖南·高一課時練習）若函數(shù)SKIPIF1<0是區(qū)間SKIPIF1<0上的減函數(shù)，求實數(shù)SKIPIF1<0的取值范圍．【答案】SKIPIF1<0函數(shù)SKIPIF1<0是區(qū)間SKIPIF1<0上的減函數(shù)，故SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.③判斷指數(shù)型復(fù)合函數(shù)單調(diào)性1．（2022·安徽省蚌埠第三中學高一開學考試）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D解：因為函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，函數(shù)SKIPIF1<0在定義域內(nèi)是單調(diào)遞減函數(shù)，所以，根據(jù)復(fù)合函數(shù)單調(diào)性法則“同增異減”得SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.故選：D2．（2022·河南·商丘市第一高級中學高一開學考試）已知函數(shù)SKIPIF1<0，且對于任意的SKIPIF1<0，都有SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B依題可知函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，∴SKIPIF1<0，解得SKIPIF1<0．故選：B．3．（2022·寧夏·吳忠中學高一期末）已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實數(shù)SKIPIF1<0的取值范圍是______．【答案】SKIPIF1<0令SKIPIF1<0，可得拋物線的開口向上，且對稱軸為SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在區(qū)間SKIPIF1<0上單調(diào)遞增，又由函數(shù)SKIPIF1<0，根據(jù)復(fù)合函數(shù)的單調(diào)性的判定方法，可得函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減，因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0，可得實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.4．（2022·河南·林州一中高一開學考試）已知函數(shù)SKIPIF1<0是奇函數(shù).(1)求SKIPIF1<0的值；(2)判斷并證明函數(shù)SKIPIF1<0的單調(diào)性.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0在R上單調(diào)遞增，證明見解析.【解析】(1)由題設(shè)，SKIPIF1<0，整理可得：SKIPIF1<0恒成立，解得SKIPIF1<0.(2)由（1）知：SKIPIF1<0，在R上單調(diào)遞增，證明如下：令SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在R上單調(diào)遞增.④比較大小1．（2022·廣東汕尾·高一期末）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ASKIPIF1<0，