新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第10講 指數(shù)與指數(shù)函數(shù)（含解析）

第10講指數(shù)與指數(shù)函數(shù)（精講）題型目錄一覽①指數(shù)冪的化簡(jiǎn)與求值②指數(shù)函數(shù)的圖像與性質(zhì)③解指數(shù)方程與不等式④指數(shù)函數(shù)的綜合應(yīng)用★【文末附錄-指數(shù)運(yùn)算和指數(shù)函數(shù)思維導(dǎo)圖】一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理1.指數(shù)及指數(shù)運(yùn)算(1)根式的定義：一般地，如果SKIPIF1<0，那么SKIPIF1<0叫做SKIPIF1<0的SKIPIF1<0次方根，其中SKIPIF1<0，SKIPIF1<0，記為SKIPIF1<0，SKIPIF1<0稱為根指數(shù)，SKIPIF1<0稱為根底數(shù)．(2)根式的性質(zhì)：當(dāng)SKIPIF1<0為奇數(shù)時(shí)，正數(shù)的SKIPIF1<0次方根是一個(gè)正數(shù)，負(fù)數(shù)的SKIPIF1<0次方根是一個(gè)負(fù)數(shù).當(dāng)SKIPIF1<0為偶數(shù)時(shí)，正數(shù)的SKIPIF1<0次方根有兩個(gè)，它們互為相反數(shù).(3)指數(shù)的概念：指數(shù)是冪運(yùn)算SKIPIF1<0中的一個(gè)參數(shù)，SKIPIF1<0為底數(shù)，SKIPIF1<0為指數(shù)，指數(shù)位于底數(shù)的右上角，冪運(yùn)算表示指數(shù)個(gè)底數(shù)相乘.(4)有理數(shù)指數(shù)冪的分類①正整數(shù)指數(shù)冪SKIPIF1<0；②零指數(shù)冪SKIPIF1<0；③負(fù)整數(shù)指數(shù)冪SKIPIF1<0，SKIPIF1<0；④SKIPIF1<0的正分?jǐn)?shù)指數(shù)冪等于SKIPIF1<0，SKIPIF1<0的負(fù)分?jǐn)?shù)指數(shù)冪沒(méi)有意義．(5)有理數(shù)指數(shù)冪的性質(zhì)①SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；②SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；③SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；④SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．2.指數(shù)函數(shù)SKIPIF1<0SKIPIF1<0SKIPIF1<0圖象性質(zhì)①定義域SKIPIF1<0，值域SKIPIF1<0②SKIPIF1<0，即時(shí)SKIPIF1<0，SKIPIF1<0，圖象都經(jīng)過(guò)SKIPIF1<0點(diǎn)③SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0等于底數(shù)SKIPIF1<0④在定義域上是單調(diào)減函數(shù)在定義域上是單調(diào)增函數(shù)⑤SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0⑥既不是奇函數(shù)，也不是偶函數(shù)【常用結(jié)論】1.指數(shù)函數(shù)常用技巧(1)當(dāng)?shù)讛?shù)大小不定時(shí)，必須分“SKIPIF1<0”和“SKIPIF1<0”兩種情形討論．(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；SKIPIF1<0的值越小，圖象越靠近SKIPIF1<0軸，遞減的速度越快．當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0；SKIPIF1<0的值越大，圖象越靠近SKIPIF1<0軸，遞增速度越快．(3)指數(shù)函數(shù)SKIPIF1<0與SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱．二、題型分類精講二、題型分類精講刷真題明導(dǎo)向刷真題明導(dǎo)向一、單選題1．（2022·北京·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0，則對(duì)任意實(shí)數(shù)x，有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】直接代入計(jì)算，注意通分不要計(jì)算錯(cuò)誤．【詳解】SKIPIF1<0，故A錯(cuò)誤，C正確；SKIPIF1<0，不是常數(shù)，故BD錯(cuò)誤；故選：C．2．（2020·全國(guó)·統(tǒng)考高考真題）設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)已知等式，利用指數(shù)對(duì)數(shù)運(yùn)算性質(zhì)即可得解【詳解】由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，所以有SKIPIF1<0，故選：B.【點(diǎn)睛】本題考查的是有關(guān)指對(duì)式的運(yùn)算的問(wèn)題，涉及到的知識(shí)點(diǎn)有對(duì)數(shù)的運(yùn)算法則，指數(shù)的運(yùn)算法則，屬于基礎(chǔ)題目.3．（2020·山東·統(tǒng)考高考真題）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．【答案】B【分析】根據(jù)偶函數(shù)，指數(shù)函數(shù)的知識(shí)確定正確選項(xiàng).【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0在SKIPIF1<0上遞增.注意到SKIPIF1<0，所以B選項(xiàng)符合.故選：B4．（2021·全國(guó)·統(tǒng)考高考真題）下列函數(shù)中最小值為4的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)二次函數(shù)的性質(zhì)可判斷SKIPIF1<0選項(xiàng)不符合題意，再根據(jù)基本不等式“一正二定三相等”，即可得出SKIPIF1<0不符合題意，SKIPIF1<0符合題意．【詳解】對(duì)于A，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以其最小值為SKIPIF1<0，A不符合題意；對(duì)于B，因?yàn)镾KIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，等號(hào)取不到，所以其最小值不為SKIPIF1<0，B不符合題意；對(duì)于C，因?yàn)楹瘮?shù)定義域?yàn)镾KIPIF1<0，而SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以其最小值為SKIPIF1<0，C符合題意；對(duì)于D，SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，而SKIPIF1<0且SKIPIF1<0，如當(dāng)SKIPIF1<0，SKIPIF1<0，D不符合題意．故選：C．【點(diǎn)睛】本題解題關(guān)鍵是理解基本不等式的使用條件，明確“一正二定三相等”的意義，再結(jié)合有關(guān)函數(shù)的性質(zhì)即可解出．5．（2022·浙江·統(tǒng)考高考真題）已知SKIPIF1<0，則SKIPIF1<0（

）A．25 B．5 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)指數(shù)式與對(duì)數(shù)式的互化，冪的運(yùn)算性質(zhì)以及對(duì)數(shù)的運(yùn)算性質(zhì)即可解出．【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．故選：C.6．（2020·全國(guó)·統(tǒng)考高考真題）若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】將不等式變?yōu)镾KIPIF1<0，根據(jù)SKIPIF1<0的單調(diào)性知SKIPIF1<0，以此去判斷各個(gè)選項(xiàng)中真數(shù)與SKIPIF1<0的大小關(guān)系，進(jìn)而得到結(jié)果.【詳解】由SKIPIF1<0得：SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0為SKIPIF1<0上的增函數(shù)，SKIPIF1<0為SKIPIF1<0上的減函數(shù)，SKIPIF1<0為SKIPIF1<0上的增函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則A正確，B錯(cuò)誤；SKIPIF1<0與SKIPIF1<0的大小不確定，故CD無(wú)法確定.故選：A.【點(diǎn)睛】本題考查對(duì)數(shù)式的大小的判斷問(wèn)題，解題關(guān)鍵是能夠通過(guò)構(gòu)造函數(shù)的方式，利用函數(shù)的單調(diào)性得到SKIPIF1<0的大小關(guān)系，考查了轉(zhuǎn)化與化歸的數(shù)學(xué)思想.7．（2022·全國(guó)·統(tǒng)考高考真題）設(shè)SKIPIF1<0，則（

）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，導(dǎo)數(shù)判斷其單調(diào)性，由此確定SKIPIF1<0的大小.【詳解】方法一：構(gòu)造法設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0,令SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0故選：C.方法二：比較法解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，①SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；②SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0故SKIPIF1<0題型一指數(shù)冪的化簡(jiǎn)與求值策略方法指數(shù)冪運(yùn)算的一般原則(1)有括號(hào)的先算括號(hào)里的，無(wú)括號(hào)的先算指數(shù)運(yùn)算．(2)先乘除后加減，負(fù)指數(shù)冪化成正指數(shù)冪的倒數(shù)．(3)底數(shù)是負(fù)數(shù)，先確定符號(hào)；底數(shù)是小數(shù)，先化成分?jǐn)?shù)；底數(shù)是帶分?jǐn)?shù)的，先化成假分?jǐn)?shù)．(4)若是根式，應(yīng)化為分?jǐn)?shù)指數(shù)冪，盡可能用冪的形式表示，運(yùn)用指數(shù)冪的運(yùn)算性質(zhì)來(lái)解答．【典例1】計(jì)算：(1)SKIPIF1<0；(2)已知：SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）利用指數(shù)冪的運(yùn)算性質(zhì)可求得所求代數(shù)式的值；（2）在等式SKIPIF1<0兩邊平方可得出SKIPIF1<0，再利用平方關(guān)系可求得SKIPIF1<0，代入計(jì)算可得出SKIPIF1<0的值.【詳解】（1）解：原式SKIPIF1<0.（2）解：因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，所以，SKIPIF1<0，可得，SKIPIF1<0，因此，SKIPIF1<0.【題型訓(xùn)練】一、單選題1．（2023春·湖南·高三校聯(lián)考階段練習(xí)）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用指數(shù)的運(yùn)算性質(zhì)可求得所求代數(shù)式的值.【詳解】SKIPIF1<0.故選：B.2．（2023·全國(guó)·高三專題練習(xí)）下列結(jié)論中，正確的是（

）A．設(shè)SKIPIF1<0則SKIPIF1<0 B．若SKIPIF1<0，則SKIPIF1<0SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)分式指數(shù)冪及根式的運(yùn)算法則，正確運(yùn)算，即可判斷出正誤．【詳解】對(duì)于A，根據(jù)分式指數(shù)冪的運(yùn)算法則，可得SKIPIF1<0，選項(xiàng)A錯(cuò)誤；對(duì)于B，SKIPIF1<0，故SKIPIF1<0，選項(xiàng)B正確；對(duì)于C，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，選項(xiàng)C錯(cuò)誤；對(duì)于D，SKIPIF1<0，選項(xiàng)D錯(cuò)誤.故選：B．二、填空題3．（2023·全國(guó)·高三專題練習(xí)）若SKIPIF1<0，則SKIPIF1<0______【答案】SKIPIF1<0【分析】在等式SKIPIF1<0兩邊平方，可得出SKIPIF1<0的值.【詳解】在等式SKIPIF1<0兩邊平方可得SKIPIF1<0，因此，SKIPIF1<0.故答案為：SKIPIF1<0.4．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，化簡(jiǎn)二次根式SKIPIF1<0的值是________【答案】SKIPIF1<0.【分析】利用根式的性質(zhì)進(jìn)行化簡(jiǎn).【詳解】由SKIPIF1<0可知，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.5．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，則SKIPIF1<0=__________【答案】SKIPIF1<0【分析】利用立方和公式化簡(jiǎn)，再代入求值即可.【詳解】SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<06．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的值為_(kāi)_________．【答案】SKIPIF1<0【分析】將SKIPIF1<0變形為SKIPIF1<0，設(shè)SKIPIF1<0，求出t的值，SKIPIF1<0可化為SKIPIF1<0，即可求得答案.【詳解】由SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，（SKIPIF1<0舍去），故SKIPIF1<0，故答案為：SKIPIF1<0三、解答題7．（2023·全國(guó)·高三專題練習(xí)）（1）計(jì)算SKIPIF1<0；（2）若SKIPIF1<0，求SKIPIF1<0的值．【答案】（1）-5；（2）14.【分析】（1）由題意利用分?jǐn)?shù)指數(shù)冪的運(yùn)算法則，計(jì)算求得結(jié)果．（2）由題意兩次利用完全平方公式，計(jì)算求得結(jié)果．【詳解】（1）SKIPIF1<00.3﹣1﹣36+33+1SKIPIF1<036+27+1SKIPIF1<05．（2）若SKIPIF1<0，∴xSKIPIF1<02＝6，xSKIPIF1<04，∴x2+x﹣2+2＝16，∴x2+x﹣2＝14．8．（2023·全國(guó)·高三專題練習(xí)）（1）計(jì)算：SKIPIF1<0；（2）已知SKIPIF1<0是方程SKIPIF1<0的兩根，求SKIPIF1<0的值.【答案】（1）16；（2）SKIPIF1<0．【分析】（1）把根式化為分?jǐn)?shù)指數(shù)冪，然后由冪的運(yùn)算法則計(jì)算．（2）由韋達(dá)定理筣出SKIPIF1<0，求出SKIPIF1<0，求值式變形后代入已知值即可得．【詳解】（1）原式＝SKIPIF1<0SKIPIF1<0；（2）由題意SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，題型二指數(shù)函數(shù)的圖像與性質(zhì)策略方法解決指數(shù)函數(shù)有關(guān)問(wèn)題，思路是從它們的圖像與性質(zhì)考慮，按照數(shù)形結(jié)合的思路分析，從圖像與性質(zhì)找到解題的突破口，但要注意底數(shù)對(duì)問(wèn)題的影響.【典例1】函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，則SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象可能為（

）A．B．C．D．【答案】B【分析】根據(jù)函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，求出SKIPIF1<0的范圍，再根據(jù)函數(shù)SKIPIF1<0的圖象是由函數(shù)SKIPIF1<0的圖象向下平移SKIPIF1<0個(gè)單位得到的，作出函數(shù)SKIPIF1<0的大致圖象，即可得解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，則在函數(shù)SKIPIF1<0中SKIPIF1<0，函數(shù)SKIPIF1<0的圖象是由函數(shù)SKIPIF1<0的圖象向下平移SKIPIF1<0個(gè)單位得到的，作出函數(shù)SKIPIF1<0的大致圖象，如圖所示，所以SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象可能為B選項(xiàng).故選：B.【典例2】已知函數(shù)SKIPIF1<0的圖像恒過(guò)一點(diǎn)P，且點(diǎn)P在直線SKIPIF1<0的圖像上，則SKIPIF1<0的最小值為（）A．4 B．6 C．7 D．8【答案】D【分析】求出函數(shù)SKIPIF1<0的圖象所過(guò)的定點(diǎn)坐標(biāo)，由此建立SKIPIF1<0的關(guān)系，再利用均值不等式“1”的妙用求解作答.【詳解】函數(shù)SKIPIF1<0中，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，恒有SKIPIF1<0，則點(diǎn)SKIPIF1<0，依題意，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，因此SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最小值為8.故選：D【典例3】比較下列幾組值的大?。?1)SKIPIF1<0和SKIPIF1<0；(2)SKIPIF1<0和SKIPIF1<0；(3)SKIPIF1<0和SKIPIF1<0；(4)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0>SKIPIF1<0(4)SKIPIF1<0【分析】（1）（2）（3）（4）利用指數(shù)函數(shù)的單調(diào)性分析比較大小即可（1）由于SKIPIF1<0，SKIPIF1<0．∵SKIPIF1<0在SKIPIF1<0上為增函數(shù)，且SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0；（2）由于SKIPIF1<0．∵SKIPIF1<0在SKIPIF1<0上為減函數(shù)，且SKIPIF1<0，∴SKIPIF1<0；（3）∵SKIPIF1<0在SKIPIF1<0上為減函數(shù)，SKIPIF1<0在SKIPIF1<0上為增函數(shù)，且SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0；（4）∵SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上為增函數(shù)，且SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0．【題型訓(xùn)練】一、單選題1．（2023·天津河?xùn)|·一模）如圖中，①②③④中不屬于函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0中一個(gè)的是（

）A．① B．② C．③ D．④【答案】B【分析】根據(jù)指數(shù)函數(shù)的圖象的特征即可得答案.【詳解】解：由指數(shù)函數(shù)的性質(zhì)可知：①是SKIPIF1<0的部分圖象；③是SKIPIF1<0的部分圖象；④是SKIPIF1<0的部分圖象；所以只有②不是指數(shù)函數(shù)的圖象.故選：B.2．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）與函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】A【分析】分析各選項(xiàng)中兩函數(shù)的單調(diào)性及其圖象與SKIPIF1<0軸的交點(diǎn)位置，即可得出合適的選項(xiàng).【詳解】A選項(xiàng)，函數(shù)SKIPIF1<0為減函數(shù)，則SKIPIF1<0，且函數(shù)SKIPIF1<0的圖象交SKIPIF1<0軸正半軸點(diǎn)SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，函數(shù)SKIPIF1<0為增函數(shù)，且函數(shù)SKIPIF1<0交SKIPIF1<0軸正半軸于點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，A滿足；對(duì)于B選項(xiàng)，函數(shù)SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，函數(shù)SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，顯然SKIPIF1<0，B不滿足；對(duì)于C選項(xiàng)，函數(shù)SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，函數(shù)SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，顯然SKIPIF1<0，C不滿足；對(duì)于D選項(xiàng)，函數(shù)SKIPIF1<0為減函數(shù)，則SKIPIF1<0，函數(shù)SKIPIF1<0為減函數(shù)，則SKIPIF1<0，D不滿足.故選：A.3．（2023·云南紅河·云南省建水第一中學(xué)?？寄M預(yù)測(cè)）函數(shù)SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0）的圖象恒過(guò)的定點(diǎn)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】令SKIPIF1<0可得定點(diǎn).【詳解】令SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，函數(shù)SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0）的圖象恒過(guò)的定點(diǎn)是SKIPIF1<0.故選：B.4．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象過(guò)定點(diǎn)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【分析】根據(jù)指數(shù)型函數(shù)的定點(diǎn)求解SKIPIF1<0，代入后再求解一元二次不等式.【詳解】當(dāng)SKIPIF1<0時(shí)，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：D5．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的圖像恒過(guò)定點(diǎn)A，若點(diǎn)A在雙曲線SKIPIF1<0上，則m-n的最大值為（

）A．6 B．-2 C．1 D．4【答案】D【分析】令SKIPIF1<0，求得SKIPIF1<0，由點(diǎn)A在雙曲線上，得到SKIPIF1<0，然后由“1”的代換，利用基本不等式求解.【詳解】令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，因?yàn)辄c(diǎn)A在雙曲線SKIPIF1<0上，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，所以m-n的最大值為4故選：D6．（2023·天津·一模）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)指數(shù)函數(shù)，冪函數(shù)的性質(zhì)即可判斷SKIPIF1<0，SKIPIF1<0，再對(duì)SKIPIF1<0，SKIPIF1<0進(jìn)行取對(duì)數(shù)，結(jié)合對(duì)數(shù)函數(shù)的性質(zhì)即可判斷SKIPIF1<0，進(jìn)而即可得到答案．【詳解】由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．故選：D．7．（2023·北京東城·統(tǒng)考二模）設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0為增函數(shù)，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】B【分析】首先分析函數(shù)在各段函數(shù)的單調(diào)性，依題意可得SKIPIF1<0且SKIPIF1<0，結(jié)合SKIPIF1<0與SKIPIF1<0的函數(shù)圖象及增長(zhǎng)趨勢(shì)求出參數(shù)的取值范圍.【詳解】因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0函數(shù)單調(diào)遞增，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，要使函數(shù)SKIPIF1<0為增函數(shù)，則SKIPIF1<0且SKIPIF1<0，又函數(shù)SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上有兩個(gè)交點(diǎn)SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0的增長(zhǎng)趨勢(shì)比SKIPIF1<0快得多，SKIPIF1<0與SKIPIF1<0的函數(shù)圖象如下所示：所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B8．（2023·浙江·高三專題練習(xí)）已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用中間值SKIPIF1<0比較a,b的大小，再讓b，c與中間值SKIPIF1<0比較，判斷b,c的大小，即可得解.【詳解】SKIPIF1<0，又因?yàn)橥ㄟ^(guò)計(jì)算知SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B二、多選題9．（2023·黑龍江哈爾濱·哈爾濱三中校考二模）點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖象上，當(dāng)SKIPIF1<0，則SKIPIF1<0可能等于（

）A．-1 B．SKIPIF1<0 C．SKIPIF1<0 D．0【答案】BC【分析】根據(jù)目標(biāo)式的幾何意義為SKIPIF1<0在SKIPIF1<0部分圖象上的動(dòng)點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0所成直線的斜率SKIPIF1<0，即可求范圍.【詳解】由SKIPIF1<0表示SKIPIF1<0與點(diǎn)SKIPIF1<0所成直線的斜率SKIPIF1<0，又SKIPIF1<0是SKIPIF1<0在SKIPIF1<0部分圖象上的動(dòng)點(diǎn)，圖象如下：如上圖，SKIPIF1<0，則SKIPIF1<0，只有B、C滿足.故選：BC三、填空題10．（2023·全國(guó)·高三專題練習(xí)）請(qǐng)寫(xiě)出一個(gè)同時(shí)滿足下列條件①②③的函數(shù)SKIPIF1<0____________．①SKIPIF1<0；②對(duì)任意SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；③SKIPIF1<0．【答案】SKIPIF1<0（答案不唯一）.【分析】根據(jù)SKIPIF1<0的圖像經(jīng)過(guò)原點(diǎn)，且在R上單調(diào)遞增，又SKIPIF1<0，利用指數(shù)函數(shù)的圖像和性質(zhì)構(gòu)造函數(shù)即可.【詳解】根據(jù)題意知SKIPIF1<0的圖像經(jīng)過(guò)原點(diǎn)，且在R上單調(diào)遞增，又SKIPIF1<0.考慮到圖像有“漸近線”的指數(shù)函數(shù)，構(gòu)造SKIPIF1<0符合題意.故答案為：SKIPIF1<0（答案不唯一）11．（2023秋·吉林松原·高三前郭爾羅斯縣第五中學(xué)?？计谀┮阎猄KIPIF1<0為SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為_(kāi)__________.【答案】SKIPIF1<0【分析】由函數(shù)的奇偶性與單調(diào)性轉(zhuǎn)化后求解，【詳解】由函數(shù)SKIPIF1<0與SKIPIF1<0均在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0等價(jià)于SKIPIF1<0，得SKIPIF1<0，故答案為：SKIPIF1<012．（2023·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則k的取值范圍為_(kāi)___________．【答案】SKIPIF1<0【分析】先畫(huà)出函數(shù)SKIPIF1<0，再根據(jù)函數(shù)在SKIPIF1<0上單調(diào)遞減求解.【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0的圖象是由函數(shù)SKIPIF1<0的圖象向下平移一個(gè)單位后，再把位于x軸下方的圖象沿x軸翻折到x軸上方得到的，函數(shù)圖象如圖所示：由圖象知，其在SKIPIF1<0上單調(diào)遞減，所以k的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0四、解答題13．（2023·全國(guó)·高三練習(xí)）已知函數(shù)SKIPIF1<0（a為常數(shù)）和函數(shù)SKIPIF1<0，且SKIPIF1<0為奇函數(shù)．(1)求實(shí)數(shù)a的值；(2)設(shè)不等式SKIPIF1<0恒成立，試求實(shí)數(shù)SKIPIF1<0的范圍．【答案】(1)1(2)SKIPIF1<0【分析】（1）根據(jù)奇函數(shù)的定義求出a；（2）運(yùn)用參數(shù)分離法，構(gòu)造函數(shù)，運(yùn)用函數(shù)的單調(diào)性求解.【詳解】（1）SKIPIF1<0為奇函數(shù)，SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0，經(jīng)檢驗(yàn)符合題意；（2）由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0；題型三解指數(shù)方程與不等式策略方法指數(shù)方程或不等式的解法(1)解指數(shù)方程或不等式的依據(jù)①af(x)＝ag(x)?f(x)＝g(x)．②af(x)＞ag(x)，當(dāng)a＞1時(shí)，等價(jià)于f(x)＞g(x)；當(dāng)0＜a＜1時(shí)，等價(jià)于f(x)＜g(x)．(2)解指數(shù)方程或不等式的方法先利用冪的運(yùn)算性質(zhì)化為同底數(shù)冪，再利用函數(shù)單調(diào)性轉(zhuǎn)化為一般不等式求解．【典例1】不等式SKIPIF1<0對(duì)于SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍是______．【答案】SKIPIF1<0【分析】由題意結(jié)合指數(shù)函數(shù)的單調(diào)性，得SKIPIF1<0對(duì)于SKIPIF1<0恒成立，設(shè)SKIPIF1<0，結(jié)合二次函數(shù)的性質(zhì)可求得答案.【詳解】由SKIPIF1<0得SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0對(duì)于SKIPIF1<0恒成立，設(shè)SKIPIF1<0，顯然SKIPIF1<0開(kāi)口向上，對(duì)稱軸為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值0，則SKIPIF1<0，即a的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.【題型訓(xùn)練】一、單選題1．（2023·海南·統(tǒng)考模擬預(yù)測(cè)）已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】先求出集合A，集合的交集運(yùn)算即可求出SKIPIF1<0．【詳解】SKIPIF1<0集合SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．故選：A．2．（2023·河北·高三學(xué)業(yè)考試）設(shè)函數(shù)SKIPIF1<0則滿足SKIPIF1<0的SKIPIF1<0取值范圍是A．[-1,2] B．[0,2] C．[1,+SKIPIF1<0） D．[0,+SKIPIF1<0）【答案】D【分析】根據(jù)函數(shù)解析式，結(jié)合指對(duì)數(shù)函數(shù)的單調(diào)性，討論不同區(qū)間對(duì)應(yīng)SKIPIF1<0的x范圍，然后取并.【詳解】由SKIPIF1<0，可得SKIPIF1<0；或SKIPIF1<0，可得SKIPIF1<0；綜上，SKIPIF1<0的SKIPIF1<0取值范圍是SKIPIF1<0.故選：D3．（2023·全國(guó)·高三專題練習(xí)）若關(guān)于x的不等式SKIPIF1<0有實(shí)數(shù)解，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】A【分析】分離參數(shù)將問(wèn)題轉(zhuǎn)化為SKIPIF1<0有解，計(jì)算即可.【詳解】由題知SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.因?yàn)殛P(guān)于SKIPIF1<0的不等式SKIPIF1<0有實(shí)數(shù)解，即SKIPIF1<0SKIPIF1<0有實(shí)數(shù)解，所以SKIPIF1<0，即SKIPIF1<0.故選：A4．（2023·全國(guó)·高三專題練習(xí)）若不等式SKIPIF1<0恒成立，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】B【詳解】分析：首先根據(jù)指數(shù)函數(shù)的性質(zhì)，將不等式恒成立轉(zhuǎn)化為SKIPIF1<0恒成立，利用判別式SKIPIF1<0，從而求得實(shí)數(shù)SKIPIF1<0的取值范圍.詳解：不等式SKIPIF1<0恒成立，即SKIPIF1<0，即SKIPIF1<0恒成立，即SKIPIF1<0恒成立，所以SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0，故選B.點(diǎn)睛：該題考查的是有關(guān)不等式恒成立，求參數(shù)的取值范圍的問(wèn)題，在解題的過(guò)程中，需要明確指數(shù)式的運(yùn)算法則，注意應(yīng)用指數(shù)函數(shù)的單調(diào)性，得到指數(shù)所滿足的大小關(guān)系，利用二次不等式恒成立問(wèn)題，結(jié)合式子的判別式，求得結(jié)果.二、填空題5．（2023·全國(guó)·高三專題練習(xí)）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)__________.【答案】SKIPIF1<0【分析】分別根據(jù)對(duì)數(shù)和指數(shù)函數(shù)的單調(diào)性解不等式，再求交集即可．【詳解】SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)SKIPIF1<0成立；當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0．所以SKIPIF1<0又SKIPIF1<0,SKIPIF1<0∴a的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<06．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，若SKIPIF1<0，則SKIPIF1<0的取值范圍為_(kāi)_______．【答案】SKIPIF1<0【分析】先求得a的值，再利用函數(shù)單調(diào)性把不等式SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，解之即可求得SKIPIF1<0的取值范圍.【詳解】定義在R上函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，則SKIPIF1<0，解之得SKIPIF1<0，經(jīng)檢驗(yàn)符合題意SKIPIF1<0均為R上增函數(shù)，則SKIPIF1<0為R上增函數(shù)，又SKIPIF1<0，則不等式SKIPIF1<0等價(jià)于SKIPIF1<0，解之得SKIPIF1<0故答案為：SKIPIF1<0三、解答題7．（2023·全國(guó)·高三練習(xí)）解下列方程：(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；(4)SKIPIF1<0．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0或SKIPIF1<0；(3)SKIPIF1<0或SKIPIF1<0；(4)SKIPIF1<0【分析】（1）（2）根據(jù)指數(shù)冪的運(yùn)算法則結(jié)合指數(shù)函數(shù)的性質(zhì)即得；（3）（4）根據(jù)對(duì)數(shù)的運(yùn)算律結(jié)合對(duì)數(shù)函數(shù)的性質(zhì)即得.【詳解】（1）由SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；（2）由SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0；（3）因?yàn)镾KIPIF1<0，所以原方程可化為SKIPIF1<0，即SKIPIF1<0，兩邊取對(duì)數(shù)可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，經(jīng)檢驗(yàn)SKIPIF1<0或SKIPIF1<0是原方程的解，所以SKIPIF1<0或SKIPIF1<0；（4）由SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，經(jīng)檢驗(yàn)滿足題意，所以SKIPIF1<0.8．（2023秋·江西鷹潭·高三貴溪市實(shí)驗(yàn)中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，與SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱的圖象過(guò)點(diǎn)SKIPIF1<0.(1)求SKIPIF1<0的值；(2)求不等式SKIPIF1<0的解集.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0且SKIPIF1<0}．【分析】（1）由對(duì)稱性知SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0，代入后可得SKIPIF1<0值；（2）結(jié)合指數(shù)函數(shù)性質(zhì)解不等式．【詳解】（1）由題意SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0；（2）由（1）SKIPIF1<0，顯然SKIPIF1<0，不等式SKIPIF1<0為SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，SKIPIF1<0，所以不等式的解集為SKIPIF1<0且SKIPIF1<0}．題型四指數(shù)函數(shù)的綜合應(yīng)用策略方法指數(shù)函數(shù)通過(guò)平移、伸縮及翻折等變換，或與其他函數(shù)進(jìn)行結(jié)合形成復(fù)合函數(shù)時(shí)，我們對(duì)這類問(wèn)題的解決方式是進(jìn)行還原分離，化繁為簡(jiǎn)，借助函數(shù)的單調(diào)性、奇偶性、對(duì)稱性及周期性解決問(wèn)題．【典例1】函數(shù)SKIPIF1<0單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)復(fù)合函數(shù)同增異減，即可判斷出單調(diào)遞增區(qū)間．【詳解】由SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0為減函數(shù)，求SKIPIF1<0的單調(diào)遞增區(qū)間，等價(jià)于求SKIPIF1<0的單調(diào)遞減區(qū)間，因?yàn)镾KIPIF1<0在SKIPIF1<0單調(diào)遞減，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0，故選：C．【典例2】當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，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】B【解析】將SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，轉(zhuǎn)化為SKIPIF1<0對(duì)一切