人教A版高中數(shù)學(xué)（必修第一冊(cè)）同步講義 第四章 指數(shù)函數(shù)與對(duì)數(shù)函數(shù) 章末重點(diǎn)題型大總結(jié)（教師版）

第第頁(yè)第四章指數(shù)函數(shù)與對(duì)數(shù)函數(shù)章末題型大總結(jié)一、思維導(dǎo)圖二、題型精講題型01有關(guān)指數(shù)、對(duì)數(shù)的運(yùn)算【典例1】（2023春·四川雅安·高二統(tǒng)考期末）計(jì)算：(1)SKIPIF1<0；(2)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（2）SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0【典例2】（2022秋·四川成都·高一石室中學(xué)校考期中）求下列各式的值：(1)計(jì)算：SKIPIF1<0；(2)若SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（2）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0.【變式1】（2023春·吉林長(zhǎng)春·高二長(zhǎng)春外國(guó)語(yǔ)學(xué)校?？计谀?）已知SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的值；（2）SKIPIF1<0【答案】（1）SKIPIF1<0；（2）SKIPIF1<0【詳解】（1）SKIPIF1<0；（2）SKIPIF1<0.【變式2】（2023·全國(guó)·高三專(zhuān)題練習(xí)）計(jì)算(1)SKIPIF1<0.(2)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)2【詳解】（1）SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0（2）SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0=2題型02數(shù)的大小比較問(wèn)題【典例1】（2023春·貴州六盤(pán)水·高一統(tǒng)考期末）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由對(duì)數(shù)函數(shù)的性質(zhì)，可得SKIPIF1<0，又由指數(shù)函數(shù)的性質(zhì)，可得SKIPIF1<0，由冪函數(shù)SKIPIF1<0在SKIPIF1<0為單調(diào)遞增函數(shù)，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：D.【典例2】（2021秋·廣東汕尾·高一海豐縣海城仁榮中學(xué)校考階段練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】∵SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，綜上，SKIPIF1<0.故選：C.【變式1】（2023春·廣西北?！じ叨y(tǒng)考期末）設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0故選：A.【變式2】（2023春·河南安陽(yáng)·高二統(tǒng)考期末）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因此SKIPIF1<0，故選:C.題型03定義域問(wèn)題【典例1】（云南省紅河哈尼族彝族自治州2022-2023學(xué)年高一下學(xué)期期末學(xué)業(yè)質(zhì)量監(jiān)測(cè)數(shù)學(xué)試題）函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題得SKIPIF1<0，解得SKIPIF1<0且SKIPIF1<0.故選：A.【典例2】（2023春·重慶永川·高一重慶市永川北山中學(xué)校?？奸_(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0是定義在R上的增函數(shù)，則實(shí)數(shù)a的取值范圍是.【答案】SKIPIF1<0【詳解】∵函數(shù)SKIPIF1<0在R上單調(diào)遞增，∴SKIPIF1<0，即實(shí)數(shù)a的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0．【變式1】（2023春·北京順義·高二牛欄山一中?？计谀┖瘮?shù)SKIPIF1<0的定義域?yàn)?【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則實(shí)數(shù)m的取值范圍是．【答案】SKIPIF1<0【詳解】由函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，得SKIPIF1<0，SKIPIF1<0恒成立．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，成立；當(dāng)SKIPIF1<0時(shí)，需滿(mǎn)足SKIPIF1<0于是SKIPIF1<0．綜上所述，m的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0.題型04值域問(wèn)題【典例1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為．【答案】SKIPIF1<0【詳解】SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0.故答案為：SKIPIF1<0【典例2】（2023春·陜西榆林·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)判斷函數(shù)SKIPIF1<0的奇偶性并予以證明；(2)若存在SKIPIF1<0使得不等式SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的最大值.【答案】(1)偶函數(shù)，證明見(jiàn)解析(2)1【詳解】（1）函數(shù)SKIPIF1<0為偶函數(shù)，證明如下：SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱(chēng)，SKIPIF1<0，SKIPIF1<0為偶函數(shù).（2）若存在SKIPIF1<0使得不等式SKIPIF1<0成立，SKIPIF1<0，而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0實(shí)數(shù)SKIPIF1<0的最大值為1.【典例3】（2023春·江蘇揚(yáng)州·高一統(tǒng)考開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0，且SKIPIF1<0．(1)求實(shí)數(shù)a的值，并用單調(diào)性定義證明SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；(2)若當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，求實(shí)數(shù)m的值．【答案】(1)a＝1，證明見(jiàn)解析(2)2【詳解】（1）由SKIPIF1<0得a＝1．任取SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．（2）由（1）知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，解得m＝2或SKIPIF1<0（舍去），所以m＝2．【典例4】（2023春·河南焦作·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在零點(diǎn)，求SKIPIF1<0的取值范圍；(2)若a＞1，且對(duì)任意SKIPIF1<0，都有SKIPIF1<0，使得SKIPIF1<0成立，求a的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0．【詳解】（1）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則由零點(diǎn)存在定理可得SKIPIF1<0，即SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0．（2）若對(duì)任意SKIPIF1<0，都有SKIPIF1<0，使得SKIPIF1<0成立，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．因?yàn)閍＞1，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．當(dāng)1＜a＜2時(shí)，SKIPIF1<0，SKIPIF1<0，不符合條件，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，符合條件，所以a的取值范圍是SKIPIF1<0．【變式1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，對(duì)SKIPIF1<0都有SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，因?yàn)閷?duì)SKIPIF1<0都有SKIPIF1<0成立，所以SKIPIF1<0，故答案為：SKIPIF1<0【變式2】（2023秋·云南紅河·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的解集；(2)求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.所以不等式SKIPIF1<0的解集為SKIPIF1<0.（2）易知SKIPIF1<0的對(duì)稱(chēng)軸為SKIPIF1<0，則①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0.②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0綜上SKIPIF1<0.【變式3】（2023春·河南新鄉(xiāng)·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0．(1)求不等式SKIPIF1<0的解集；(2)若不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）令SKIPIF1<0，則不等式SKIPIF1<0即為SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0時(shí)SKIPIF1<0，即不等式SKIPIF1<0的解集為SKIPIF1<0.（2）由SKIPIF1<0對(duì)SKIPIF1<0恒成立，可得SKIPIF1<0對(duì)SKIPIF1<0恒成立，令SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.【變式4】（2023春·新疆昌吉·高二統(tǒng)考期末）已知SKIPIF1<0（實(shí)數(shù)b為常數(shù)）．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的定義域D；(2)若不等式SKIPIF1<0當(dāng)SKIPIF1<0時(shí)恒成立，求實(shí)數(shù)b的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，解之得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0．（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為單調(diào)遞增函數(shù)，故SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0．令SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即b的取值范圍為SKIPIF1<0．題型05指數(shù)（型）函數(shù)的圖象與性質(zhì)【典例1】（2023秋·廣西河池·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0,若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】作出SKIPIF1<0的圖象，如圖所示：

由SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0．故選：D．【典例2】（2023春·陜西寶雞·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0.(1)試判斷函數(shù)的單調(diào)性，并加以證明；(2)若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有解，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)在SKIPIF1<0上單調(diào)遞增，證明見(jiàn)解析(2)SKIPIF1<0【詳解】（1）SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，證明如下：任取SKIPIF1<0，則SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.（2）由（1）可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)镾KIPIF1<0，所以m的取值范圍是SKIPIF1<0.【典例3】（2023春·福建·高二統(tǒng)考學(xué)業(yè)考試）函數(shù)SKIPIF1<0，SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的定義域；(2)若SKIPIF1<0為奇函數(shù)，求m的值；(3)當(dāng)SKIPIF1<0時(shí)，不等SKIPIF1<0在SKIPIF1<0恒成立，求k的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【詳解】（1）依題意可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0.（2）若SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.（3）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0,所以不等式SKIPIF1<0在SKIPIF1<0恒成立，即SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0取等，所以SKIPIF1<0.故k的取值范圍為SKIPIF1<0.【變式1】（多選）（2023·全國(guó)·高三專(zhuān)題練習(xí)）（多選）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象如圖所示，則下列結(jié)論正確的是（）

A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【詳解】由圖象可知，函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0.對(duì)于A選項(xiàng)，SKIPIF1<0，A對(duì)；對(duì)于B選項(xiàng)，SKIPIF1<0，B對(duì)；對(duì)于C選項(xiàng)，SKIPIF1<0，C錯(cuò)；對(duì)于D選項(xiàng)，由題意可知，SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，D對(duì).故選：ABD.【變式2】（2023春·湖北荊州·高一校聯(lián)考期中）已知函數(shù)SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0的定義域是SKIPIF1<0時(shí)，此時(shí)值域也是SKIPIF1<0.(1)求SKIPIF1<0的值；(2)若SKIPIF1<0，證明SKIPIF1<0為奇函數(shù)，并求不等式SKIPIF1<0的解集.【答案】(1)SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0(2)證明見(jiàn)解析，SKIPIF1<0【詳解】（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞減，且SKIPIF1<0.又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，根據(jù)復(fù)合函數(shù)的單調(diào)性可知，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增，且SKIPIF1<0.又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，根據(jù)復(fù)合函數(shù)的單調(diào)性可知，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0.綜上，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0.（2）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，定義域?yàn)镾KIPIF1<0，且函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.因?yàn)镾KIPIF1<0SKIPIF1<0，所以SKIPIF1<0為奇函數(shù).則不等式SKIPIF1<0，可化為SKIPIF1<0.又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，即SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0.【變式3】（2023春·江蘇南京·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，解不等式SKIPIF1<0；(2)當(dāng)SKIPIF1<0對(duì)SKIPIF1<0恒成立時(shí)，求整數(shù)SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0(2)4【詳解】（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，解得SKIPIF1<0.即SKIPIF1<0（2）因?yàn)镾KIPIF1<0，所以可得SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0,所以整數(shù)SKIPIF1<0的最小值為4.題型06對(duì)數(shù)（型）函數(shù)的圖象與性質(zhì)【典例1】（2023春·湖北荊門(mén)·高二統(tǒng)考期末）設(shè)函數(shù)SKIPIF1<0在定義域SKIPIF1<0上滿(mǎn)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<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0，可得SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0.因?yàn)镾KIPIF1<0在SKIPIF1<0上是減函數(shù)，所以SKIPIF1<0在SKIPIF1<0上是減函數(shù).又SKIPIF1<0，所以SKIPIF1<0.由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.所以不等式SKIPIF1<0的解集為SKIPIF1<0.故選:D.【典例2】（2023春·陜西寶雞·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0的最小值為0，則實(shí)數(shù)SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0最小值為0，設(shè)SKIPIF1<0，所以只要滿(mǎn)足SKIPIF1<0恒成立，函數(shù)對(duì)稱(chēng)軸為SKIPIF1<0，且SKIPIF1<0，①SKIPIF1<0，即SKIPIF1<0時(shí)，滿(mǎn)足題意；②SKIPIF1<0，即SKIPIF1<0時(shí)，需滿(mǎn)足SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，此時(shí)實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.綜上，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0故答案為：SKIPIF1<0.【典例3】（2023春·浙江麗水·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0是偶函數(shù)．(1)若SKIPIF1<0，求SKIPIF1<0的值；(2)若實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0【詳解】（1）由已知可得，SKIPIF1<0SKIPIF1<0.因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.由SKIPIF1<0可得，SKIPIF1<0，化簡(jiǎn)可得SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0．（2）SKIPIF1<0.令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，根據(jù)對(duì)勾函數(shù)的單調(diào)性可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.又SKIPIF1<0單調(diào)遞增，根據(jù)復(fù)合函數(shù)的單調(diào)性可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.又函數(shù)SKIPIF1<0單調(diào)遞增，根據(jù)復(fù)合函數(shù)的單調(diào)性可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.又因?yàn)楹瘮?shù)SKIPIF1<0為偶函數(shù)，所以有SKIPIF1<0，SKIPIF1<0.所以，由SKIPIF1<0即可得出SKIPIF1<0，所以，SKIPIF1<0.平方整理可得，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．【典例4】（2023秋·重慶長(zhǎng)壽·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0為奇函數(shù).(1)求實(shí)數(shù)SKIPIF1<0的值，判斷函數(shù)SKIPIF1<0的單調(diào)性并用定義證明；(2)求關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集.【答案】(1)SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上是增函數(shù)，證明見(jiàn)解析(2)SKIPIF1<0或SKIPIF1<0.【詳解】（1）解：因?yàn)镾KIPIF1<0的定義域是SKIPIF1<0且SKIPIF1<0是奇函數(shù)，可得SKIPIF1<0，可得SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，證明如下：任取SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0為增函數(shù)，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上是增函數(shù).（2）解：由（1）知SKIPIF1<0在SKIPIF1<0上是增函數(shù)，且SKIPIF1<0，則不等式SKIPIF1<0，即為SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以不等式的解集為SKIPIF1<0或SKIPIF1<0.【變式1】（2023春·廣西北?！じ叨y(tǒng)考期末）設(shè)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最大值為.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0，根據(jù)基本不等式有SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0.則SKIPIF1<0，所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023春·山東日照·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0是偶函數(shù)．(1)求k的值；(2)若方程SKIPIF1<0有解，求實(shí)數(shù)m的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）由已知可得，SKIPIF1<0SKIPIF1<0.因?yàn)镾KIPIF1<0為R上的偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0恒成立，所以SKIPIF1<0，解得SKIPIF1<0，經(jīng)檢驗(yàn)，SKIPIF1<0滿(mǎn)足題意，故SKIPIF1<0.（2）由（1）知，SKIPIF1<0SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.因?yàn)榉匠蘏KIPIF1<0有解，即SKIPIF1<0有解，所以SKIPIF1<0，即SKIPIF1<0.【變式3】（2023·全國(guó)·高一假期作業(yè)）已知函數(shù)SKIPIF1<0為定義在SKIPIF1<0上的偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0．(1)求a的值：(2)求SKIPIF1<0的解析式；(3)求不等式SKIPIF1<0的解集．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【詳解】（1）解：因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．（2）設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0．

因?yàn)镾KIPIF1<0為定義在SKIPIF1<0上的偶函數(shù)，則SKIPIF1<0．

綜上所述，SKIPIF1<0（3）由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0

解得SKIPIF1<0或SKIPIF1<0．

故不等式SKIPIF1<0的解集為SKIPIF1<0．題型07函數(shù)與方程【典例1】（2023春·河南商丘·高二商丘市第一高級(jí)中學(xué)?？计谀┮阎瘮?shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)的個(gè)數(shù)是（

）A．3 B．4 C．5 D．6【答案】D【詳解】令SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由于SKIPIF1<0，由零點(diǎn)存在定理可知，存在SKIPIF1<0，使得SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0．作出函數(shù)SKIPIF1<0，直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖象如下圖所示：

由圖象可知，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有三個(gè)交點(diǎn)；直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)；直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有且只有一個(gè)交點(diǎn)．綜上所述，函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為SKIPIF1<0．故選：D．【典例2】（多選）（2023春·陜西西安·高一西安市鐵一中學(xué)校考期末）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則下列結(jié)論正確的是（

）A．m的取值范圍為SKIPIF1<0 B．SKIPIF1<0的取值范圍為SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0最大值為1【答案】AC【詳解】函數(shù)SKIPIF1<0圖象如圖所示：

由圖可得SKIPIF1<0，A正確；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故SKIPIF1<0，B錯(cuò)誤；又SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0,可得SKIPIF1<0，C正確又SKIPIF1<0可得SKIPIF1<0，又SKIPIF1<0，故等號(hào)不成立,即SKIPIF1<0，D錯(cuò)誤,故選：AC.【典例3】（2023春·山東煙臺(tái)·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，則實(shí)數(shù)a的取值范圍為．【答案】SKIPIF1<0【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)榉匠蘏KIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，如圖，

所以SKIPIF1<0，即SKIPIF1<0.若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)方程SKIPIF1<0有1個(gè)解，如圖，

當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有1個(gè)解需滿(mǎn)足SKIPIF1<0，即SKIPIF1<0.綜上所述，實(shí)數(shù)a的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（多選）（2023春·山東德州·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0，下列說(shuō)法正確的是（

）A．若SKIPIF1<0是偶函數(shù)，則SKIPIF1<0B．SKIPIF1<0的單調(diào)減區(qū)間是SKIPIF1<0C．SKIPIF1<0的值域是SKIPIF1<0D．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)【答案】ABD【詳解】對(duì)于A，若SKIPIF1<0是偶函數(shù)，定義域?yàn)镾KIPIF1<0，對(duì)于任意的SKIPIF1<0，由SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，A正確，對(duì)于B,SKIPIF1<0由SKIPIF1<0復(fù)合而成，由于SKIPIF1<0在SKIPIF1<0單調(diào)遞減，開(kāi)口向上的二次函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以由復(fù)合函數(shù)單調(diào)性的判斷可知SKIPIF1<0的單調(diào)減區(qū)間是SKIPIF1<0，B正確，對(duì)于C，由B可知，SKIPIF1<0的單調(diào)減區(qū)間是SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值，故SKIPIF1<0，故SKIPIF1<0值域?yàn)镾KIPIF1<0，故C錯(cuò)誤，對(duì)于D，由C可知SKIPIF1<0值域?yàn)镾KIPIF1<0，如圖：當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，所以SKIPIF1<0有兩個(gè)交點(diǎn)，故D正確，

故選：ABD【變式2】（2023秋·云南紅河·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.函數(shù)SKIPIF1<0(SKIPIF1<0且SKIPIF1<0)，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有20個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為.【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有20個(gè)零點(diǎn)，則函數(shù)SKIPIF1<0圖象與函數(shù)SKIPIF1<0圖象在區(qū)間SKIPIF1<0上有20個(gè)交點(diǎn)，由SKIPIF1<0知，SKIPIF1<0是周期為2的函數(shù)，作函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的部分圖象如下：

易知當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0圖象與函數(shù)SKIPIF1<0圖象有17個(gè)交點(diǎn)，故在SKIPIF1<0上有3個(gè)交點(diǎn)，顯然SKIPIF1<0不滿(mǎn)足題意，所以則需SKIPIF1<0，解得SKIPIF1<0故答案為：SKIPIF1<0【變式3】（2023秋·山西運(yùn)城·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0.(1)若函數(shù)SKIPIF1<0，SKIPIF1<0，求函數(shù)SKIPIF1<0的最小值；(2)設(shè)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0圖象有SKIPIF1<0個(gè)公共點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）解：因?yàn)镾KIPIF1<0，則SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，則二次函數(shù)SKIPIF1<0的對(duì)稱(chēng)軸方程為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.綜上：SKIPIF1<0.（2）解：SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0與SKIPIF1<0圖象有SKIPIF1<0個(gè)公共點(diǎn)，由SKIPIF1<0可得SKIPIF1<0且SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，又因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以方程SKIPIF1<0有兩個(gè)不等的正根，所以，SKIPIF1<0，解得SKIPIF1<0，因此，實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.題型08函數(shù)模型及其應(yīng)用【典例1】（2023春·湖南長(zhǎng)沙·高一湖南師大附中校考期末）某教學(xué)軟件在剛發(fā)布時(shí)有100名教師用戶(hù)，發(fā)布5天后有1000名教師用戶(hù)，如果教師用戶(hù)人數(shù)SKIPIF1<0與天數(shù)t之間滿(mǎn)足關(guān)系式：SKIPIF1<0，其中SKIPIF1<0為常數(shù)，SKIPIF1<0是剛發(fā)布的時(shí)間，則教師用戶(hù)超過(guò)30000名至少經(jīng)過(guò)的天數(shù)為（

）（參考數(shù)據(jù)：SKIPIF1<0）A．11 B．12 C．13 D．14【答案】C【詳解】由題意得SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，所以教師用戶(hù)超過(guò)20000名至少經(jīng)過(guò)SK