高考數(shù)學(xué)二輪復(fù)習(xí)壓軸題專題09 一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（利用導(dǎo)數(shù)研究函數(shù)圖象及性質(zhì)全題型壓軸題）（教師版）

專題09一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（利用導(dǎo)數(shù)研究函數(shù)圖象及性質(zhì)）（全題型壓軸題）目錄TOC\o"1-1"\h\u①圖象識(shí)別題 1②函數(shù)切線條數(shù)問題 5③不等式整數(shù)解問題 9④函數(shù)零點(diǎn)，方程根，兩個(gè)函數(shù)圖象交點(diǎn)問題 14⑤不等式恒成立問題 21①圖象識(shí)別題1．（2023·吉林通化·梅河口市第五中學(xué)校考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的圖像大致為（

）A．

B．

C．

D．

【答案】C【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故排除A；注意到SKIPIF1<0，則SKIPIF1<0為奇函數(shù)，故可排除B；又注意到SKIPIF1<0時(shí)，SKIPIF1<0，故可排除D.故選：C2．（2023·河北·統(tǒng)考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的大致圖象是（

）A．

B．

C．

D．

【答案】D【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)椋篠KIPIF1<0，且SKIPIF1<0，所以函數(shù)SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極小值，故選：D3．（2023·廣東珠?！ぶ楹Ｊ卸烽T區(qū)第一中學(xué)校考三模）曲線是造型中的精靈，以曲線為元素的LOGO給人簡約而不簡單的審美感受，某數(shù)學(xué)興趣小組設(shè)計(jì)了如圖所示的雙SKIPIF1<0型曲線LOGO，以下4個(gè)函數(shù)中最能擬合該曲線的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由函數(shù)SKIPIF1<0，其定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，所以排除B；由函數(shù)SKIPIF1<0，可得SKIPIF1<0，故排除C；由函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，故排除D．由函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，圖象關(guān)于原點(diǎn)對(duì)稱，由SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，所以A項(xiàng)符合題意.故選：A.4．（2023·新疆·校聯(lián)考二模）函數(shù)SKIPIF1<0，SKIPIF1<0的圖像大致為（

）A． B．C． D．【答案】B【詳解】對(duì)于A，因?yàn)镾KIPIF1<0關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，排除A；對(duì)于D，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，排除D；對(duì)于B，C，關(guān)鍵看SKIPIF1<0還是SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，所以排除C．故選：B5．（多選）（2023·福建泉州·統(tǒng)考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的圖象可以是（

）A． B．C． D．【答案】AD【詳解】因?yàn)镾KIPIF1<0與SKIPIF1<0均為偶函數(shù)，所以SKIPIF1<0為偶函數(shù)，函數(shù)圖象關(guān)于SKIPIF1<0軸對(duì)稱，故排除B；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)SKIPIF1<0，由于SKIPIF1<0為定義域上的偶函數(shù)，只需考慮SKIPIF1<0的情況即可，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，方程SKIPIF1<0的兩根為SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，故A正確；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由于SKIPIF1<0為定義域上的偶函數(shù)，只需考慮SKIPIF1<0的情況即可，即SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故D正確；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由于SKIPIF1<0為定義域上的偶函數(shù)，只需考慮SKIPIF1<0的情況即可，此時(shí)SKIPIF1<0，對(duì)于函數(shù)SKIPIF1<0，與SKIPIF1<0軸交于正半軸SKIPIF1<0，對(duì)稱軸為SKIPIF1<0，開口向上，無論是否與SKIPIF1<0軸有交點(diǎn)，函數(shù)在靠近SKIPIF1<0處函數(shù)值均大于SKIPIF1<0，即SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，故C錯(cuò)誤；故選：AD②函數(shù)切線條數(shù)問題1．（2023春·陜西漢中·高二校聯(lián)考期中）過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0切線有且只有兩條，則b的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0，所以過SKIPIF1<0的切線方程為SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0有且僅有兩根，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的圖象如下：故SKIPIF1<0有且僅有兩根，則b的取值范圍為SKIPIF1<0．故選：A．2．（2023·全國·高二專題練習(xí)）已知函數(shù)SKIPIF1<0，若過點(diǎn)SKIPIF1<0可以作出三條直線與曲線SKIPIF1<0相切，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)過點(diǎn)SKIPIF1<0的切線與SKIPIF1<0相切于點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則切線方程為：SKIPIF1<0，又切線過點(diǎn)SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則問題等價(jià)于SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn)，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，由此可得SKIPIF1<0圖象如下圖所示，由圖象可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn)，即當(dāng)SKIPIF1<0時(shí)，過點(diǎn)SKIPIF1<0可以作出三條直線與曲線SKIPIF1<0相切.故選：A.3．（2023春·陜西西安·高二陜西師大附中?？计谀┤羟€SKIPIF1<0有三條過點(diǎn)SKIPIF1<0的切線，則實(shí)數(shù)SKIPIF1<0的取值范圍為.【答案】SKIPIF1<0【詳解】設(shè)點(diǎn)SKIPIF1<0為曲線SKIPIF1<0上一點(diǎn)，則SKIPIF1<0又SKIPIF1<0，則SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，又切線過點(diǎn)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞增；SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞減，則SKIPIF1<0時(shí)SKIPIF1<0取得極小值SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0取得極大值SKIPIF1<0，又SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0時(shí)，SKIPIF1<0，又由題意得方程SKIPIF1<0有3個(gè)根，則SKIPIF1<0與SKIPIF1<0圖像有3個(gè)交點(diǎn)，則SKIPIF1<0.則曲線SKIPIF1<0有三條過點(diǎn)SKIPIF1<0的切線時(shí)實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.

故答案為：SKIPIF1<04．（2023·山東煙臺(tái)·統(tǒng)考三模）若曲線SKIPIF1<0與曲線SKIPIF1<0有兩條公切線，則SKIPIF1<0的值為．【答案】SKIPIF1<0【詳解】令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處切線為SKIPIF1<0，設(shè)SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處切線為SKIPIF1<0，由題意SKIPIF1<0，消去SKIPIF1<0得SKIPIF1<0，由題意，方程SKIPIF1<0有兩個(gè)不同的實(shí)數(shù)根，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極大值SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極小值SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，根據(jù)以上信息作出SKIPIF1<0的大致圖象，

由圖可知當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，從而方程SKIPIF1<0有兩個(gè)不同的實(shí)數(shù)根，所以，曲線SKIPIF1<0與曲線SKIPIF1<0有兩條公切線時(shí)，SKIPIF1<0的值為SKIPIF1<0．故答案為：SKIPIF1<0．5．（2023春·湖北襄陽·高二校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)若函數(shù)SKIPIF1<0在R上單調(diào)遞減，求實(shí)數(shù)a的取值范圍；(2)若過點(diǎn)SKIPIF1<0可作三條直線與曲線SKIPIF1<0相切，求實(shí)數(shù)a的取值范圍.【答案】(1)證明見解析(2)SKIPIF1<0【詳解】（1）因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上恒成立，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.（2）設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0所以切線方程為SKIPIF1<0將點(diǎn)SKIPIF1<0代入得SKIPIF1<0，整理得SKIPIF1<0，即關(guān)于SKIPIF1<0的方程SKIPIF1<0有三個(gè)不同根，等價(jià)于SKIPIF1<0的圖象與直線SKIPIF1<0有三個(gè)交點(diǎn).因?yàn)镾KIPIF1<0，所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.因?yàn)镾KIPIF1<0，SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.③不等式整數(shù)解問題1．（多選）（2023·山東泰安·?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若不等式SKIPIF1<0有且只有三個(gè)整數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值可以為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AB【詳解】因?yàn)镾KIPIF1<0定義域?yàn)镾KIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，即不等式SKIPIF1<0有且只有三個(gè)整數(shù)解，令SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，易知函數(shù)SKIPIF1<0SKIPIF1<0的圖象恒過點(diǎn)SKIPIF1<0，在同一平面直角坐標(biāo)系中作出SKIPIF1<0SKIPIF1<0與SKIPIF1<0的圖象如下圖所示：

由題意及圖象可知SKIPIF1<0，要使不等式SKIPIF1<0有且只有三個(gè)整數(shù)解，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故符合題意的有A、B.故選：AB2．（2023春·安徽滁州·高二校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0，若不等式SKIPIF1<0有且只有2個(gè)整數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0．不等式SKIPIF1<0有且只有2個(gè)整數(shù)解即不等式SKIPIF1<0有且只有2個(gè)整數(shù)解，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為增函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為減函數(shù)，又SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，設(shè)SKIPIF1<0，則直線SKIPIF1<0恒過點(diǎn)SKIPIF1<0，在同一直角坐標(biāo)系中，作出函數(shù)SKIPIF1<0與直線SKIPIF1<0的圖像，如圖所示．由圖像可知，SKIPIF1<0不滿足條件，則SKIPIF1<0，要使不等式SKIPIF1<0有且只有2個(gè)整數(shù)解，則這兩個(gè)整數(shù)解是2和3，則有SKIPIF1<0解得SKIPIF1<0，故答案為：SKIPIF1<0．3．（2023春·山東濟(jì)南·高二山東省濟(jì)南市萊蕪第一中學(xué)?？计谥校┮阎瘮?shù)SKIPIF1<0，關(guān)于SKIPIF1<0的不等式SKIPIF1<0有且只有四個(gè)整數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】由SKIPIF1<0SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0的遞增區(qū)間為SKIPIF1<0，遞減區(qū)間為SKIPIF1<0，故SKIPIF1<0的最大值為SKIPIF1<0，當(dāng)SKIPIF1<0趨于SKIPIF1<0時(shí)，SKIPIF1<0趨于SKIPIF1<0；當(dāng)SKIPIF1<0趨于SKIPIF1<0時(shí)，SKIPIF1<0趨于SKIPIF1<0，且SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0的圖象如圖，

①當(dāng)SKIPIF1<0時(shí)，由不等式SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有無數(shù)多個(gè)整數(shù)解；當(dāng)SKIPIF1<0時(shí)，其解集為SKIPIF1<0的子集，不含有整數(shù)解；所以SKIPIF1<0不合題意；②當(dāng)SKIPIF1<0時(shí)，由不等式SKIPIF1<0，當(dāng)?shù)肧KIPIF1<0，得SKIPIF1<0，則解集為SKIPIF1<0，整數(shù)解有無數(shù)多個(gè)，不合題意；③當(dāng)SKIPIF1<0時(shí)，由不等式SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0，無整數(shù)解；當(dāng)SKIPIF1<0時(shí)，因?yàn)椴坏仁絊KIPIF1<0有且僅有四個(gè)整數(shù)解，又SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，又因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以四個(gè)整數(shù)解只能為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.4．（2023春·安徽安慶·高二安慶市第二中學(xué)?？茧A段練習(xí)）已知關(guān)于x的不等式SKIPIF1<0的解集中恰有兩個(gè)正整數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值范圍為.【答案】SKIPIF1<0【詳解】由題意知，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恰有兩個(gè)正整數(shù)解．設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，作出SKIPIF1<0的大致圖象，如圖．

設(shè)SKIPIF1<0的圖象恒過定點(diǎn)SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，直線AM，BM在SKIPIF1<0圖象下方，由題意可知，SKIPIF1<0和SKIPIF1<0是不等式SKIPIF1<0的兩個(gè)正整數(shù)解，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．故答案為：SKIPIF1<0．5．（2023春·吉林長春·高二長春十一高?？计谀┮阎坏仁絊KIPIF1<0恰有1個(gè)整數(shù)解，則實(shí)數(shù)a的取值范圍為.【答案】SKIPIF1<0【詳解】原不等式SKIPIF1<0等價(jià)于SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極大值，又SKIPIF1<0，且SKIPIF1<0時(shí)，SKIPIF1<0，因此SKIPIF1<0的圖像如下，

直線SKIPIF1<0恒過點(diǎn)SKIPIF1<0.當(dāng)SKIPIF1<0有無數(shù)個(gè)整數(shù)解，不滿足條件；當(dāng)SKIPIF1<0時(shí)，只需要滿足SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.則實(shí)數(shù)a的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.④函數(shù)零點(diǎn)，方程根，兩個(gè)函數(shù)圖象交點(diǎn)問題1．（2023春·四川宜賓·高二校考期中）已知函數(shù)SKIPIF1<0恰有三個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為【答案】SKIPIF1<0【詳解】令SKIPIF1<0，SKIPIF1<0有三個(gè)零點(diǎn)即SKIPIF1<0與SKIPIF1<0的圖象有三個(gè)交點(diǎn)，SKIPIF1<0，當(dāng)SKIPIF1<0，或SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0的極大值為SKIPIF1<0，極小值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，結(jié)合圖象SKIPIF1<0與SKIPIF1<0有三個(gè)交點(diǎn)，即SKIPIF1<0.故答案為：SKIPIF1<0.

2．（2023春·云南大理·高二統(tǒng)考期末）若二次函數(shù)SKIPIF1<0的圖象與曲線SKIPIF1<0的圖象有3個(gè)公共點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】由題意SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，由題意，直線SKIPIF1<0與SKIPIF1<0的圖象有3個(gè)公共點(diǎn)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，所以，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極小值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極大值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，作出SKIPIF1<0的大致圖象，如圖，由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0的圖象有3個(gè)公共點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.3．（2023春·江西撫州·高二江西省樂安縣第二中學(xué)校考期末）已知函數(shù)SKIPIF1<0與SKIPIF1<0有兩個(gè)不同的交點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為.【答案】SKIPIF1<0【詳解】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0；SKIPIF1<0，SKIPIF1<0在SKIPIF1<0處的切線方程為：SKIPIF1<0；SKIPIF1<0恒過定點(diǎn)SKIPIF1<0，SKIPIF1<0若SKIPIF1<0與SKIPIF1<0有兩個(gè)不同交點(diǎn)，則SKIPIF1<0與SKIPIF1<0圖象如下圖所示，由圖象可知：當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0有兩個(gè)不同交點(diǎn)；即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.4．（2023春·安徽滁州·高二校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的單調(diào)區(qū)間和極值；(2)若SKIPIF1<0有兩個(gè)零點(diǎn)，求SKIPIF1<0的取值范圍．【答案】(1)單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0；極大值為SKIPIF1<0，無極小值；(2)SKIPIF1<0【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0．SKIPIF1<0極大值為SKIPIF1<0，無極小值．（2）SKIPIF1<0，定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0不是函數(shù)SKIPIF1<0的零點(diǎn)，要使SKIPIF1<0有兩個(gè)零點(diǎn)，則方程SKIPIF1<0有兩個(gè)不同的實(shí)數(shù)根．令SKIPIF1<0且SKIPIF1<0，則直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像有兩個(gè)不同的交點(diǎn)．SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，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在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0的取值接近0時(shí)，SKIPIF1<0的值接近SKIPIF1<0，SKIPIF1<0的極大值為SKIPIF1<0，又知SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．又知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0的極小值為SKIPIF1<0，當(dāng)SKIPIF1<0的取值接近SKIPIF1<0時(shí)，SKIPIF1<0的值趨向SKIPIF1<0，當(dāng)SKIPIF1<0的取值趨向SKIPIF1<0時(shí)，SKIPIF1<0的值趨向SKIPIF1<0，所以函數(shù)SKIPIF1<0的大致圖像如圖所示．由圖像可知，要滿足題設(shè)條件，則SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0的取值范圍為SKIPIF1<0．5．（2023春·浙江衢州·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0(1)若過點(diǎn)SKIPIF1<0作函數(shù)SKIPIF1<0的切線有且僅有兩條，求SKIPIF1<0的值；(2)若對(duì)于任意SKIPIF1<0，直線SKIPIF1<0與曲線SKIPIF1<0都有唯一交點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）設(shè)過點(diǎn)SKIPIF1<0作函數(shù)SKIPIF1<0切線的切點(diǎn)為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以切線方程為SKIPIF1<0，即SKIPIF1<0，又因?yàn)榍芯€過點(diǎn)SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0遞減；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0遞增；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0遞減.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極小值SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取極小值SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0；SKIPIF1<0時(shí)SKIPIF1<0，根據(jù)以上信息作出SKIPIF1<0的大致圖象，

由題意，直線SKIPIF1<0與SKIPIF1<0的圖象有且僅有兩個(gè)交點(diǎn)，所以SKIPIF1<0.（2）由題可得SKIPIF1<0有唯一解，即SKIPIF1<0有唯一解．令SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0與題設(shè)SKIPIF1<0，矛盾，故SKIPIF1<0.又因?yàn)镾KIPIF1<0，SKIPIF1<0；SKIPIF1<0，SKIPIF1<0，結(jié)合題意可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0，所以SKIPIF1<0，結(jié)合（1）可得SKIPIF1<0，所以SKIPIF1<0.6．（2023春·陜西咸陽·高二校考期中）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的極小值；(2)討論關(guān)于SKIPIF1<0的方程SKIPIF1<0的解的個(gè)數(shù).【答案】(1)極小值為SKIPIF1<0(2)答案見解析【詳解】（1）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0.令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0隨SKIPIF1<0的變化的情況如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0極小值SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上的極小值是SKIPIF1<0.（2）當(dāng)SKIPIF1<0，SKIPIF1<0單調(diào)遞減且SKIPIF1<0的取值范圍是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增且SKIPIF1<0的取值范圍是SKIPIF1<0.令SKIPIF1<0，SKIPIF1<0，兩函數(shù)圖象交點(diǎn)的橫坐標(biāo)是SKIPIF1<0的解，由（1）知當(dāng)SKIPIF1<0時(shí)，原方程無解.由SKIPIF1<0的單調(diào)區(qū)間上函數(shù)值的范圍知，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，原方程有唯一解；當(dāng)SKIPIF1<0時(shí)，原方程有兩解.7．（2023春·貴州·高二校聯(lián)考階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處取得極值．(1)求實(shí)數(shù)a的值；(2)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(3)令函數(shù)SKIPIF1<0，是否存在實(shí)數(shù)k使得SKIPIF1<0沒有零點(diǎn)？若存在，請(qǐng)求出實(shí)數(shù)k的范圍；若不存在，請(qǐng)說明理由．【答案】(1)SKIPIF1<0(2)函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0；(3)SKIPIF1<0【詳解】（1）SKIPIF1<0，因?yàn)榍€SKIPIF1<0在點(diǎn)SKIPIF1<0處取得極值，所以SKIPIF1<0，解得SKIPIF1<0，經(jīng)檢驗(yàn)符合題意；（2）由（1）SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，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)遞減，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0；（3）存在，理由如下，由（2）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0；所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0的大致圖象如下，

若函數(shù)SKIPIF1<0沒有零點(diǎn)，則函數(shù)SKIPIF1<0與SKIPIF1<0的圖象沒有交點(diǎn)，所以SKIPIF1<0.⑤不等式恒成立問題1．（2023·全國·高二專題練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的最大值為．【答案】2e【詳解】由題意，可得SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，若SKIPIF1<0與直線SKIPIF1<0相切，設(shè)切點(diǎn)為SKIPIF1<0，則切線斜率SKIPIF1<0，所以該切線方程為SKIPIF1<0，注意到切線過點(diǎn)SKIPIF1<0，則SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，結(jié)合圖象，可得實(shí)數(shù)a的取值范圍為SKIPIF1<0，即實(shí)數(shù)a的最大值為2e．2．（2023·全國·高三專題練習(xí)）對(duì)任意的SKIPIF1<0，若關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，則SKIPIF1<0的最小值為.【答案】SKIPIF1<0/0.5【詳解】因?yàn)殛P(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，所以SKIPIF1<0的圖象在函數(shù)SKIPIF1<0的圖象上方相切.當(dāng)m>0時(shí),SKIPIF1<0的圖象與x軸的交點(diǎn)在x軸的負(fù)半軸上.由圖可知當(dāng)正數(shù)m最小時(shí),直線SKIPIF1<0與SKIPIF1<0在SKIPIF1<0內(nèi)相切.對(duì)函數(shù)SKIPIF1<0求導(dǎo)得到SKIPIF1<0.令SKIPIF1<0,解得x=0.所以SKIPIF1<0,所以切點(diǎn)的坐標(biāo)為SKIPIF1<0，把點(diǎn)SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0.故答案為：SKIPIF1<0.3．（2023秋·云南·高三云南民族大學(xué)附屬中學(xué)校考期末）已知不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】依題意，SKIPIF1<0變形為SKIPIF1<0，對(duì)任意SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，設(shè)SKIPIF1<0，則直線SKIPIF1<0應(yīng)介于函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0之間，由SKIPIF1<0易知，函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，SKIPIF1<0在SKIPIF1<0處取得極小值，也是最小值SKIPIF1<0，由SKIPIF1<0及雙勾函數(shù)的性質(zhì)可知，函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，SKIPIF1<0在SKIPIF1<0處取得最小值SKIPIF1<0，作出函數(shù)SKIPIF1<0及函數(shù)SKIPIF1<0的圖象如下，由圖象可知，滿足條件的實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0，SKIPIF1<0．4．（2023春·江西上饒·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0．(1)若關(guān)于SKIPIF1<0的方程SKIPIF1<0在區(qū)間SKIPIF1<0上恰有2個(gè)不同的實(shí)數(shù)解，求SKIPIF1<0的取值范圍；(2)設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，對(duì)SKIPIF1<0總有SKIPIF1<0成立，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.【詳解】（1）函數(shù)SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，依題意，曲線SKIPIF1<0與直線SKIPIF1<0在區(qū)間SKIPIF1<0上恰有2個(gè)交點(diǎn)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí),SKIPIF1<0，當(dāng)SKIPIF1<0時(shí),