新高考數(shù)學(xué)一輪復(fù)習(xí)提升訓(xùn)練3.6 零點(diǎn)定理（精講）（解析版）

3.6零點(diǎn)定理（精講）（提升版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一零點(diǎn)的區(qū)間【例1】（2022·河南開封·）函數(shù)SKIPIF1<0的一個(gè)零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，SKIPIF1<0，由零點(diǎn)存在性定理知：SKIPIF1<0，函數(shù)SKIPIF1<0的一個(gè)零點(diǎn)所在的區(qū)間是SKIPIF1<0.故選：D.【一隅三反】1．（2022·湖南）函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0是SKIPIF1<0上的增函數(shù)，且SKIPIF1<0，所以SKIPIF1<0的零點(diǎn)在區(qū)間SKIPIF1<0內(nèi).故選：B2．（2022·四川攀枝花）已知函數(shù)SKIPIF1<0的零點(diǎn)在區(qū)間SKIPIF1<0上，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且在SKIPIF1<0上單調(diào)遞增，故其至多一個(gè)零點(diǎn)；又SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的零點(diǎn)在區(qū)間SKIPIF1<0，故SKIPIF1<0.故選：SKIPIF1<0．3．（2022·云南德宏）方程SKIPIF1<0的解所在的區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)SKIPIF1<0，易知SKIPIF1<0在定義域SKIPIF1<0內(nèi)是增函數(shù)，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的零點(diǎn)在SKIPIF1<0上，即題中方程的根屬于SKIPIF1<0．故選：B．考點(diǎn)二零點(diǎn)的個(gè)數(shù)【例2-1】（2022·陜西）函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）A．0 B．1 C．2 D．3【答案】D【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0，SKIPIF1<0的交點(diǎn)個(gè)數(shù)作出兩個(gè)函數(shù)的圖象如下圖所示，由圖可知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的零點(diǎn)有兩個(gè)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的零點(diǎn)有一個(gè).綜上，函數(shù)SKIPIF1<0的零點(diǎn)有三個(gè).故選：D【例2-2】（2022·山西）已知SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0內(nèi)的零點(diǎn)個(gè)數(shù)為（

）A．8 B．9 C．10 D．11【答案】B【解析】作出SKIPIF1<0的圖像，則SKIPIF1<0在SKIPIF1<0內(nèi)的零點(diǎn)個(gè)數(shù)為曲線SKIPIF1<0與直線SKIPIF1<0在SKIPIF1<0內(nèi)的交點(diǎn)個(gè)數(shù)9.選：B.【一隅三反】1．（2022·安徽）已知函數(shù)SKIPIF1<0則方程SKIPIF1<0的解的個(gè)數(shù)是（

）A．0 B．1 C．2 D．3【答案】C【解析】令SKIPIF1<0，得SKIPIF1<0，則函數(shù)SKIPIF1<0零點(diǎn)的個(gè)數(shù)即函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的交點(diǎn)個(gè)數(shù)．作出函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖像，可知兩個(gè)函數(shù)圖像的交點(diǎn)的個(gè)數(shù)為2，故方程SKIPIF1<0的解的個(gè)數(shù)為2個(gè)．故選：C．2．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的圖像與函數(shù)SKIPIF1<0的圖像的交點(diǎn)個(gè)數(shù)為（

）A．2 B．3 C．4 D．0【答案】C【解析】SKIPIF1<0在SKIPIF1<0上是增函數(shù)，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上是減函數(shù)，在SKIPIF1<0和SKIPIF1<0上是增函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，作出函數(shù)SKIPIF1<0SKIPIF1<0的圖像，如圖，由圖像可知它們有4個(gè)交點(diǎn)．故選：C．3．（2022·海南?。┰O(shè)函數(shù)SKIPIF1<0定義域?yàn)镽，SKIPIF1<0為奇函數(shù)，SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0有（

）個(gè)零點(diǎn)A．4 B．5 C．6 D．7【答案】C【解析】SKIPIF1<0的零點(diǎn)個(gè)數(shù)即SKIPIF1<0的圖象交點(diǎn)個(gè)數(shù).因?yàn)镾KIPIF1<0為奇函數(shù)，故SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，故SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，又SKIPIF1<0為偶函數(shù)，故SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，畫出圖象，易得函數(shù)SKIPIF1<0的圖象有6個(gè)交點(diǎn)故選：C考點(diǎn)三比較零點(diǎn)的大小【例3】（2022·安徽）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的零點(diǎn)分別為a，b，c則a，b，c的大小順序?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0.在同一平面直角坐標(biāo)系中畫出SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的圖象，由圖象知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：D【一隅三反】1．（2022·河南）若實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】畫出SKIPIF1<0與SKIPIF1<0三個(gè)函數(shù)的圖象，如圖可得SKIPIF1<0的與SKIPIF1<0交點(diǎn)的橫坐標(biāo)依次為SKIPIF1<0，故SKIPIF1<0故選：B2．（2022·安徽）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)函數(shù)SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，由零點(diǎn)存在定理可知．SKIPIF1<0；設(shè)函數(shù)SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，由零點(diǎn)存在定理可知，SKIPIF1<0；設(shè)函數(shù)SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，由函數(shù)單調(diào)性可知，SKIPIF1<0，即SKIPIF1<0.故選:A．3．（2022·山西）正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0之間的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn)，則SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn)，則SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0；SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn)，則SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0故選：A.考點(diǎn)四已知零點(diǎn)求參數(shù)【例4-1】（2022·山東濰坊）已知函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0有3個(gè)不同的交點(diǎn)，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】對(duì)函數(shù)SKIPIF1<0求導(dǎo)得：SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0處取得極大值SKIPIF1<0，在SKIPIF1<0處取得極小值SKIPIF1<0，在同一坐標(biāo)系內(nèi)作出函數(shù)SKIPIF1<0的圖像和直線SKIPIF1<0，如圖，觀察圖象知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0有3個(gè)不同的交點(diǎn)，所以實(shí)數(shù)m的取值范圍是SKIPIF1<0.故選：B【例4-2】（2022·吉林）已知SKIPIF1<0若關(guān)于x的方程SKIPIF1<0有3個(gè)不同實(shí)根，則實(shí)數(shù)SKIPIF1<0取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞減；且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞減；且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0；作出SKIPIF1<0在SKIPIF1<0上的圖象，如圖：由圖可知要使SKIPIF1<0有3個(gè)不同的實(shí)根，則SKIPIF1<0.故選：D.【例4-3】（2022·安徽·合肥市）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有4個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】根據(jù)題意，函數(shù)SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0，必有SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0和SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有4個(gè)交點(diǎn)，又由于SKIPIF1<0，必有SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0，故選：B.【一隅三反】1．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0若關(guān)于x的方程SKIPIF1<0恰有三個(gè)不相等的實(shí)數(shù)解，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】函數(shù)SKIPIF1<0的圖像如下圖所示：若關(guān)于x的方程SKIPIF1<0恰有三個(gè)不相等的實(shí)數(shù)解，則函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0有三個(gè)交點(diǎn)，若直線SKIPIF1<0經(jīng)過原點(diǎn)時(shí)，m＝0，若直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像相切，令SKIPIF1<0，令SKIPIF1<0.故SKIPIF1<0.故選：D．2．（2022·河南·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0為定義在SKIPIF1<0上的單調(diào)函數(shù)，且SKIPIF1<0．若函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0為定義在R上的單調(diào)函數(shù)，所以存在唯一的SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為增函數(shù)，且SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0.結(jié)合函數(shù)的圖象可知，若SKIPIF1<0有3個(gè)零點(diǎn)，則SKIPIF1<0．故選：A3．（2022·廣西·貴港市高級(jí)中學(xué)三模）已知SKIPIF1<0在SKIPIF1<0有且僅有6個(gè)實(shí)數(shù)根，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0.設(shè)SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0有且僅有6個(gè)實(shí)數(shù)根，因?yàn)镾KIPIF1<0，故只需SKIPIF1<0，解得SKIPIF1<0，故選：D．4．（2022·山西）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰好有兩個(gè)零點(diǎn)，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意知，畫出函數(shù)SKIPIF1<0的簡(jiǎn)圖，如圖所示由SKIPIF1<0恰好有兩個(gè)零點(diǎn)轉(zhuǎn)化為SKIPIF1<0與直線SKIPIF1<0有兩個(gè)不同的交點(diǎn)，由圖知，當(dāng)直線經(jīng)過點(diǎn)SKIPIF1<0兩點(diǎn)的斜率為SKIPIF1<0，則SKIPIF1<0.所以實(shí)數(shù)k的取值范圍為SKIPIF1<0.故選：C.考點(diǎn)五零點(diǎn)的綜合運(yùn)用【例5-1】（2022·新疆克拉瑪依）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的所有零點(diǎn)之和為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因?yàn)镾KIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)顯然不成立，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，作出SKIPIF1<0和SKIPIF1<0的圖象，如圖，它們關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，由圖象可知它們?cè)赟KIPIF1<0上有4個(gè)交點(diǎn)，且關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，每對(duì)稱的兩個(gè)點(diǎn)的橫坐標(biāo)和為SKIPIF1<0，所以4個(gè)點(diǎn)的橫坐標(biāo)之和為SKIPIF1<0．故選：C．【例5-2】（2022·甘肅）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有2個(gè)零點(diǎn)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】A【解析】函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有2個(gè)零點(diǎn)SKIPIF1<0即方程SKIPIF1<0在區(qū)間SKIPIF1<0上有2個(gè)實(shí)數(shù)根SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0為偶函數(shù).且SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的大致圖像如圖.所以方程SKIPIF1<0在區(qū)間SKIPIF1<0上有2個(gè)實(shí)數(shù)根SKIPIF1<0滿足SKIPIF1<0則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恒成立所以SKIPIF1<0故選：A【例5-3】（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，則下列不等式中成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】令SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0、SKIPIF1<0，在同一坐標(biāo)系中分別繪出函數(shù)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的圖像，因?yàn)楹瘮?shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，解方程組SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0與SKIPIF1<0互為反函數(shù)，所以由反函數(shù)性質(zhì)知SKIPIF1<0、SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，A、B、D錯(cuò)誤，因?yàn)镾KIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0在直線SKIPIF1<0上，所以SKIPIF1<0，SKIPIF1<0，故C正確，故選：C．【一隅三反】1．（2022·安徽·合肥一六八中學(xué)）若SKIPIF1<0為奇函數(shù)，且SKIPIF1<0是SKIPIF1<0的一個(gè)零點(diǎn)，則SKIPIF1<0一定是下列哪個(gè)函數(shù)的零點(diǎn)（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0是奇函數(shù)，SKIPIF1<0且SKIPIF1<0是SKIPIF1<0的一個(gè)零點(diǎn)，所以SKIPIF1<0，把SKIPIF1<0分別代入下面四個(gè)選項(xiàng)，對(duì)于A，SKIPIF1<0，不一定為0，故A錯(cuò)誤；對(duì)于B，SKIPIF1<0SKIPIF1<0，所以SK