新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)專題06 導(dǎo)數(shù)與函數(shù)的零點(diǎn)問(wèn)題（講）（解析版）

第一篇熱點(diǎn)、難點(diǎn)突破篇專題06導(dǎo)數(shù)與函數(shù)的零點(diǎn)問(wèn)題（講）真題體驗(yàn)感悟高考1．（2021·北京·高考真題）已知函數(shù)SKIPIF1<0，給出下列四個(gè)結(jié)論：①若SKIPIF1<0，SKIPIF1<0恰有2個(gè)零點(diǎn)；②存在負(fù)數(shù)SKIPIF1<0，使得SKIPIF1<0恰有1個(gè)零點(diǎn)；③存在負(fù)數(shù)SKIPIF1<0，使得SKIPIF1<0恰有3個(gè)零點(diǎn)；④存在正數(shù)SKIPIF1<0，使得SKIPIF1<0恰有3個(gè)零點(diǎn)．其中所有正確結(jié)論的序號(hào)是_______．【答案】①②④【分析】由SKIPIF1<0可得出SKIPIF1<0，考查直線SKIPIF1<0與曲線SKIPIF1<0的左、右支分別相切的情形，利用方程思想以及數(shù)形結(jié)合可判斷各選項(xiàng)的正誤.【詳解】對(duì)于①，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，①正確；對(duì)于②，考查直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，所以，存在SKIPIF1<0，使得SKIPIF1<0只有一個(gè)零點(diǎn)，②正確；對(duì)于③，當(dāng)直線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)交點(diǎn)，若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，則直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)交點(diǎn)，直線SKIPIF1<0與曲線SKIPIF1<0有一個(gè)交點(diǎn)，所以，SKIPIF1<0，此不等式無(wú)解，因此，不存在SKIPIF1<0，使得函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，③錯(cuò)誤；對(duì)于④，考查直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，④正確.故答案為：①②④.【點(diǎn)睛】思路點(diǎn)睛：已知函數(shù)的零點(diǎn)或方程的根的情況，求解參數(shù)的取值范圍問(wèn)題的本質(zhì)都是研究函數(shù)的零點(diǎn)問(wèn)題，求解此類問(wèn)題的一般步驟：（1）轉(zhuǎn)化，即通過(guò)構(gòu)造函數(shù)，把問(wèn)題轉(zhuǎn)化成所構(gòu)造函數(shù)的零點(diǎn)問(wèn)題；（2）列式，即根據(jù)函數(shù)的零點(diǎn)存在定理或結(jié)合函數(shù)的圖象列出關(guān)系式；（3）得解，即由列出的式子求出參數(shù)的取值范圍．2．（2019·全國(guó)·高考真題（文））已知函數(shù)SKIPIF1<0.證明：（1）SKIPIF1<0存在唯一的極值點(diǎn)；（2）SKIPIF1<0有且僅有兩個(gè)實(shí)根，且兩個(gè)實(shí)根互為倒數(shù).【答案】（1）見(jiàn)詳解；（2）見(jiàn)詳解【分析】（1）先對(duì)函數(shù)SKIPIF1<0求導(dǎo)，根據(jù)導(dǎo)函數(shù)的單調(diào)性，得到存在唯一SKIPIF1<0，使得SKIPIF1<0，進(jìn)而可得判斷函數(shù)SKIPIF1<0的單調(diào)性，即可確定其極值點(diǎn)個(gè)數(shù)，證明出結(jié)論成立；（2）先由（1）的結(jié)果，得到SKIPIF1<0，SKIPIF1<0，得到SKIPIF1<0在SKIPIF1<0內(nèi)存在唯一實(shí)根，記作SKIPIF1<0，再求出SKIPIF1<0，即可結(jié)合題意，說(shuō)明結(jié)論成立.【詳解】（1）由題意可得，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，顯然SKIPIF1<0單調(diào)遞增；又SKIPIF1<0，SKIPIF1<0，故存在唯一SKIPIF1<0，使得SKIPIF1<0；又當(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；因此，SKIPIF1<0存在唯一的極值點(diǎn)；（2）由（1）知，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0內(nèi)存在唯一實(shí)根，記作SKIPIF1<0.由SKIPIF1<0得SKIPIF1<0，又SKIPIF1<0，故SKIPIF1<0是方程SKIPIF1<0在SKIPIF1<0內(nèi)的唯一實(shí)根；綜上，SKIPIF1<0有且僅有兩個(gè)實(shí)根，且兩個(gè)實(shí)根互為倒數(shù).【點(diǎn)睛】本題主要考查導(dǎo)數(shù)的應(yīng)用，通常需要對(duì)函數(shù)求導(dǎo)，用導(dǎo)數(shù)的方法研究函數(shù)的單調(diào)性、極值、以及函數(shù)零點(diǎn)的問(wèn)題，屬于?？碱}型.3．（2021·浙江·高考真題）設(shè)a，b為實(shí)數(shù)，且SKIPIF1<0，函數(shù)SKIPIF1<0（1）求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；（2）若對(duì)任意SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，求a的取值范圍；（3）當(dāng)SKIPIF1<0時(shí)，證明：對(duì)任意SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)SKIPIF1<0，滿足SKIPIF1<0.(注：SKIPIF1<0是自然對(duì)數(shù)的底數(shù))【答案】(1)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；SKIPIF1<0時(shí)，函數(shù)的單調(diào)減區(qū)間為SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0；(2)SKIPIF1<0；(3)證明見(jiàn)解析.【分析】(1)首先求得導(dǎo)函數(shù)的解析式，然后分類討論即可確定函數(shù)的單調(diào)性；(2)將原問(wèn)題進(jìn)行等價(jià)轉(zhuǎn)化，然后構(gòu)造新函數(shù)，利用導(dǎo)函數(shù)研究函數(shù)的性質(zhì)并進(jìn)行放縮即可確定實(shí)數(shù)a的取值范圍；(3)方法一：結(jié)合(2)的結(jié)論將原問(wèn)題進(jìn)行等價(jià)變形，然后利用分析法即可證得題中的結(jié)論成立.【詳解】(1)SKIPIF1<0，①若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②若SKIPIF1<0，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.綜上可得，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；SKIPIF1<0時(shí)，函數(shù)的單調(diào)減區(qū)間為SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0.(2)SKIPIF1<0有2個(gè)不同零點(diǎn)SKIPIF1<0有2個(gè)不同解SKIPIF1<0有2個(gè)不同的解，令SKIPIF1<0，則SKIPIF1<0，記SKIPIF1<0，記SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，SKIPIF1<0，SKIPIF1<0.即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.(3)[方法一]【最優(yōu)解】：SKIPIF1<0有2個(gè)不同零點(diǎn)，則SKIPIF1<0，故函數(shù)的零點(diǎn)一定為正數(shù).由(2)可知有2個(gè)不同零點(diǎn)，記較大者為SKIPIF1<0，較小者為SKIPIF1<0，SKIPIF1<0，注意到函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，在區(qū)間SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，又由SKIPIF1<0知SKIPIF1<0，SKIPIF1<0，要證SKIPIF1<0，只需SKIPIF1<0，SKIPIF1<0且關(guān)于SKIPIF1<0的函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以只需證SKIPIF1<0，只需證SKIPIF1<0，只需證SKIPIF1<0，SKIPIF1<0，只需證SKIPIF1<0在SKIPIF1<0時(shí)為正，由于SKIPIF1<0，故函數(shù)SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0時(shí)為正，從而題中的不等式得證.[方法二]：分析+放縮法SKIPIF1<0有2個(gè)不同零點(diǎn)SKIPIF1<0，不妨設(shè)SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0（其中SKIPIF1<0）．且SKIPIF1<0．要證SKIPIF1<0，只需證SKIPIF1<0，即證SKIPIF1<0，只需證SKIPIF1<0．又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．所以只需證SKIPIF1<0．而SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以只需證SKIPIF1<0．所以SKIPIF1<0，原命題得證．[方法三]：若SKIPIF1<0且SKIPIF1<0，則滿足SKIPIF1<0且SKIPIF1<0，由（Ⅱ）知SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0且SKIPIF1<0．又SKIPIF1<0，故進(jìn)一步有SKIPIF1<0．由SKIPIF1<0可得SKIPIF1<0且SKIPIF1<0，從而SKIPIF1<0．．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故只需證SKIPIF1<0．又因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，故只需證SKIPIF1<0，即SKIPIF1<0，注意SKIPIF1<0時(shí)有SKIPIF1<0，故不等式成立．【整體點(diǎn)評(píng)】本題第二、三問(wèn)均涉及利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)問(wèn)題，其中第三問(wèn)難度更大，涉及到三種不同的處理方法，方法一：直接分析零點(diǎn)SKIPIF1<0，將要證明的不等式消元，代換為關(guān)于SKIPIF1<0的函數(shù)，再利用零點(diǎn)反代法，換為關(guān)于SKIPIF1<0的不等式，移項(xiàng)作差構(gòu)造函數(shù)，利用導(dǎo)數(shù)分析范圍.方法二：通過(guò)分析放縮，找到使得結(jié)論成立的充分條件，方法比較冒險(xiǎn)！方法三：利用兩次零點(diǎn)反代法，將不等式化簡(jiǎn)，再利用函數(shù)的單調(diào)性，轉(zhuǎn)化為SKIPIF1<0與0比較大小，代入函數(shù)放縮得到結(jié)論.總結(jié)規(guī)律預(yù)測(cè)考向（一）規(guī)律與預(yù)測(cè)1.高考對(duì)導(dǎo)數(shù)的考查要求一般有三個(gè)層次：第一層次主要考查求導(dǎo)公式，求導(dǎo)法則與導(dǎo)數(shù)的幾何意義；第二層次是導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用，包括求函數(shù)的單調(diào)區(qū)間、極值、最值等；第三層次是綜合考查，如研究函數(shù)零點(diǎn)、證明不等式、恒成立問(wèn)題、求參數(shù)等，包括解決應(yīng)用問(wèn)題，將導(dǎo)數(shù)內(nèi)容和傳統(tǒng)內(nèi)容中有關(guān)不等式、數(shù)列及函數(shù)單調(diào)性有機(jī)結(jié)合，設(shè)計(jì)綜合題.2.涉及導(dǎo)數(shù)與零點(diǎn)問(wèn)題，主要有：函數(shù)零點(diǎn)個(gè)數(shù)的判斷與證明、根據(jù)函數(shù)的零點(diǎn)個(gè)數(shù)或零點(diǎn)情況求參數(shù)的取值范圍、與零點(diǎn)相關(guān)的不等式恒成立或證明問(wèn)題等（二）本專題考向展示考點(diǎn)突破典例分析考向一函數(shù)零點(diǎn)個(gè)數(shù)的判斷與證明【核心知識(shí)】解函數(shù)零點(diǎn)問(wèn)題的一般思路(1)對(duì)函數(shù)求導(dǎo)．(2)分析函數(shù)的單調(diào)性，極值情況．(3)結(jié)合函數(shù)性質(zhì)畫函數(shù)的草圖．(4)依據(jù)函數(shù)草圖確定函數(shù)零點(diǎn)情況．【典例分析】典例1.（2022·河南·駐馬店市第二高級(jí)中學(xué)高三階段練習(xí)（文））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【分析】根據(jù)給定條件，構(gòu)造函數(shù)SKIPIF1<0，探討函數(shù)單調(diào)性，借助零點(diǎn)存在性定理判斷作答.【詳解】令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，則存在SKIPIF1<0，使得SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求導(dǎo)得SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上遞增，而SKIPIF1<0，則存在SKIPIF1<0，使得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0遞增，SKIPIF1<0，而SKIPIF1<0，則存在SKIPIF1<0，使得SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，又函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)，因此函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，所以函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為2，即方程SKIPIF1<0的解的個(gè)數(shù)是2.故選：C典例2.（2022·吉林長(zhǎng)春·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的值域；(2)討論函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)．【答案】(1)SKIPIF1<0(2)答案見(jiàn)解析【分析】（1）利用導(dǎo)數(shù)求得SKIPIF1<0的單調(diào)區(qū)間，進(jìn)而求得SKIPIF1<0的值域.（2）利用多次求導(dǎo)的方法，結(jié)合對(duì)SKIPIF1<0進(jìn)行分類討論，由此求得SKIPIF1<0零點(diǎn)的個(gè)數(shù).【詳解】（1）由SKIPIF1<0可知SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0，xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0減極小值增SKIPIF1<0，SKIPIF1<0無(wú)最大值．即SKIPIF1<0的值域?yàn)镾KIPIF1<0．（2）SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0單調(diào)遞增，即此時(shí)SKIPIF1<0有唯一零點(diǎn)．當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．即SKIPIF1<0，①當(dāng)k＝1時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0有唯一零點(diǎn)．②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0存在一個(gè)零點(diǎn)，此時(shí)SKIPIF1<0共有2個(gè)零點(diǎn)．③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0存在一個(gè)零點(diǎn)，此時(shí)SKIPIF1<0共有2個(gè)零點(diǎn)．綜上，當(dāng)SKIPIF1<0或k＝1時(shí)，SKIPIF1<0有唯一零點(diǎn)．當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0有2個(gè)零點(diǎn)．典例3.（2019·全國(guó)·高考真題（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的導(dǎo)數(shù)．證明：（1）SKIPIF1<0在區(qū)間SKIPIF1<0存在唯一極大值點(diǎn)；（2）SKIPIF1<0有且僅有2個(gè)零點(diǎn)．【答案】（1）見(jiàn)解析；（2）見(jiàn)解析【分析】（1）求得導(dǎo)函數(shù)后，可判斷出導(dǎo)函數(shù)在SKIPIF1<0上單調(diào)遞減，根據(jù)零點(diǎn)存在定理可判斷出SKIPIF1<0，使得SKIPIF1<0，進(jìn)而得到導(dǎo)函數(shù)在SKIPIF1<0上的單調(diào)性，從而可證得結(jié)論；（2）由（1）的結(jié)論可知SKIPIF1<0為SKIPIF1<0在SKIPIF1<0上的唯一零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，首先可判斷出在SKIPIF1<0上無(wú)零點(diǎn)，再利用零點(diǎn)存在定理得到SKIPIF1<0在SKIPIF1<0上的單調(diào)性，可知SKIPIF1<0，不存在零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，利用零點(diǎn)存在定理和SKIPIF1<0單調(diào)性可判斷出存在唯一一個(gè)零點(diǎn)；當(dāng)SKIPIF1<0，可證得SKIPIF1<0；綜合上述情況可證得結(jié)論.【詳解】（1）由題意知：SKIPIF1<0定義域?yàn)椋篠KIPIF1<0且SKIPIF1<0令SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減SKIPIF1<0在SKIPIF1<0上單調(diào)遞減又SKIPIF1<0，SKIPIF1<0SKIPIF1<0，使得SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0上單調(diào)遞減則SKIPIF1<0為SKIPIF1<0唯一的極大值點(diǎn)即：SKIPIF1<0在區(qū)間SKIPIF1<0上存在唯一的極大值點(diǎn)SKIPIF1<0.（2）由（1）知：SKIPIF1<0，SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，由（1）可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增SKIPIF1<0

SKIPIF1<0在SKIPIF1<0上單調(diào)遞減又SKIPIF1<0SKIPIF1<0為SKIPIF1<0在SKIPIF1<0上的唯一零點(diǎn)②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減又SKIPIF1<0

SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，此時(shí)SKIPIF1<0，不存在零點(diǎn)又SKIPIF1<0SKIPIF1<0，使得SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減又SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0上恒成立，此時(shí)不存在零點(diǎn)③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞減SKIPIF1<0在SKIPIF1<0上單調(diào)遞減又SKIPIF1<0，SKIPIF1<0即SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞減SKIPIF1<0SKIPIF1<0在SKIPIF1<0上存在唯一零點(diǎn)④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0SKIPIF1<0即SKIPIF1<0在SKIPIF1<0上不存在零點(diǎn)綜上所述：SKIPIF1<0有且僅有SKIPIF1<0個(gè)零點(diǎn)【規(guī)律方法】1.利用導(dǎo)數(shù)判斷或證明函數(shù)零點(diǎn)個(gè)數(shù)的策略:借助導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值后，通過(guò)極值的正負(fù)以及函數(shù)的單調(diào)性判斷函數(shù)圖象的走勢(shì)，從而判斷零點(diǎn)個(gè)數(shù).2.常用方法：（1）直接法：直接研究函數(shù)，求出極值以及最值，畫出草圖．函數(shù)零點(diǎn)的個(gè)數(shù)問(wèn)題即是函數(shù)圖象與x軸交點(diǎn)的個(gè)數(shù)問(wèn)題．（2）構(gòu)造新函數(shù)法：將問(wèn)題轉(zhuǎn)化為研究?jī)珊瘮?shù)圖象的交點(diǎn)問(wèn)題；（3）分離參數(shù)法：分離出參數(shù)，轉(zhuǎn)化為a＝g(x)，根據(jù)導(dǎo)數(shù)的知識(shí)求出函數(shù)g(x)在某區(qū)間的單調(diào)性，求出極值以及最值，畫出草圖．函數(shù)零點(diǎn)的個(gè)數(shù)問(wèn)題即是直線y＝a與函數(shù)y＝g(x)圖象交點(diǎn)的個(gè)數(shù)問(wèn)題．只需要用a與函數(shù)g(x)的極值和最值進(jìn)行比較即可．考向二根據(jù)函數(shù)零點(diǎn)的情況求參數(shù)取值范圍【核心知識(shí)】利用函數(shù)零點(diǎn)的情況求參數(shù)范圍的方法(1)分離參數(shù)(a＝g(x))后，將原問(wèn)題轉(zhuǎn)化為y＝g(x)的值域(最值)問(wèn)題或轉(zhuǎn)化為直線y＝a與y＝g(x)的圖象的交點(diǎn)個(gè)數(shù)問(wèn)題(優(yōu)選分離、次選分類)求解；(2)利用零點(diǎn)的存在性定理構(gòu)建不等式求解；(3)轉(zhuǎn)化為兩個(gè)熟悉的函數(shù)圖象的位置關(guān)系問(wèn)題，從而構(gòu)建不等式求解.【典例分析】典例4.（2022·青海玉樹(shù)·高二期末（理））已知SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間與極值；(2)若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，求實(shí)數(shù)SKIPIF1<0的取值范圍．參考數(shù)據(jù)：SKIPIF1<0【答案】(1)答案見(jiàn)解析；(2)SKIPIF1<0【分析】（1）當(dāng)SKIPIF1<0時(shí)，利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0的單調(diào)性，可得出函數(shù)SKIPIF1<0的單調(diào)區(qū)間與極值；（2）分析可知SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性，數(shù)形結(jié)合可得出實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】（1）解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，該函數(shù)的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0增極大值減極小值增所以，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，減區(qū)間為SKIPIF1<0，極大值為SKIPIF1<0，即小值為SKIPIF1<0.（2）解：由題意可知，關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，即關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0；由SKIPIF1<0可得SKIPIF1<0.所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.所以，SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，作出函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象如下圖所示：由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象有兩個(gè)交點(diǎn).因此，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.典例5.（2022·山東菏澤·高三期中）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0的極值；(2)若SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有零點(diǎn)，求實(shí)數(shù)a的取值范圍．【答案】(1)極小值為SKIPIF1<0，無(wú)極大值(2)SKIPIF1<0【分析】（1）當(dāng)SKIPIF1<0時(shí)，對(duì)SKIPIF1<0求導(dǎo)，得出SKIPIF1<0的單調(diào)性，即可求出SKIPIF1<0的極值；（2）方法一：分類討論SKIPIF1<0，SKIPIF1<0和SKIPIF1<0，得出SKIPIF1<0的單調(diào)性，利用單調(diào)性列出不等式即可求出實(shí)數(shù)a的取值范圍；方法二：分離參數(shù)，構(gòu)造新函數(shù)，研究SKIPIF1<0的單調(diào)性，求出SKIPIF1<0在SKIPIF1<0的值域，進(jìn)而求出實(shí)數(shù)a的取值范圍.【詳解】（1）由函數(shù)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0得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)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0的極小值為SKIPIF1<0，無(wú)極大值．（2）方法一：由SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，要使SKIPIF1<0在SKIPIF1<0內(nèi)有零點(diǎn)，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0得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)遞減，在SKIPIF1<0上單調(diào)遞增，此時(shí)SKIPIF1<0，所以需SKIPIF1<0，所以SKIPIF1<0．③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，此時(shí)SKIPIF1<0，所以SKIPIF1<0恒成立，不符合條件．綜上可知，a的取值范圍為SKIPIF1<0．方法二：令SKIPIF1<0得SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．典例6.（2022·遼寧·高三期中）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)當(dāng)a＝1時(shí)，若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，求實(shí)數(shù)t的取值范圍．【答案】(1)答案見(jiàn)解析(2)SKIPIF1<0【分析】（1）分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0討論求解即可；（2）由題意可知關(guān)于x的方程SKIPIF1<0有兩個(gè)不同的實(shí)根，進(jìn)而SKIPIF1<0，令SKIPIF1<0，要使SKIPIF1<0有兩個(gè)不同的實(shí)根，則需SKIPIF1<0有兩個(gè)不同的實(shí)根．令SKIPIF1<0，利用導(dǎo)數(shù)法研究SKIPIF1<0的零點(diǎn)即可【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．綜上可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以關(guān)于x的方程SKIPIF1<0有兩個(gè)不同的實(shí)根，即關(guān)于x的方程SKIPIF1<0有兩個(gè)不同的實(shí)根．因?yàn)閤＞0，所以SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．要使SKIPIF1<0有兩個(gè)不同的實(shí)根，則需SKIPIF1<0有兩個(gè)不同的實(shí)根．令SKIPIF1<0，則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，所以SKIPIF1<0．當(dāng)t＜1時(shí)，SKIPIF1<0，SKIPIF1<0沒(méi)有零點(diǎn)；當(dāng)t＝1時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)x＝1時(shí)，等號(hào)成立，SKIPIF1<0只有一個(gè)零點(diǎn)；當(dāng)t＞1時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0．所以SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)，在SKIPIF1<0上有一個(gè)零點(diǎn)，符合條件．綜上，實(shí)數(shù)t的取值范圍是SKIPIF1<0．【總結(jié)提升】已知函數(shù)有零點(diǎn)(方程有根)求參數(shù)值(取值范圍)常用的方法：（1）直接法：直接求解方程得到方程的根，再通過(guò)解不等式確定參數(shù)范圍；（2）分離參數(shù)法：先將參數(shù)分離，轉(zhuǎn)化成求函數(shù)的值域問(wèn)題加以解決；（3）數(shù)形結(jié)合法：先對(duì)解析式變形，進(jìn)而構(gòu)造兩個(gè)函數(shù)，然后在同一平面直角坐標(biāo)系中畫出函數(shù)的圖象，利用數(shù)形結(jié)合的方法求解考向三與零點(diǎn)相關(guān)的不等式恒成立或證明問(wèn)題【核心知識(shí)】1.不等式恒成立問(wèn)題常見(jiàn)方法：①分離參數(shù)恒成立(即可)或恒成立（即可）；②數(shù)形結(jié)合(圖象在上方即可)；③討論最值或恒成立；④討論參數(shù)，排除不合題意的參數(shù)范圍，篩選出符合題意的參數(shù)范圍.2.含參數(shù)的不等式恒成立的處理方法：①的圖象永遠(yuǎn)落在圖象的上方；②構(gòu)造函數(shù)法，一般構(gòu)造，；③參變分離法，將不等式等價(jià)變形為，或，進(jìn)而轉(zhuǎn)化為求函數(shù)的最值.3.利用參變量分離法求解函數(shù)不等式恒（能）成立，可根據(jù)以下原則進(jìn)行求解：（1），；（2），；（3），；（4），.【典例分析】典例7.（2022·貴州·頂效開(kāi)發(fā)區(qū)頂興學(xué)校高三期中（理））已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有3個(gè)不同的實(shí)根SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】對(duì)SKIPIF1<0求導(dǎo)，研究函數(shù)SKIPIF1<0的單調(diào)性、極值等性質(zhì)，利用SKIPIF1<0的圖象求得SKIPIF1<0的范圍，以及SKIPIF1<0與SKIPIF1<0的關(guān)系，將問(wèn)題轉(zhuǎn)化為關(guān)于SKIPIF1<0的函數(shù)的值域的問(wèn)題進(jìn)行求解即可．【詳解】因?yàn)镾KIPIF1<0，故可得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增．則SKIPIF1<0的極大值為SKIPIF1<0，SKIPIF1<0的極小值為SKIPIF1<0，∵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，根據(jù)以上信息，作出SKIPIF1<0的大致圖象如圖所示：由圖可知，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有3個(gè)交點(diǎn)時(shí)，方程SKIPIF1<0有3個(gè)不同的實(shí)根，則SKIPIF1<0，因?yàn)榉匠蘏KIPIF1<0的3個(gè)不同的實(shí)根為SKIPIF1<0，則SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0故SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得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，所以SKIPIF1<0，又SKIPIF1<0SKIPIF1<0，SKIPIF1<0，故可得SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0SKIPIF1<0．故選：A．典例8.【多選題】（2022·山東·青島二中高三期中）已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有四個(gè)不同的零點(diǎn)：SKIPIF1<0，且SKIPIF1<0,則以下結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【分析】設(shè)SKIPIF1<0，利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0的單調(diào)性與極值，數(shù)形結(jié)合可判斷B的正誤；分析可知SKIPIF1<0，結(jié)合基本不等式可判斷A的正誤；構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性，可判斷CD的正誤.【詳解】設(shè)SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減，所以，函數(shù)SKIPIF1<0的極大值為SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出函數(shù)SKIPIF1<0與SKIPIF1<0的大致圖象，由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有四個(gè)交點(diǎn)，B對(duì)；因?yàn)镾KIPIF1<0，則SKIPIF1<0，由圖可知SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，A對(duì)；令SKIPIF1<0，其中SKIPIF1<0，由圖可知SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減，所以，SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，且函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，C錯(cuò)D對(duì).故選：ABD.典例9.（貴州省六盤水市2021-2022學(xué)年高二下期末）已知函數(shù)SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0有兩個(gè)不相同的零點(diǎn)SKIPIF1<0，證明SKIPIF1<0.【答案】(1)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)；SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)；(2)證明見(jiàn)解析.【分析】（1）求出函數(shù)導(dǎo)數(shù)，對(duì)SKIPIF1<0分類討論，根據(jù)導(dǎo)數(shù)的正負(fù)確定函數(shù)的單調(diào)性即可得解；（2）由函數(shù)的單調(diào)性可確定函數(shù)零點(diǎn)在SKIPIF1<0兩側(cè)，要證原不等式可轉(zhuǎn)化為證SKIPIF1<0，再由函數(shù)的單調(diào)性轉(zhuǎn)化為證SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)即可得證.【詳解】（1）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立，所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)；SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為減函數(shù).綜上，SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)；SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù).（2）由（1）知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，至多有1個(gè)零點(diǎn)，不符合題意；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)，所以SKIPIF1<0為函數(shù)的極小值，函數(shù)有兩個(gè)零點(diǎn)則SKIPIF1<0，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，要證SKIPIF1<0，即證SKIPIF1<0，因?yàn)镾KIPIF1<0SKIPIF1<0在SKIPIF1<0上為減函數(shù)，所以只要證SKIPIF1<0，又SKIPIF1<0，即證SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0成立，從而SKIPIF1<0成立.典例10.（遼寧省名校聯(lián)盟2022-2023學(xué)年高三上學(xué)期11月份聯(lián)合考試數(shù)學(xué)試題）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)設(shè)SKIPIF1<0是SKIPIF1<0的兩個(gè)零點(diǎn)，證明：SKIPIF1<0．【答案】(1)答案見(jiàn)解析(2)證明見(jiàn)解析【分析】（1）先求導(dǎo)，然后分類討論即可求解；（2）先利用導(dǎo)數(shù)法可得SKIPIF1<0是SKIPIF1<0上的減函數(shù)，SKIPIF1<0上的增函數(shù)，從而可知SKIPIF1<0，由（1）可知，SKIPIF1<0，SKIPIF1<0，則有SKIPIF1<0，由此即可求證【詳解】（1）SKIPIF1<0定義域?yàn)镾KIPIF1<0，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)．議SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0符號(hào)相同，若SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；若SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0有兩個(gè)不等正根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．SKIPIF1<0與SKIPIF1<0隨x的變化情況如下表：xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的增區(qū)間為SKIPIF1<0，沒(méi)有減區(qū)間；當(dāng)SKIPIF1<0吋，SKIPIF1<0的減區(qū)間為SKIPIF1<