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（精練）（解析版）

3.6零點(diǎn)定理（精練）（提升版）題組一題組一零點(diǎn)的區(qū)間1．（2022·甘肅·天水市第一中學(xué)）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.當(dāng)SKIPIF1<0時，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.由零點(diǎn)存在定理可得：函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是SKIPIF1<0.故選：C2（2022·江蘇揚(yáng)州）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】函數(shù)SKIPIF1<0,SKIPIF1<0是單調(diào)遞增函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0故函數(shù)的零點(diǎn)所在的區(qū)間為SKIPIF1<0，故選：B3．（2022·天津紅橋·一模）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】函數(shù)SKIPIF1<0是SKIPIF1<0上的連續(xù)增函數(shù),SKIPIF1<0,可得SKIPIF1<0,所以函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是SKIPIF1<0.故選：C4．（2022·廣東中山）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的零點(diǎn)在區(qū)間SKIPIF1<0.故選：A5．（2022·北京師大附中）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（

）A．（0，1） B．(1，2) C．(2，3) D．（3，4）【答案】B【解析】因為函數(shù)SKIPIF1<0均為SKIPIF1<0上的單調(diào)遞減函數(shù)，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，因為SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是SKIPIF1<0.故選：B6．（2022·云南玉溪·高一期末）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由解析式知：在SKIPIF1<0上SKIPIF1<0恒成立，在SKIPIF1<0上SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，SKIPIF1<0，綜上，零點(diǎn)所在的區(qū)間為SKIPIF1<0.故選：B7．（2022·寧夏·青銅峽市寧朔中學(xué)高二學(xué)業(yè)考試）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】易知SKIPIF1<0為增函數(shù)，又SKIPIF1<0，SKIPIF1<0，故零點(diǎn)所在的區(qū)間是SKIPIF1<0.故選：B.8．（2022·新疆維吾爾自治區(qū)喀什第二中學(xué)）函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意可知，SKIPIF1<0的定義域為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在定義域內(nèi)單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減.所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故SKIPIF1<0，根據(jù)零點(diǎn)的存在性定理，可得函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間為SKIPIF1<0.故選：B.9．（2022·海南·嘉積中學(xué)高一期末）SKIPIF1<0零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意知：SKIPIF1<0在SKIPIF1<0上連續(xù)且單調(diào)遞增；對于A，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0內(nèi)不存在零點(diǎn)，A錯誤；對于B，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0內(nèi)不存在零點(diǎn)，B錯誤；對于C，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0內(nèi)存在零點(diǎn)，C正確；對于D，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0內(nèi)不存在零點(diǎn)，D錯誤.故選：C.10．（2022·四川·德陽五中）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】函數(shù)SKIPIF1<0在R上單調(diào)遞增，而SKIPIF1<0，SKIPIF1<0，由零點(diǎn)存在性定理知，函數(shù)SKIPIF1<0的唯一零點(diǎn)在區(qū)間SKIPIF1<0內(nèi).故選：B11．（2022·安徽·池州市第一中學(xué)）函數(shù)SKIPIF1<0的零點(diǎn)所在的一個區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0，且SKIPIF1<0是單調(diào)遞減函數(shù)，故函數(shù)SKIPIF1<0的零點(diǎn)所在的一個區(qū)間是SKIPIF1<0，故選：B12．（2022·廣東汕尾）函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵f(x)定義域為R，且f(x)在R上單調(diào)遞增，又∵f(1)＝－10＜0，f(2)＝19＞0，∴f(x)在(1，2)上存在唯一零點(diǎn).故選：B.題組二題組二零點(diǎn)的個數(shù)1．（2022·四川省瀘縣第二中學(xué)）函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)為（

）A．0 B．1 C．2 D．3【答案】B【解析】由于函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，且SKIPIF1<0，故函數(shù)在SKIPIF1<0上有唯一零點(diǎn)，也即在SKIPIF1<0上有唯一零點(diǎn).故選：B.2．（2022·重慶）函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為（

）A．1 B．2 C．3 D．4【答案】B【解析】函數(shù)SKIPIF1<0，x＞0，則SKIPIF1<0，令SKIPIF1<0，解得x∈（0，3），此時函數(shù)是增函數(shù)，x∈（3，＋∞）時，SKIPIF1<0，f（x）是減函數(shù)，所以x＝3時，函數(shù)取得最大值，又f（3）＝ln3－1＞0，SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為2，故選：B.3．（2022·重慶·三模）已知函數(shù)SKIPIF1<0則函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為（

）A．0個 B．1個 C．2個 D．3個【答案】C【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，因為SKIPIF1<0，所以舍去；當(dāng)SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0，滿足SKIPIF1<0.所以SKIPIF1<0或SKIPIF1<0.函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為2個.故選：C4．（2022·新疆·三模（理））函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為___________.【答案】2【解析】當(dāng)SKIPIF1<0時，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，此時有1個零點(diǎn)；當(dāng)SKIPIF1<0時，SKIPIF1<0，顯然SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，由零點(diǎn)存在定理知此時有1個零點(diǎn)；綜上共有2個零點(diǎn).故答案為：2.5．（2022·新疆）函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為_________．【答案】1【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0有一個零點(diǎn)SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，無零點(diǎn)，故函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為1個故答案為：1題組三題組三比較零點(diǎn)的大小1．（2022·山西·二模（理））已知SKIPIF1<0是SKIPIF1<0的一個零點(diǎn)，SKIPIF1<0是SKIPIF1<0的一個零點(diǎn)，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】A【解析】因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上是減函數(shù)，在SKIPIF1<0上是增函數(shù)，在SKIPIF1<0上是減函數(shù)，因為SKIPIF1<0，所以SKIPIF1<0僅有1個零點(diǎn)，因為SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0是增函數(shù)，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A．2．（2022·湖南·益陽市箴言中學(xué)）已知三個函數(shù)SKIPIF1<0的零點(diǎn)依次為SKIPIF1<0，則SKIPIF1<0的大小關(guān)系（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】∵函數(shù)SKIPIF1<0為增函數(shù)，又SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.故選：D.3．（2022·陜西·長安一中模擬預(yù)測）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的零點(diǎn)分別為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的大小順序為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因為函數(shù)SKIPIF1<0、SKIPIF1<0均為SKIPIF1<0上的增函數(shù)，故函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，因為SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，因為函數(shù)SKIPIF1<0、SKIPIF1<0在SKIPIF1<0上均為增函數(shù)，故函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因為SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，因此，SKIPIF1<0.故選：A.4．（2022·安徽·蚌埠二中模擬預(yù)測（文））已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)函數(shù)SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因為SKIPIF1<0，所以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上單調(diào)遞增，因為SKIPIF1<0，所以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，因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，由函數(shù)單調(diào)性可知，SKIPIF1<0，所以SKIPIF1<0，故選：SKIPIF1<0.5．（2022·河南河南·三模）若實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0，SKIPIF1<0，對于函數(shù)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0，所以SKIPIF1<0存在唯一零點(diǎn)SKIPIF1<0，SKIPIF1<0，使SKIPIF1<0，所以對于SKIPIF1<0，有SKIPIF1<0，所以SKIPIF1<0.故選：A題組四題組四已知零點(diǎn)求參數(shù)1．（2022·湖北宜昌）函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有兩個零點(diǎn)，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【解析】作出函數(shù)SKIPIF1<0的圖象，令SKIPIF1<0，可得SKIPIF1<0，畫出直線SKIPIF1<0，可得當(dāng)SKIPIF1<0時，直線SKIPIF1<0和函數(shù)SKIPIF1<0的圖象有兩個交點(diǎn)，則SKIPIF1<0有兩個零點(diǎn)．故選：B．2．（2022·首都師范大學(xué)附屬中學(xué)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0有三個不同的零點(diǎn)，則實(shí)數(shù)k的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，故當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，故SKIPIF1<0，且SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，由此作出函數(shù)的大致圖象如圖：由SKIPIF1<0有三個不同的零點(diǎn)，即函數(shù)SKIPIF1<0的圖象與SKIPIF1<0有三個不同的交點(diǎn)，結(jié)合圖象，可得SKIPIF1<0，故選：C3．（2022·河北唐山）已知函數(shù)SKIPIF1<0，若SKIPIF1<0有3個零點(diǎn)，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)SKIPIF1<0，令SKIPIF1<0，令SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.所以SKIPIF1<0.令SKIPIF1<0有三個零點(diǎn).作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象如圖所示，所以a的取值范圍為SKIPIF1<0.故選：B4．（2022·安徽）已知函數(shù)SKIPIF1<0在（0，+∞）上有3個不同的零點(diǎn)，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】D【解析】解法一：因為函數(shù)在（0，+∞）上有3個不同的零點(diǎn)，所以SKIPIF1<0，和SKIPIF1<0的圖像在（0，+∞）上有3個交點(diǎn)，代入SKIPIF1<0，不合題意，排除A、C，又k取+∞顯然不合題意，排除B；解法二：因為函數(shù)SKIPIF1<0在SKIPIF1<0上有3個不同的零點(diǎn)，所以SKIPIF1<0|和SKIPIF1<0的圖像在SKIPIF1<0上有3個交點(diǎn)，畫出函數(shù)g（x）的圖像，如圖.SKIPIF1<0的圖像恒過點(diǎn)（0，2），且當(dāng)SKIPIF1<0時與x軸的交點(diǎn)為（SKIPIF1<0，0），當(dāng)SKIPIF1<0時，SKIPIF1<0與g（x）的圖像在SKIPIF1<0上有3個不同的交點(diǎn)，如圖.當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0與g（x）的圖像在SKIPIF1<0上僅有2個不同的交點(diǎn)，如圖.當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0與g（x）的圖像在（0，SKIPIF1<0）上有1個交點(diǎn)，在（SKIPIF1<0，∞）上有2個交點(diǎn)，如圖.當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0與g（x）的圖像在（0，SKIPIF1<0）上有3個交點(diǎn)，在SKIPIF1<0上有0個交點(diǎn)，如圖，當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0與g（x）的圖像在（0，+∞）上有2個交點(diǎn)，如圖.當(dāng)SKIPIF1<0時，SKIPIF1<0的左支與g（x）的圖像無交點(diǎn)，當(dāng)直線SKIPIF1<0與SKIPIF1<0相切時，聯(lián)立方程得SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0舍去），所以SKIPIF1<0當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0與g（x）的圖像在SKIPIF1<0上有3個交點(diǎn).綜上，可得k的取值范圍為SKIPIF1<0故選：D.5．（2022·江蘇泰州·模擬預(yù)測）已知定義在R上的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0，已知當(dāng)SKIPIF1<0時，SKIPIF1<0，若SKIPIF1<0恰有六個不相等的零點(diǎn)，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0.所以當(dāng)SKIPIF1<0時，SKIPIF1<0.因為SKIPIF1<0，則SKIPIF1<0關(guān)于SKIPIF1<0對稱，因為SKIPIF1<0關(guān)于SKIPIF1<0對稱，SKIPIF1<0有6個不相同的根，∴SKIPIF1<0在SKIPIF1<0有三個不同的根，SKIPIF1<0表示過定點(diǎn)SKIPIF1<0的直線系，SKIPIF1<0SKIPIF1<0.作出SKIPIF1<0在SKIPIF1<0上的圖象,如圖所示，SKIPIF1<0時，SKIPIF1<0，又SKIPIF1<0,則SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0;SKIPIF1<0時，顯然不滿足題意.∴m的取值范圍SKIPIF1<0.故選：D．6．（2022·內(nèi)蒙古·滿洲里市教研培訓(xùn)中心三模（文））已知函數(shù)SKIPIF1<0有唯一零點(diǎn)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】令SKIPIF1<0，則SKIPIF1<0，記SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，圖象關(guān)于SKIPIF1<0軸對稱，因為SKIPIF1<0只有唯一的零點(diǎn)，所以零點(diǎn)只能是SKIPIF1<0于是SKIPIF1<0故選：C7．（2022·全國·高三專題練習(xí)）已知SKIPIF1<0是以2為周期的偶函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0，那么在區(qū)間SKIPIF1<0內(nèi)，關(guān)于x的方程SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）有4個根，則k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】本題考查函數(shù)性質(zhì)的綜合應(yīng)用及數(shù)形結(jié)合的數(shù)學(xué)思想．由函數(shù)性質(zhì)作圖如下：令SKIPIF1<0其圖像為通過定點(diǎn)SKIPIF1<0斜率為k的直線，要使SKIPIF1<0有四解，即SKIPIF1<0和SKIPIF1<0有四個交點(diǎn)，由圖知當(dāng)SKIPIF1<0在SKIPIF1<0與SKIPIF1<0之間轉(zhuǎn)動時滿足題意．易得SKIPIF1<0的斜率為0，SKIPIF1<0的斜率為SKIPIF1<0.所以SKIPIF1<0.故選：C．8．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個不同的交點(diǎn)，則實(shí)數(shù)m取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意得：SKIPIF1<0，則SKIPIF1<0，問題轉(zhuǎn)化為y＝m和SKIPIF1<0有2個交點(diǎn)，而SKIPIF1<0，在SKIPIF1<0和SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞增，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞減，當(dāng)x趨于正無窮大時，SKIPIF1<0無限接近于0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象，如圖所示：觀察圖象得：函數(shù)SKIPIF1<0和SKIPIF1<0的圖象有2個不同的交點(diǎn)時，實(shí)數(shù)SKIPIF1<0.故選：D．9．（2022·天津·南開中學(xué)模擬預(yù)測）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰有兩個零點(diǎn)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】A【解析】SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，作出h(x)的圖象：如圖y=h(x)與y=a的圖象有兩個交點(diǎn)時，SKIPIF1<0，故選：A．10．（2022·天津南開·二模）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0恰有2個零點(diǎn)，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【解析】SKIPIF1<0，故SKIPIF1<0，則函數(shù)SKIPIF1<0恰有2個零點(diǎn)等價于SKIPIF1<0有兩個不同的解，故SKIPIF1<0的圖象有兩個不同的交點(diǎn)，設(shè)SKIPIF1<0又SKIPIF1<0的圖象如圖所示，由圖象可得兩個函數(shù)的圖象均過原點(diǎn)，若SKIPIF1<0，此時兩個函數(shù)的圖象有兩個不同的交點(diǎn)，當(dāng)SKIPIF1<0時，考慮直線SKIPIF1<0與SKIPIF1<0的圖象相切，則由SKIPIF1<0可得SKIPIF1<0即SKIPIF1<0，考慮直線SKIPIF1<0與SKIPIF1<0的圖象相切，由SKIPIF1<0可得SKIPIF1<0，則SKIPIF1<0即SKIPIF1<0.考慮直線SKIPIF1<0與SKIPIF1<0的圖象相切，由SKIPIF1<0可得SKIPIF1<0即SKIPIF1<0，結(jié)合圖象可得當(dāng)SKIPIF1<0或SKIPIF1<0時，兩個函數(shù)的圖象有兩個不同的交點(diǎn)，綜上，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故選：B.11．（2022·陜西·模擬預(yù)測（理））已知函數(shù)SKIPIF1<0SKIPIF1<0在SKIPIF1<0上有且只有5個零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0在SKIPIF1<0上有且只有5個零點(diǎn)，因為SKIPIF1<0，所以SKIPIF1<0，所以，如圖，由正弦函數(shù)圖像，要使SKIPIF1<0在SKIPIF1<0上有且只有5個零點(diǎn)，則SKIPIF1<0，即SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的范圍是SKIPIF1<0.

故選：C12．（2022·陜西寶雞·二模（文））已知函數(shù)SKIPIF1<0（SKIPIF1<0是自然對數(shù)的底數(shù)）在定義域SKIPIF1<0上有三個零點(diǎn)，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【解析】當(dāng)SKIPIF1<0時，令SKIPIF1<0，解得：SKIPIF1<0；當(dāng)SKIPIF1<0時，令SKIPIF1<0，解得：SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，則當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0；SKIPIF1<0在定義域SKIPIF1<0上有三個零點(diǎn)，SKIPIF1<0為SKIPIF1<0一個零點(diǎn)且SKIPIF1<0有兩個解，SKIPIF1<0，解得：SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B.13．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上既有極大值又有極小值，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】C【解析】函數(shù)SKIPIF1<0，導(dǎo)函數(shù)SKIPIF1<0.因為SKIPIF1<0在SKIPIF1<0上既有極大值又有極小值，所以SKIPIF1<0在SKIPIF1<0內(nèi)應(yīng)有兩個不同的異號實(shí)數(shù)根．SKIPIF1<0，解得：SKIPIF1<0，實(shí)數(shù)a的取值范圍SKIPIF1<0.故選：C．14．（2022·河南·模擬預(yù)測（理））已知函數(shù)SKIPIF1<0至多有2個不同的零點(diǎn)，則實(shí)數(shù)a的最大值為（

）．A．0 B．1 C．2 D．e【答案】C【解析】令SKIPIF1<0，得到SKIPIF1<0，函數(shù)SKIPIF1<0至多有2個不同的零點(diǎn)，等價于SKIPIF1<0至多有兩個不同的根，即函數(shù)SKIPIF1<0與SKIPIF1<0至多有2個不同的交點(diǎn)令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0與SKIPIF1<0為函數(shù)SKIPIF1<0的極值點(diǎn)，且SKIPIF1<0，且SKIPIF1<0在R上恒成立，畫出SKIPIF1<0的圖象如下：有圖可知：SKIPIF1<0或SKIPIF1<0時，符合題意，其中SKIPIF1<0，解得：SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，由SKIPIF1<0可得：SKIPIF1<0，所以SKIPIF1<0，綜上：實(shí)數(shù)a的最大值為2故選：C15（2022·天津市武清區(qū)楊村第一中學(xué)二模）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0．若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有三個交點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0，又因函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有三個交點(diǎn)，所以在SKIPIF1<0上函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有兩個交點(diǎn)，即方程SKIPIF1<0在SKIPIF1<0上有兩個不同的實(shí)數(shù)根，即方程SKIPIF1<0在SKIPIF1<0上有兩個不同的實(shí)數(shù)根，所以SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時，當(dāng)SKIPIF1<0時，令SKIPIF1<0，由SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，此時，SKIPIF1<0，結(jié)合圖象，所以SKIPIF1<0時，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象只有一個交點(diǎn)，綜上所述，SKIPIF1<0.故選：B.16．（2022·江西·南昌市八一中學(xué)三模（文））已知函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0存在零點(diǎn)，則實(shí)數(shù)SKIPIF1<0值可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】根據(jù)題意，令SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上存在零點(diǎn)等價于SKIPIF1<0與SKIPIF1<0的圖像有交點(diǎn)．SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因為SKIPIF1<0，SKIPIF1<0，所以存在唯一的SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，又SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故選：D．題組五題組五零點(diǎn)的綜合運(yùn)用1．（2022·江西師大附中三模）定義在R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0.則函數(shù)SKIPIF1<0的所有零點(diǎn)之和為（

）A．7 B．14 C．21 D．28【答案】B【解析】依題意，SKIPIF1<0是奇函數(shù).又由SKIPIF1<0知，SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱.SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是周期為4的周期函數(shù).SKIPIF1<0，所以SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對稱.由于SKIPIF1<0從而函數(shù)SKIPIF1<0的所有零點(diǎn)之和即為函數(shù)SKIPIF1<0與SKIPIF1<0的圖像的交點(diǎn)的橫坐標(biāo)之和.而函數(shù)SKIPIF1<0的圖像也關(guān)于點(diǎn)SKIPIF1<0