新高考數(shù)學(xué)二輪復(fù)習(xí)講義+分層訓(xùn)練專題25 函數(shù)與導(dǎo)數(shù)專項(xiàng)訓(xùn)練（解析版）

函數(shù)與導(dǎo)數(shù)專項(xiàng)測試卷考試時(shí)間：120分鐘滿分：150分一、單選題：本大題共8小題，每個(gè)小題5分，共40分.在每小題給出的選項(xiàng)中，只有一項(xiàng)是符合題目要求的.1．（2023·遼寧·遼寧實(shí)驗(yàn)中學(xué)校考模擬預(yù)測）下列函數(shù)不是偶函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)函數(shù)的奇偶性的定義求解.【詳解】對(duì)于A項(xiàng)，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)；對(duì)于B項(xiàng)，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)；對(duì)于C項(xiàng)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0不是偶函數(shù)；對(duì)于D項(xiàng)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是偶函數(shù).故選:C.2．（2022·吉林·東北師大附中?？寄M預(yù)測）一種藥在病人血液中的量不低于1800mg時(shí)才有療效，如果用藥前，病人血液中該藥的量為0mg，用藥后，藥在血液中以每小時(shí)20%的比例衰減．現(xiàn)給某病人靜脈注射了3600mg的此藥，為了持續(xù)保持療效，則最長需要在多少小時(shí)后再次注射此藥（SKIPIF1<0，結(jié)果精確到0.1）（

）A．2.7 B．2.9 C．3.1 D．3.3【答案】C【分析】根據(jù)指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的運(yùn)算法則即可求解.【詳解】設(shè)注射后經(jīng)過的時(shí)間為SKIPIF1<0，血液中藥物的含量為SKIPIF1<0，則有SKIPIF1<0，因?yàn)樗幵诓∪搜褐械牧坎坏陀?800mg時(shí)才有療效，所以令SKIPIF1<0，解得SKIPIF1<0.故選：C.3．（2023·全國·模擬預(yù)測）高斯是德國著名的數(shù)學(xué)家，近代數(shù)學(xué)奠基者之一，享有“數(shù)學(xué)王子”的稱號(hào)，設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)，SKIPIF1<0也被稱為“高斯函數(shù)”，例如SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0為函數(shù)SKIPIF1<0的零點(diǎn)，則SKIPIF1<0（

）A．2 B．3 C．4 D．5【答案】A【分析】首先判斷函數(shù)的單調(diào)性，再根據(jù)零點(diǎn)存在性定理判斷SKIPIF1<0所在區(qū)間，最后根據(jù)高斯函數(shù)的定義計(jì)算可得.【詳解】解：因?yàn)镾KIPIF1<0與SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上存在唯一零點(diǎn)SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：A4．（2021·天津薊州·天津市薊州區(qū)第一中學(xué)?？寄M預(yù)測）已知函數(shù)SKIPIF1<0是SKIPIF1<0上的單調(diào)函數(shù)，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【分析】由函數(shù)可得SKIPIF1<0且SKIPIF1<0，故可得函數(shù)只能是SKIPIF1<0上的單調(diào)遞減函數(shù)，然后列不等式即可【詳解】由SKIPIF1<0可得SKIPIF1<0且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0不可能是增函數(shù)，所以函數(shù)SKIPIF1<0在SKIPIF1<0上不可能是增函數(shù)，則函數(shù)SKIPIF1<0是SKIPIF1<0上的單調(diào)遞減函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，綜上：實(shí)數(shù)a的取值范圍為SKIPIF1<0，故選：B5．（2023·河南信陽·河南省信陽市第二高級(jí)中學(xué)校聯(lián)考一模）已知函數(shù)SKIPIF1<0對(duì)SKIPIF1<0均滿足SKIPIF1<0，其中SKIPIF1<0是SKIPIF1<0的導(dǎo)數(shù)，則下列不等式恒成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)給定的等式，構(gòu)造函數(shù)并探討其單調(diào)性，再逐項(xiàng)計(jì)算判斷作答.【詳解】SKIPIF1<0，令SKIPIF1<0，求導(dǎo)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上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，對(duì)于A，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，A正確；對(duì)于B，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，B錯(cuò)誤；對(duì)于C，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，C錯(cuò)誤；對(duì)于D，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，D錯(cuò)誤．故選：A6．（2023·青海海東·統(tǒng)考一模）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)SKIPIF1<0，求導(dǎo)可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，再根據(jù)SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，再結(jié)合SKIPIF1<0的單調(diào)性求解即可.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.不等式SKIPIF1<0等價(jià)于不等式SKIPIF1<0，即SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，解得SKIPIF1<0故選：A7．（2021·陜西漢中·統(tǒng)考模擬預(yù)測）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0，其導(dǎo)函數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)題意當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，結(jié)合導(dǎo)數(shù)的運(yùn)算法則可構(gòu)造函數(shù)SKIPIF1<0，由此判斷其單調(diào)性，利用函數(shù)的單調(diào)性，即可判斷SKIPIF1<0的大小.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，由題意知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0時(shí)單調(diào)遞增，故SKIPIF1<0，即SKIPIF1<0，故選：D.8．（2022·江蘇南京·模擬預(yù)測）已知函數(shù)SKIPIF1<0（SKIPIF1<0），且SKIPIF1<0在SKIPIF1<0有兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)給定條件，利用零點(diǎn)的意義等價(jià)轉(zhuǎn)化，構(gòu)造函數(shù)SKIPIF1<0，再借助導(dǎo)數(shù)探討函數(shù)SKIPIF1<0在SKIPIF1<0有兩個(gè)零點(diǎn)作答.【詳解】SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，依題意，函數(shù)SKIPIF1<0在SKIPIF1<0有兩個(gè)零點(diǎn)，顯然SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則有SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增或單調(diào)遞減，即有函數(shù)SKIPIF1<0在SKIPIF1<0只有一個(gè)零點(diǎn)1，因此SKIPIF1<0，此時(shí)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，則SKIPIF1<0，要函數(shù)SKIPIF1<0在SKIPIF1<0有兩個(gè)零點(diǎn)，當(dāng)且僅當(dāng)SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)，即有SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的取值范圍SKIPIF1<0.故選：C二、多選題：本大題共4小題，每個(gè)小題5分，共20分.在每小題給出的選項(xiàng)中，只有一項(xiàng)或者多項(xiàng)是符合題目要求的.9．（2023·安徽淮南·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的值域?yàn)镾KIPIF1<0B．直線SKIPIF1<0是曲線SKIPIF1<0的一條切線C．SKIPIF1<0圖象的對(duì)稱中心為SKIPIF1<0D．方程SKIPIF1<0有三個(gè)實(shí)數(shù)根【答案】BD【分析】A.分SKIPIF1<0兩種情況求函數(shù)的值域；B.利用導(dǎo)數(shù)求函數(shù)的切線，判斷選項(xiàng)；C.利用平移判斷函數(shù)的對(duì)稱中心；D.首先求SKIPIF1<0的值，再求解方程的實(shí)數(shù)根.【詳解】A.SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等號(hào)成立，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故A錯(cuò)誤；B.令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以圖象在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以圖象在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，得SKIPIF1<0，故B正確；C.SKIPIF1<0的對(duì)稱中心是SKIPIF1<0，所以SKIPIF1<0的對(duì)稱中心是SKIPIF1<0，向右平移1個(gè)單位得SKIPIF1<0，對(duì)稱中心是SKIPIF1<0，故C錯(cuò)誤；D.SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，1個(gè)實(shí)根，當(dāng)SKIPIF1<0時(shí)，得SKIPIF1<0或SKIPIF1<0，2個(gè)實(shí)根，所以共3個(gè)實(shí)根，故D正確.故選：BD10．（2023·安徽馬鞍山·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的可能的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AD【分析】根據(jù)SKIPIF1<0轉(zhuǎn)化成SKIPIF1<0恒成立，構(gòu)造函數(shù)SKIPIF1<0利用導(dǎo)數(shù)求解SKIPIF1<0的單調(diào)性，問題進(jìn)一步轉(zhuǎn)化成SKIPIF1<0恒成立，構(gòu)造SKIPIF1<0，求解最值即可.【詳解】SKIPIF1<0，故SKIPIF1<0恒成立，轉(zhuǎn)化成SKIPIF1<0恒成立，記SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞增，故由SKIPIF1<0得SKIPIF1<0，故SKIPIF1<0恒成立，記SKIPIF1<0，故當(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取最大值SKIPIF1<0，故由SKIPIF1<0恒成立，即SKIPIF1<0,故SKIPIF1<0，故選：AD【點(diǎn)睛】本題主要考查導(dǎo)數(shù)在函數(shù)中的應(yīng)用，著重考查了轉(zhuǎn)化與化歸思想、邏輯推理能力與計(jì)算能力，對(duì)導(dǎo)數(shù)的應(yīng)用的考查主要從以下幾個(gè)角度進(jìn)行：(1)考查導(dǎo)數(shù)的幾何意義，求解曲線在某點(diǎn)處的切線方程；(2)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間，判斷單調(diào)性；已知單調(diào)性，求參數(shù)；(3)利用導(dǎo)數(shù)求函數(shù)的最值(極值)，解決函數(shù)的恒成立與有解問題，同時(shí)注意數(shù)形結(jié)合思想的應(yīng)用.11．（2023·廣東肇慶·統(tǒng)考二模）函數(shù)SKIPIF1<0的部分圖像如圖所示，SKIPIF1<0，則下列選項(xiàng)中正確的有（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．SKIPIF1<0是奇函數(shù)C．SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0D．SKIPIF1<0，其中SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)【答案】AD【分析】根據(jù)題意可求得函數(shù)的周期，即可判斷A，進(jìn)而可求得SKIPIF1<0，再根據(jù)待定系數(shù)法可求得SKIPIF1<0，再根據(jù)三角函數(shù)的奇偶性可判斷B，根據(jù)余弦函數(shù)的單調(diào)性即可判斷C，求導(dǎo)計(jì)算即可判斷D.【詳解】解：由題意可得SKIPIF1<0，所以SKIPIF1<0，故A正確；則SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0為偶函數(shù)，故B錯(cuò)誤；令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，故C錯(cuò)誤；SKIPIF1<0，則SKIPIF1<0，故D正確.故選：AD.12．（2023·湖南岳陽·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0是周期函數(shù) B．函數(shù)SKIPIF1<0在定義域上是單調(diào)遞增函數(shù)C．函數(shù)SKIPIF1<0是偶函數(shù) D．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】ABD【分析】根據(jù)正弦函數(shù)周期判斷A，由指數(shù)函數(shù)、反比例函數(shù)的單調(diào)性判斷B，根據(jù)奇偶性定義判斷C，由函數(shù)中心對(duì)稱充要條件判斷D.【詳解】令SKIPIF1<0，則SKIPIF1<0，所以函數(shù)為周期函數(shù)，故A正確；因?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0在定義域上單調(diào)遞減，且SKIPIF1<0，所以由復(fù)合函數(shù)的單調(diào)性質(zhì)可得SKIPIF1<0在定義域上是單調(diào)遞增函數(shù)，故B正確；令SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，故C錯(cuò)誤；因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，故D正確.故選：ABD

三、填空題：本大題共4小題，每小題5分，共20分.把答案填在答題卡中的橫線上.13．（2022·四川樂山·統(tǒng)考一模）函數(shù)SKIPIF1<0在SKIPIF1<0上所有零點(diǎn)之和為__________________.【答案】4【分析】利用數(shù)形結(jié)合，將函數(shù)零點(diǎn)問題轉(zhuǎn)化為函數(shù)SKIPIF1<0和SKIPIF1<0的交點(diǎn)問題，利用函數(shù)的對(duì)稱性，可求零點(diǎn)的和.【詳解】函數(shù)SKIPIF1<0，即SKIPIF1<0，函數(shù)SKIPIF1<0和SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，如圖畫出兩個(gè)函數(shù)在區(qū)間SKIPIF1<0的函數(shù)圖象，兩個(gè)函數(shù)圖象有4個(gè)交點(diǎn)，利用對(duì)稱性可知，交點(diǎn)橫坐標(biāo)的和SKIPIF1<0.故答案為：414．（2023·江西景德鎮(zhèn)·統(tǒng)考模擬預(yù)測）已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則滿足SKIPIF1<0的SKIPIF1<0的取值范圍是_________.【答案】SKIPIF1<0【分析】首先判斷函數(shù)的單調(diào)性，根據(jù)偶函數(shù)的性質(zhì)及單調(diào)性原不等式等價(jià)于SKIPIF1<0，解得即可.【詳解】解：因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則不等式SKIPIF1<0等價(jià)于SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<015．（2021·陜西榆林·?？寄M預(yù)測）函數(shù)SKIPIF1<0的定義域?yàn)開_____.【答案】SKIPIF1<0【分析】根據(jù)函數(shù)的解析式，列出函數(shù)有意義時(shí)滿足的不等式，求得答案.【詳解】函數(shù)SKIPIF1<0需滿足SKIPIF1<0，解得SKIPIF1<0且SKIPIF1<0，故函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故答案為：SKIPIF1<016．（2022·上海徐匯·統(tǒng)考一模）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0有四個(gè)交點(diǎn)，且這些交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0，則SKIPIF1<0的取值范圍為___________.【答案】SKIPIF1<0【分析】根據(jù)題意，利用韋達(dá)定理，求得SKIPIF1<0，SKIPIF1<0和SKIPIF1<0的關(guān)系，以及SKIPIF1<0的范圍，將目標(biāo)式轉(zhuǎn)化為關(guān)于SKIPIF1<0的函數(shù)，借助對(duì)勾函數(shù)的單調(diào)性，即可求得結(jié)果.【詳解】根據(jù)題意，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，不妨設(shè)SKIPIF1<0作圖如下：又直線SKIPIF1<0的斜率為SKIPIF1<0，數(shù)形結(jié)合可知，要滿足題意，SKIPIF1<0；且SKIPIF1<0為方程SKIPIF1<0，即SKIPIF1<0的兩根，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0；SKIPIF1<0為方程SKIPIF1<0，即SKIPIF1<0的兩根，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0；則SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，由對(duì)勾函數(shù)單調(diào)性可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，故SKIPIF1<0SKIPIF1<0，即SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題考查函數(shù)與方程；處理問題的關(guān)鍵是能夠數(shù)形結(jié)合求得SKIPIF1<0，SKIPIF1<0和SKIPIF1<0的關(guān)系，從而借助函數(shù)單調(diào)性求值域，屬綜合中檔題.

四、解答題：本大題共6小題，共70分.解答應(yīng)寫出必要的文字說明、證明過程或演算步驟．17．（2023·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處的切線方程；(2)判斷函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，并說明理由.【答案】(1)SKIPIF1<0(2)有兩個(gè)零點(diǎn)，理由見解析【分析】（1）根據(jù)導(dǎo)數(shù)的幾何意義，結(jié)合導(dǎo)數(shù)的運(yùn)算進(jìn)行求解即可；（2）令SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0圖象交點(diǎn)的個(gè)數(shù)，利用導(dǎo)數(shù)得到SKIPIF1<0單調(diào)性，結(jié)合兩個(gè)函數(shù)的圖象判斷可得答案.【詳解】（1）SKIPIF1<0，所以切線斜率為SKIPIF1<0，SKIPIF1<0，所以切點(diǎn)坐標(biāo)為SKIPIF1<0，函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處的切線方程為SKIPIF1<0；（2）有兩個(gè)零點(diǎn)，理由如下，令SKIPIF1<0，可得SKIPIF1<0，判斷函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)即判斷SKIPIF1<0與SKIPIF1<0圖象交點(diǎn)的個(gè)數(shù)，因?yàn)镾KIPIF1<0為單調(diào)遞增函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0無限接近于SKIPIF1<0時(shí)SKIPIF1<0無限接近于SKIPIF1<0，且SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，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SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且當(dāng)SKIPIF1<0無限接近于2時(shí)SKIPIF1<0無限接近于SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0時(shí)有一個(gè)交點(diǎn)，在SKIPIF1<0時(shí)有一個(gè)交點(diǎn)，綜上函數(shù)SKIPIF1<0有2個(gè)零點(diǎn).【點(diǎn)睛】方法點(diǎn)睛：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18．（2023·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)是否存在正整數(shù)m，使得SKIPIF1<0恒成立，若存在求出m的最小值，若不存在說明理由.【答案】(1)函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0.(2)存在正整數(shù)SKIPIF1<0【分析】（1）當(dāng)SKIPIF1<0時(shí)，對(duì)函數(shù)SKIPIF1<0求導(dǎo)，令SKIPIF1<0和SKIPIF1<0，即可求出函數(shù)SKIPIF1<0的單調(diào)區(qū)間；（2）要使SKIPIF1<0恒成立，即SKIPIF1<0恒成立，討論SKIPIF1<0和SKIPIF1<0，求出SKIPIF1<0的單調(diào)性，即可知要使SKIPIF1<0，令SKIPIF1<0，對(duì)SKIPIF1<0求導(dǎo)，得出SKIPIF1<0的單調(diào)性，即可得解.【詳解】（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0；令SKIPIF1<0，解得：SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0.（2）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，無最大值；若SKIPIF1<0，即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.當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最大值，且SKIPIF1<0，要使SKIPIF1<0恒成立，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0趨近于2時(shí)，SKIPIF1<0，SKIPIF1<0，所以存在最小正整數(shù)SKIPIF1<0，使得SKIPIF1<0，即是使得SKIPIF1<0恒成立.19．（2023·四川綿陽·統(tǒng)考模擬預(yù)測）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的極值；(2)設(shè)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0，求SKIPIF1<0及SKIPIF1<0的最大值．【答案】(1)極大值SKIPIF1<0，極小值0(2)SKIPIF1<0，SKIPIF1<0的最大值為0，【分析】（1）由極值的概念求解，（2）根據(jù)SKIPIF1<0的取值分類討論求解SKIPIF1<0的單調(diào)區(qū)間后得SKIPIF1<0，再由導(dǎo)數(shù)判斷單調(diào)性后求解最大值，【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，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，（2）SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，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和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，同理得SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，若SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0綜上，SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0的最大值為0，20．（2023·湖南湘潭·統(tǒng)考二模）已知SKIPIF1<0，曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0．(1)求a，b的值；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．【答案】(1)SKIPIF1<0(2)證明見解析【分析】（1）根據(jù)切點(diǎn)和斜率求得SKIPIF1<0.（2）化簡SKIPIF1<0，利用構(gòu)造函數(shù)法，結(jié)合導(dǎo)數(shù)證得不等式成立.【詳解】（1）由題可知SKIPIF1<0，即SKIPIF1<0．又SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0．（2）SKIPIF1<0，SKIPIF1<0，要證SKIPIF1<0，SKIPIF1<0，只需證SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以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上單調(diào)遞增，則SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．21．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0．【答案】(1)答案見解析；(2)證明見解析.【分析】（1）利用函數(shù)的導(dǎo)數(shù)判斷函數(shù)的單調(diào)性，按照SKIPIF1<0和SKIPIF1<0的大小關(guān)系分類討論；（2）先轉(zhuǎn)化需證明的結(jié)論，構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)的符號(hào)，推得SKIPIF1<0，進(jìn)而證明結(jié)論.【詳解】（1）因?yàn)楹瘮?shù)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減；綜上可得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減；（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，要證SKIPIF1<0，即證SKIPIF1<0，即證SKIPIF1<0，令SKIPIF1<0，SKIPIF