(新高考)高考數(shù)學(xué)二輪復(fù)習(xí)核心考點(diǎn)重難點(diǎn)練習(xí)02《五種導(dǎo)數(shù)及其應(yīng)用中的數(shù)學(xué)思想》(解析版)

重難點(diǎn)02五種導(dǎo)數(shù)及其應(yīng)用中的數(shù)學(xué)思想（核心考點(diǎn)講與練）題型一：函數(shù)與方程思想一、單選題1．（2022·廣西柳州·三模（理））若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用導(dǎo)數(shù)的幾何意義求出切線方程，結(jié)合題設(shè)可得SKIPIF1<0，再根據(jù)目標(biāo)式構(gòu)造SKIPIF1<0，利用導(dǎo)數(shù)求其最大值即可.【詳解】由題設(shè)，SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0遞減；所以SKIPIF1<0，故SKIPIF1<0的最大值SKIPIF1<0.故選：A2．（2022·浙江·寧波市鄞州高級(jí)中學(xué)高三開(kāi)學(xué)考試）已知實(shí)數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】設(shè)SKIPIF1<0是SKIPIF1<0的兩個(gè)零點(diǎn)且SKIPIF1<0，應(yīng)用根與系數(shù)關(guān)系求得SKIPIF1<0，SKIPIF1<0，進(jìn)而代換目標(biāo)式得到以SKIPIF1<0為參數(shù)、SKIPIF1<0為自變量的二次函數(shù)，由二次函數(shù)的性質(zhì)可得SKIPIF1<0，構(gòu)造函數(shù)并應(yīng)用導(dǎo)數(shù)研究單調(diào)性，即可求最大值.【詳解】令SKIPIF1<0是SKIPIF1<0的兩個(gè)零點(diǎn)，由題設(shè)若SKIPIF1<0，由根與系數(shù)關(guān)系有：SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，在SKIPIF1<0上SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，則SKIPIF1<0.綜上，SKIPIF1<0，此時(shí)SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0的最大值SKIPIF1<0.故選：B.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：設(shè)SKIPIF1<0的零點(diǎn)并注意SKIPIF1<0，由根與系數(shù)關(guān)系用零點(diǎn)表示m、n，進(jìn)而轉(zhuǎn)化為以SKIPIF1<0為自變量的二次函數(shù)形式，根據(jù)其開(kāi)口方向及其最值得到不等關(guān)系，最后構(gòu)造函數(shù)并應(yīng)用導(dǎo)數(shù)求不等式中關(guān)于SKIPIF1<0表達(dá)式的值域.二、多選題3．（2022·全國(guó)·高三專題練習(xí)）在平面直角坐標(biāo)系中，我們把橫縱坐標(biāo)相等的點(diǎn)稱之為“完美點(diǎn)”，下列函數(shù)的圖象中存在完美點(diǎn)的是（

）A．y=﹣2x B．y=x﹣6 C．y=SKIPIF1<0 D．y=x2﹣3x+4【答案】AC【分析】橫縱坐標(biāo)相等的函數(shù)即SKIPIF1<0，與SKIPIF1<0有交點(diǎn)即存在完美點(diǎn)，依次計(jì)算即可.【詳解】橫縱坐標(biāo)相等的函數(shù)即SKIPIF1<0，與SKIPIF1<0有交點(diǎn)即存在完美點(diǎn)，對(duì)于A,SKIPIF1<0，解得SKIPIF1<0，即存在完美點(diǎn)SKIPIF1<0，對(duì)于B,SKIPIF1<0，無(wú)解，即不存在完美點(diǎn)，對(duì)于C,SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即存在完美點(diǎn)SKIPIF1<0，SKIPIF1<0對(duì)于D,SKIPIF1<0，SKIPIF1<0,即SKIPIF1<0，解得SKIPIF1<0，即不存在完美點(diǎn)，故選：AC.三、雙空題4．（2022·云南師大附中高三階段練習(xí)（文））如圖，某城市公園內(nèi)有一矩形空地SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，現(xiàn)規(guī)劃在邊SKIPIF1<0上分別取點(diǎn)E，F(xiàn)，G，且滿足SKIPIF1<0，在△SKIPIF1<0內(nèi)建造噴泉瀑布，在△SKIPIF1<0內(nèi)種植花卉，其余區(qū)域鋪設(shè)草坪，并修建棧道SKIPIF1<0作為觀光路線（不考慮寬度），則當(dāng)SKIPIF1<0______時(shí)，棧道SKIPIF1<0最短，此時(shí)SKIPIF1<0_______．【答案】

SKIPIF1<0

SKIPIF1<0【分析】由題設(shè)有SKIPIF1<0△SKIPIF1<0SKIPIF1<0SKIPIF1<0△SKIPIF1<0，設(shè)SKIPIF1<0，根據(jù)圖形中邊角關(guān)系，結(jié)合三角函數(shù)可得SKIPIF1<0SKIPIF1<0，注意SKIPIF1<0的范圍，進(jìn)而應(yīng)用換元法并構(gòu)造函數(shù)，利用導(dǎo)數(shù)求最值.【詳解】由題意，SKIPIF1<0△SKIPIF1<0SKIPIF1<0SKIPIF1<0△SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0．在SKIPIF1<0△SKIPIF1<0中SKIPIF1<0，得SKIPIF1<0SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．由于SKIPIF1<0，解得SKIPIF1<0.令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0遞增；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0遞減；所以SKIPIF1<0，SKIPIF1<0有最大值，則SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0．【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：注意根據(jù)SKIPIF1<0，SKIPIF1<0的長(zhǎng)度判斷SKIPIF1<0對(duì)應(yīng)三角函數(shù)值的范圍.四、解答題5．（2021·全國(guó)·模擬預(yù)測(cè)）SKIPIF1<0．（Ⅰ）若函數(shù)SKIPIF1<0在定義域內(nèi)有兩個(gè)極值點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍；（Ⅱ）若函數(shù)SKIPIF1<0有三個(gè)不相同的零點(diǎn)，求證：SKIPIF1<0．【答案】（Ⅰ）SKIPIF1<0；（Ⅱ）證明見(jiàn)解析.【分析】（Ⅰ）利用導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系及零點(diǎn)的性質(zhì)，即可求解；（Ⅱ）構(gòu)造函數(shù)，利用零點(diǎn)存在性定理及導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系即可得證．【詳解】（Ⅰ）由題得SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0．∵SKIPIF1<0有兩個(gè)極值點(diǎn)，∴SKIPIF1<0在SKIPIF1<0內(nèi)有兩個(gè)零點(diǎn)．設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0的最小值為SKIPIF1<0，∴SKIPIF1<0．（Ⅱ）證明：SKIPIF1<0，設(shè)SKIPIF1<0的兩根為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，依題意有三個(gè)不同的零點(diǎn)，∴SKIPIF1<0，SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，則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，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，根據(jù)零點(diǎn)存在性定理得SKIPIF1<0，使SKIPIF1<0；SKIPIF1<0，使SKIPIF1<0．令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．6．（2021·河南平頂山·高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0在SKIPIF1<0處的切線與直線SKIPIF1<0平行（1）求實(shí)數(shù)SKIPIF1<0的值，并求SKIPIF1<0的極值；（2）若方程SKIPIF1<0有兩個(gè)不相等的實(shí)根SKIPIF1<0，SKIPIF1<0，求證：SKIPIF1<0.【答案】（1）SKIPIF1<0，極小值為SKIPIF1<0；（2）證明見(jiàn)解析.【分析】（1）求出函數(shù)的導(dǎo)數(shù)，利用切線的斜率求出SKIPIF1<0的值，解關(guān)于導(dǎo)函數(shù)的不等式，分析函數(shù)的單調(diào)區(qū)間即可得到極值；（2）令SKIPIF1<0，SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，原題轉(zhuǎn)化為SKIPIF1<0有兩個(gè)實(shí)數(shù)根SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)可得SKIPIF1<0，再構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)可得SKIPIF1<0，利用單調(diào)性，可得SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0即可求證.【詳解】（1）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，由題意知SKIPIF1<0，SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0.SKIPIF1<0的極小值為SKIPIF1<0（2）由（1）知SKIPIF1<0，由SKIPIF1<0得，SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0.SKIPIF1<0，不妨設(shè)SKIPIF1<0令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0則原題轉(zhuǎn)化為SKIPIF1<0有兩個(gè)實(shí)數(shù)根SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得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時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0圖象可知，SKIPIF1<0，SKIPIF1<0.設(shè)SKIPIF1<0，SKIPIF1<0則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.又SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0，得到SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0.7．（2022·全國(guó)·高三專題練習(xí)）某地打算修建一條公路，但設(shè)計(jì)路線正好經(jīng)過(guò)一個(gè)野生動(dòng)物遷徙路線，為了保護(hù)野生動(dòng)物，決定修建高架橋，為野生動(dòng)物的遷徙提供安全通道.若高架橋的兩端及兩端的橋墩已建好，兩端的橋墩相距1200米，余下的工程只需要建兩端橋墩之間的橋面和橋墩.經(jīng)預(yù)測(cè)，一個(gè)橋墩的工程費(fèi)用為500萬(wàn)元，距離為x米的相鄰兩橋墩之間的橋面工程費(fèi)用為SKIPIF1<0萬(wàn)元，假設(shè)橋墩等距離分布，所有橋墩都視為點(diǎn)，且不考慮其它因素，記余下工程的費(fèi)用為y萬(wàn)元.(1)試寫(xiě)出y關(guān)于x的函數(shù)關(guān)系式；(2)需新建多少個(gè)橋墩才能使y最?。坎⑶蟪銎渥钚≈?參考數(shù)據(jù)：SKIPIF1<0，SKIPIF1<0【答案】(1)SKIPIF1<0(2)需新建SKIPIF1<0個(gè)橋墩才能使y最小，最小值為SKIPIF1<0萬(wàn)元.【分析】（1）利用題中的已知條件設(shè)出需要建設(shè)橋墩的個(gè)數(shù)，進(jìn)而表示出工程的費(fèi)用即可；（2）利用（1）的結(jié)果，再利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性即可求出最值.(1)由已知兩端的橋墩相距1200米，且相鄰兩橋墩相距x米，故需要建橋墩SKIPIF1<0個(gè)，則SKIPIF1<0SKIPIF1<0SKIPIF1<0所以y關(guān)于x的函數(shù)關(guān)系式為SKIPIF1<0，SKIPIF1<0(2)由（1）知SKIPIF1<0SKIPIF1<0令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0（舍）或SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞增；所以當(dāng)SKIPIF1<0時(shí)，y有最小值，且SKIPIF1<0又SKIPIF1<0SKIPIF1<0（萬(wàn)元）所以需新建SKIPIF1<0個(gè)橋墩才能使y最小，最小值為SKIPIF1<0萬(wàn)元.題型二：數(shù)形結(jié)合思想一、單選題1．（2022·全國(guó)·高三專題練習(xí)）如圖所示，已知直線SKIPIF1<0與曲線SKIPIF1<0相切于兩點(diǎn)，函數(shù)SKIPIF1<0，則對(duì)函數(shù)SKIPIF1<0描述正確的是（

）A．有極小值點(diǎn)，沒(méi)有極大值點(diǎn) B．有極大值點(diǎn)，沒(méi)有極小值點(diǎn)C．至少有兩個(gè)極小值點(diǎn)和一個(gè)極大值點(diǎn) D．至少有一個(gè)極小值點(diǎn)和兩個(gè)極大值點(diǎn)【答案】C【分析】由題設(shè)SKIPIF1<0，令SKIPIF1<0與SKIPIF1<0切點(diǎn)橫坐標(biāo)為SKIPIF1<0且SKIPIF1<0，由圖存在SKIPIF1<0使SKIPIF1<0，則SKIPIF1<0有三個(gè)不同零點(diǎn)SKIPIF1<0，結(jié)合圖象判斷SKIPIF1<0的符號(hào)，進(jìn)而確定SKIPIF1<0單調(diào)性，即可確定答案.【詳解】由題設(shè)，SKIPIF1<0，則SKIPIF1<0，又直線SKIPIF1<0與曲線SKIPIF1<0相切于兩點(diǎn)且橫坐標(biāo)為SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0，由圖知：存在SKIPIF1<0使SKIPIF1<0，綜上，SKIPIF1<0有三個(gè)不同零點(diǎn)SKIPIF1<0，由圖：SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增，SKIPIF1<0上遞減，SKIPIF1<0上遞增.故SKIPIF1<0至少有兩個(gè)極小值點(diǎn)和一個(gè)極大值點(diǎn).故選：C.2．（2021·河南·西南大學(xué)附中高三期中（文））已知函數(shù)SKIPIF1<0，若存在實(shí)數(shù)SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，滿足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】D【分析】由正弦函數(shù)的性質(zhì)可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，利用導(dǎo)數(shù)得到函數(shù)SKIPIF1<0的單調(diào)性和極值，求出SKIPIF1<0在SKIPIF1<0時(shí)的值域，從而得到SKIPIF1<0的取值范圍．【詳解】由正弦函數(shù)的性質(zhì)可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，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，又SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的取值范圍為SKIPIF1<0，故選：D．3．（2022·全國(guó)·高三專題練習(xí)）如圖，函數(shù)SKIPIF1<0的圖象SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作其切線SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作其切線SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作其切線SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0，則SKIPIF1<0的取值（

）A．與SKIPIF1<0有關(guān)，且存在最大值 B．與SKIPIF1<0有關(guān)，且存在最小值C．與SKIPIF1<0有關(guān)，但無(wú)最值 D．與SKIPIF1<0無(wú)關(guān)，為定值【答案】D【分析】先證明一個(gè)結(jié)論：函數(shù)SKIPIF1<0的圖象SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，SKIPIF1<0.過(guò)點(diǎn)SKIPIF1<0作其切線SKIPIF1<0交于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于另兩個(gè)點(diǎn)SKIPIF1<0，則SKIPIF1<0；利用該結(jié)論即可求出SKIPIF1<0的橫坐標(biāo)關(guān)于SKIPIF1<0的表達(dá)式，進(jìn)而求出直線SKIPIF1<0與SKIPIF1<0的方程，聯(lián)立直線SKIPIF1<0與SKIPIF1<0的方程，即可求出點(diǎn)SKIPIF1<0的橫坐標(biāo)，再根據(jù)SKIPIF1<0，即可求出結(jié)果.【詳解】先證函數(shù)SKIPIF1<0的圖象SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，SKIPIF1<0.過(guò)點(diǎn)SKIPIF1<0作其切線SKIPIF1<0交于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于另兩個(gè)點(diǎn)SKIPIF1<0，則SKIPIF1<0.證明:設(shè)過(guò)點(diǎn)SKIPIF1<0的直線為SKIPIF1<0，聯(lián)立得:SKIPIF1<0，得方程SKIPIF1<0則方程SKIPIF1<0必有一根SKIPIF1<0，于是方程SKIPIF1<0可改寫(xiě)為SKIPIF1<0，其中SKIPIF1<0，當(dāng)SKIPIF1<0與SKIPIF1<0相切于SKIPIF1<0點(diǎn)時(shí)，方程SKIPIF1<0有重根SKIPIF1<0，韋達(dá)定理知SKIPIF1<0；當(dāng)SKIPIF1<0與相交SKIPIF1<0于SKIPIF1<0點(diǎn)時(shí)，方程SKIPIF1<0有另兩個(gè)根SKIPIF1<0，韋達(dá)定理知SKIPIF1<0.故SKIPIF1<0.由于函數(shù)SKIPIF1<0的圖象SKIPIF1<0關(guān)于原點(diǎn)SKIPIF1<0對(duì)稱，設(shè)SKIPIF1<0，連結(jié)SKIPIF1<0，交SKIPIF1<0于另一點(diǎn)SKIPIF1<0，由對(duì)稱性，則SKIPIF1<0，由上述結(jié)論，則SKIPIF1<0，所以SKIPIF1<0；設(shè)SKIPIF1<0，連結(jié)SKIPIF1<0交SKIPIF1<0于另一點(diǎn)SKIPIF1<0由對(duì)稱性，則SKIPIF1<0，由上述結(jié)論，則SKIPIF1<0，所以SKIPIF1<0.于是直線SKIPIF1<0為SKIPIF1<0，直線SKIPIF1<0為SKIPIF1<0，聯(lián)立得:SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0的取值與SKIPIF1<0無(wú)關(guān)，為定值SKIPIF1<0.故選：D.二、多選題4．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，下列說(shuō)法正確的有（

）A．函數(shù)SKIPIF1<0是周期函數(shù) B．函數(shù)SKIPIF1<0有唯一零點(diǎn)C．函數(shù)SKIPIF1<0有無(wú)數(shù)個(gè)極值點(diǎn) D．函數(shù)SKIPIF1<0在SKIPIF1<0上不是單調(diào)函數(shù)【答案】CD【分析】根據(jù)SKIPIF1<0不是周期函數(shù)，從而可判斷選項(xiàng)A錯(cuò)誤；令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，作出SKIPIF1<0與SKIPIF1<0的圖象，由圖象可判斷選項(xiàng)B；作出SKIPIF1<0與SKIPIF1<0的圖象，由圖可判斷選項(xiàng)C；通過(guò)圖象可判斷SKIPIF1<0在SKIPIF1<0不單調(diào)，從而可判斷選項(xiàng)D.【詳解】SKIPIF1<0，因?yàn)镾KIPIF1<0不是周期函數(shù)，則SKIPIF1<0不是周期函數(shù)，A錯(cuò)；令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，作出SKIPIF1<0與SKIPIF1<0的圖象，由圖可知，SKIPIF1<0與SKIPIF1<0的圖象至少有兩個(gè)交點(diǎn)，SKIPIF1<0至少有兩個(gè)零點(diǎn)，SKIPIF1<0至少有兩個(gè)零點(diǎn)，B錯(cuò)誤；作出SKIPIF1<0與SKIPIF1<0的圖象，由圖可知，SKIPIF1<0有無(wú)數(shù)個(gè)零點(diǎn)SKIPIF1<0有無(wú)數(shù)個(gè)極值點(diǎn)，即SKIPIF1<0有無(wú)數(shù)個(gè)極值點(diǎn)，C正確；因?yàn)镾KIPIF1<0在SKIPIF1<0有零點(diǎn)，所以SKIPIF1<0在SKIPIF1<0不單調(diào)，SKIPIF1<0在SKIPIF1<0不單調(diào)，D正確；故選：CD.三、雙空題5．（2022·湖南·長(zhǎng)沙一中高三階段練習(xí)）已知函數(shù)SKIPIF1<0，則方程SKIPIF1<0的根為_(kāi)_______．若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是________．【答案】

SKIPIF1<0或2；

SKIPIF1<0.【分析】（1）當(dāng)SKIPIF1<0時(shí)，運(yùn)用導(dǎo)數(shù)求得函數(shù)單調(diào)區(qū)間，可得SKIPIF1<0，可得一根，當(dāng)SKIPIF1<0時(shí)，直接求解可得.（2）先運(yùn)用導(dǎo)數(shù)求得函數(shù)單調(diào)區(qū)間，并作出函數(shù)的圖象，再根據(jù)圖象列出函數(shù)有3個(gè)零點(diǎn)所需要的條件，即可求得結(jié)果.【詳解】解：（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，并且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有唯一根SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0（舍去）或2，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的根為2，綜上，SKIPIF1<0根為SKIPIF1<0或2；（2）因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由（1）SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，且僅當(dāng)SKIPIF1<0，且SKIPIF1<0，因?yàn)楫?dāng)SKIPIF1<0時(shí)，則有SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，由圖象得，要使函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，且SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0或SKIPIF1<0解得實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0故答案是：SKIPIF1<0或2；SKIPIF1<0.6．（2022·廣東·金山中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0則函數(shù)SKIPIF1<0的最小值為_(kāi)_______；若關(guān)于SKIPIF1<0的方程SKIPIF1<0有四個(gè)不同的實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值范圍是________．【答案】

SKIPIF1<0

SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)求出函數(shù)在每段上的最小值，比較大小即可；求出過(guò)點(diǎn)SKIPIF1<0且與SKIPIF1<0相切的直線的斜率，再由數(shù)形結(jié)合可得出SKIPIF1<0與SKIPIF1<0有4個(gè)交點(diǎn)時(shí)的斜率取值范圍，即可得解.【詳解】令SKIPIF1<0，SKIPIF1<0表示過(guò)定點(diǎn)SKIPIF1<0，斜率為SKIPIF1<0的動(dòng)直線，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0，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，故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，SKIPIF1<0，在同一坐標(biāo)系內(nèi)作出函數(shù)SKIPIF1<0圖象與直線SKIPIF1<0，如圖所示，關(guān)于SKIPIF1<0的方程SKIPIF1<0有四個(gè)不同的實(shí)根，等價(jià)于函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有四個(gè)不同的交點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處切線斜率為SKIPIF1<0，該切線過(guò)點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0滿足SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0的切線斜SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線斜率為SKIPIF1<0，該切線過(guò)點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0的切線斜率為2，由函數(shù)圖象知，當(dāng)動(dòng)直線SKIPIF1<0在直線SKIPIF1<0與SKIPIF1<0所夾不含SKIPIF1<0軸的對(duì)頂角區(qū)域內(nèi)轉(zhuǎn)動(dòng)（不含邊界直線）時(shí)，函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有四個(gè)不同的交點(diǎn)，此時(shí)SKIPIF1<0的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0；SKIPIF1<0四、填空題7．（2022·河南·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0在區(qū)間SKIPIF1<0上恰有四個(gè)不同的實(shí)數(shù)根，則實(shí)數(shù)SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【分析】將問(wèn)題轉(zhuǎn)化為SKIPIF1<0在區(qū)間SKIPIF1<0上恰有四個(gè)不同的實(shí)數(shù)根，進(jìn)而設(shè)SKIPIF1<0，然后先通過(guò)導(dǎo)數(shù)的方法探討函數(shù)SKIPIF1<0的圖象和性質(zhì)，再討論關(guān)于t的方程SKIPIF1<0的根的分布，最后求得答案.【詳解】問(wèn)題SKIPIF1<0SKIPIF1<0即SKIPIF1<0在區(qū)間SKIPIF1<0上恰有四個(gè)不同的實(shí)數(shù)根.設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞增，SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0且SKIPIF1<0.如示意圖：

由圖可知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有2個(gè)零點(diǎn)，于是問(wèn)題SKIPIF1<0關(guān)于t的的方程SKIPIF1<0即SKIPIF1<0在SKIPIF1<0上有2個(gè)不等實(shí)根.設(shè)SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0，易知SKIPIF1<0.于是，SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】本題較難，首先直接處理較為麻煩，因此對(duì)原方程進(jìn)行恒等變形，進(jìn)而采用“換元法”降低試題的難度.另外，我們經(jīng)常采用“數(shù)形結(jié)合法”進(jìn)行輔助解題，這樣更加形象和直觀.題型三：分類與整合思想一、多選題1．（2022·重慶南開(kāi)中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，其中常數(shù)SKIPIF1<0，SKIPIF1<0，則下列說(shuō)法正確的有（

）A．函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0B．當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)C．不存在實(shí)數(shù)SKIPIF1<0和m，使得函數(shù)SKIPIF1<0恰好只有一個(gè)極值點(diǎn)D．若SKIPIF1<0，則“SKIPIF1<0”是“函數(shù)SKIPIF1<0是增函數(shù)”的充分不必要條件【答案】BC【分析】A判斷SKIPIF1<0時(shí)的定義域情況即可；B利用導(dǎo)數(shù)研究SKIPIF1<0的單調(diào)性，判斷是否有兩個(gè)變號(hào)零點(diǎn)即可；C、D對(duì)SKIPIF1<0求導(dǎo)，構(gòu)造SKIPIF1<0結(jié)合二次函數(shù)性質(zhì)討論SKIPIF1<0和m，應(yīng)用零點(diǎn)存在性定理判斷SKIPIF1<0變號(hào)零點(diǎn)的個(gè)數(shù)，進(jìn)而判斷SKIPIF1<0極值點(diǎn)個(gè)數(shù)及單調(diào)性.【詳解】A：當(dāng)SKIPIF1<0時(shí)定義域?yàn)镾KIPIF1<0，錯(cuò)誤；B：SKIPIF1<0且定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上各有一個(gè)變號(hào)零點(diǎn)，則SKIPIF1<0有兩個(gè)極值點(diǎn)，正確；C：SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則圖象開(kāi)口向上，對(duì)稱軸SKIPIF1<0且SKIPIF1<0，要使SKIPIF1<0有極值點(diǎn)，SKIPIF1<0必有變號(hào)零點(diǎn)，則SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0定義域?yàn)镾KIPIF1<0，又SKIPIF1<0，此時(shí)SKIPIF1<0則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上遞增，又SKIPIF1<0，即SKIPIF1<0，無(wú)極值點(diǎn)；此時(shí)SKIPIF1<0則SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0遞減，SKIPIF1<0遞增，故SKIPIF1<0、SKIPIF1<0各有一個(gè)零點(diǎn)，即SKIPIF1<0有兩個(gè)變號(hào)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上遞增，又SKIPIF1<0，即SKIPIF1<0，無(wú)極值點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0SKIPIF1<0上遞減，SKIPIF1<0遞增，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0趨向正無(wú)窮SKIPIF1<0趨于正無(wú)窮，故SKIPIF1<0在SKIPIF1<0、SKIPIF1<0各有一個(gè)變號(hào)零點(diǎn)，即SKIPIF1<0有兩個(gè)變號(hào)零點(diǎn)；此時(shí)SKIPIF1<0則SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0遞增，又SKIPIF1<0，即SKIPIF1<0，無(wú)極值點(diǎn)；綜上，不存在實(shí)數(shù)SKIPIF1<0和m，使得函數(shù)SKIPIF1<0恰好只有一個(gè)極值點(diǎn)，正確；D：結(jié)合C分析：當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)有SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恒正，即SKIPIF1<0，此時(shí)SKIPIF1<0是增函數(shù)；當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)有SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0，SKIPIF1<0各有一個(gè)零點(diǎn)，易得SKIPIF1<0有兩個(gè)變號(hào)零點(diǎn)，此時(shí)SKIPIF1<0不單調(diào)，命題的充分性不成立，錯(cuò)誤.故選：BC【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：C、D首先對(duì)SKIPIF1<0求導(dǎo)，構(gòu)造SKIPIF1<0，結(jié)合二次函數(shù)性質(zhì)討論參數(shù)判斷SKIPIF1<0變號(hào)零點(diǎn)的個(gè)數(shù)及SKIPIF1<0單調(diào)性.二、解答題2．（2022·四川南充·三模（理））已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，求證：對(duì)于任意SKIPIF1<0，函數(shù)SKIPIF1<0有唯一零點(diǎn)．【分析】（1）求導(dǎo)，通過(guò)討論SKIPIF1<0的范圍研究導(dǎo)函數(shù)的符號(hào)變化，進(jìn)而研究函數(shù)的單調(diào)區(qū)間；（2）求導(dǎo)，構(gòu)造函數(shù)SKIPIF1<0，再次求導(dǎo)研究.SKIPIF1<0的單調(diào)性，再利用放縮法進(jìn)行轉(zhuǎn)化求證.(1)解：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；②當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0、SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增；綜上所述：①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增；③SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0、SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增；(2)解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0．則SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減．所以SKIPIF1<0所以SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減．當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0存在唯一SKIPIF1<0，使得函數(shù)SKIPIF1<0．所以對(duì)于任意SKIPIF1<0，函數(shù)SKIPIF1<0有唯一零點(diǎn)．3．（2022·云南·二模（文））已知e是自然對(duì)數(shù)的底數(shù)，SKIPIF1<0，常數(shù)a是實(shí)數(shù).(1)設(shè)SKIPIF1<0，求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)SKIPIF1<0，都有SKIPIF1<0，求a的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）先求出SKIPIF1<0，再求導(dǎo)得到SKIPIF1<0，即可求出切線方程；（2）令SKIPIF1<0，求導(dǎo)后令SKIPIF1<0，通過(guò)SKIPIF1<0得到SKIPIF1<0在SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，符合題意，當(dāng)SKIPIF1<0時(shí)，說(shuō)明SKIPIF1<0，使SKIPIF1<0，不合題意，即可求解.(1)設(shè)SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程頭SKIPIF1<0，即SKIPIF1<0．∴曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0．(2)設(shè)SKIPIF1<0,則SKIPIF1<0．設(shè)SKIPIF1<0，則SKIPIF1<0．∴函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．∴SKIPIF1<