2025年高考數(shù)學(xué)一輪復(fù)習(xí) 講練測第06講 函數(shù)的圖象（九大題型）（練習(xí)）（含解析）

第06講函數(shù)的圖象目錄TOC\o"1-2"\h\z\u模擬基礎(chǔ)練 2題型一：由解析式選圖（識圖） 2題型二：由圖象選表達(dá)式 4題型三：表達(dá)式含參數(shù)的圖象問題 6題型四：函數(shù)圖象應(yīng)用題 10題型五：函數(shù)圖象的變換 12題型六：利用函數(shù)的圖像研究函數(shù)的性質(zhì)、最值 14題型七：利用函數(shù)的圖像解不等式 17題型八：利用函數(shù)的圖像求恒成立問題 18題型九：利用函數(shù)的圖像判斷零點(diǎn)的個(gè)數(shù) 20重難創(chuàng)新練 22真題實(shí)戰(zhàn)練 32題型一：由解析式選圖（識圖）1．（2024·全國·模擬預(yù)測）函數(shù)SKIPIF1<0的大致圖象為（

）A． B．C． D．【答案】C【解析】由題可知，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0是偶函數(shù)，排除A，B，又SKIPIF1<0，排除D，故選：C．2．（2024·全國·模擬預(yù)測）函數(shù)SKIPIF1<0的部分圖象為（

）A． B．C． D．【答案】B【解析】由題意可知：SKIPIF1<0的定義域?yàn)镽，關(guān)于原點(diǎn)對稱，且SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，其圖象關(guān)于原點(diǎn)對稱，排除A；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除D；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除C．故選：B．3．（2024·全國·模擬預(yù)測）函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B．C． D．【答案】A【解析】依題意得SKIPIF1<0，函數(shù)的定義域?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0為偶函數(shù)，圖象關(guān)于y軸對稱，排除B,D兩項(xiàng)，又SKIPIF1<0，排除C項(xiàng)，所以只有A選項(xiàng)符合.故選:A．4．（2024·河北保定·二模）函數(shù)SKIPIF1<0的部分圖象大致為（

）A．

B．

C．

D．

【答案】A【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，設(shè)SKIPIF1<0SKIPIF1<0，可知SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0為奇函數(shù)，則B，C錯(cuò)誤，易知SKIPIF1<0，所以A正確，D錯(cuò)誤．故選：A.題型二：由圖象選表達(dá)式5．（2024·天津河?xùn)|·一模）如圖中，圖象對應(yīng)的函數(shù)解析式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由圖象可知函數(shù)關(guān)于原點(diǎn)對稱，故為奇函數(shù)，對于A，SKIPIF1<0，故函數(shù)為偶函數(shù)，不符合，對于B,SKIPIF1<0，根據(jù)圖象可知，4處的函數(shù)值不超過5，故B不符合，對于C，由于SKIPIF1<0，顯然不符合，故選：D6．（2024·陜西西安·二模）已知函數(shù)SKIPIF1<0的圖象如圖所示，則函數(shù)SKIPIF1<0的解析式可能為（

）

A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】對于A，函數(shù)SKIPIF1<0的定義域?yàn)镽，而題設(shè)函數(shù)的圖象中在自變量為0時(shí)無意義，不符合題意，排除；對于C，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不符合圖象，排除；對于D，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不符合圖象，排除.故選：B7．（2024·廣東廣州·一模）已知函數(shù)SKIPIF1<0的部分圖像如圖所示，則SKIPIF1<0的解析式可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】觀察圖象可知函數(shù)為偶函數(shù)，對于A，SKIPIF1<0，為奇函數(shù)，排除；對于B，SKIPIF1<0，為奇函數(shù)，排除；同理，C、D選項(xiàng)為偶函數(shù)，而對于C項(xiàng)，其定義域?yàn)镾KIPIF1<0，不是R，舍去，故D正確.故選：D8．（2024·黑龍江哈爾濱·模擬預(yù)測）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0的解析式可能為（

）

A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】對于B，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，不滿足圖象，故B錯(cuò)誤;對于C，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，而SKIPIF1<0，關(guān)于SKIPIF1<0軸對稱，故C錯(cuò)誤;對于D,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由反比例函數(shù)的性質(zhì)可知SKIPIF1<0在SKIPIF1<0單調(diào)遞減，故D錯(cuò)誤;利用排除法可以得到，SKIPIF1<0在滿足題意，A正確.故選：A題型三：表達(dá)式含參數(shù)的圖象問題9．（多選題）函數(shù)SKIPIF1<0的圖象可能是（

）A．

B．

C．

D．

【答案】ABC【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則選項(xiàng)C符合；當(dāng)SKIPIF1<0，故排除D；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號，由于SKIPIF1<0在SKIPIF1<0為減函數(shù)，SKIPIF1<0為增函數(shù)，則函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，在SKIPIF1<0為減函數(shù)，SKIPIF1<0是奇函數(shù)，則奇偶性可得SKIPIF1<0在SKIPIF1<0上的單調(diào)性，故選項(xiàng)B符合；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，由于SKIPIF1<0在SKIPIF1<0，SKIPIF1<0為增函數(shù)，則SKIPIF1<0在SKIPIF1<0，SKIPIF1<0為減函數(shù)，SKIPIF1<0是奇函數(shù)，則由奇偶性可得SKIPIF1<0在SKIPIF1<0上的單調(diào)性，故A符合.故選：ABC.10．（多選題）（2024·高三·河北衡水·開學(xué)考試）已知SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象可能是（

）A．

B．

C．

D．

【答案】AD【解析】由于當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除B，C，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)圖象對應(yīng)的圖形可能為A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)圖象對應(yīng)的的圖形可能為D.故選：AD.11．（多選題）對數(shù)函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）與二次函數(shù)SKIPIF1<0在同一坐標(biāo)系內(nèi)的圖象不可能是（

）A．

B．

C．

D．

【答案】BCD【解析】選項(xiàng)A，B中，由對數(shù)函數(shù)圖象得SKIPIF1<0，則二次函數(shù)中二次項(xiàng)系數(shù)SKIPIF1<0，其對應(yīng)方程的兩個(gè)根為0，SKIPIF1<0，選項(xiàng)A中，由圖象得SKIPIF1<0，從而SKIPIF1<0，選項(xiàng)A可能；選項(xiàng)B中，由圖象得SKIPIF1<0，與SKIPIF1<0相矛盾，選項(xiàng)B不可能．選項(xiàng)C，D中，由對數(shù)函數(shù)的圖象得SKIPIF1<0，則SKIPIF1<0，二次函數(shù)圖象開口向下，D不可能；選項(xiàng)C中，由圖象與x軸的交點(diǎn)的位置得SKIPIF1<0，與SKIPIF1<0相矛盾，選項(xiàng)C不可能．故選：BCD．12．（多選題）函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象可能為（

）A． B．C． D．【答案】BCD【解析】SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，函數(shù)SKIPIF1<0最多有兩個(gè)零點(diǎn)，故A錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0顯然為偶函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0單調(diào)遞增，單調(diào)性結(jié)合奇偶性可知，B選項(xiàng)正確；當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0或SKIPIF1<0，記SKIPIF1<0，則SKIPIF1<0因?yàn)镾KIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0單調(diào)遞增又SKIPIF1<0，SKIPIF1<0，所以存在SKIPIF1<0使得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以，當(dāng)SKIPIF1<0時(shí)，可知圖象如選項(xiàng)C，故C選項(xiàng)正確；當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0的圖象如D選項(xiàng)，故D選項(xiàng)正確；故選：BCD題型四：函數(shù)圖象應(yīng)用題13．（2024·海南省直轄縣級單位·三模）小李在如圖所示的跑道（其中左、右兩邊分別是兩個(gè)半圓）上勻速跑步，他從點(diǎn)SKIPIF1<0處出發(fā)，沿箭頭方向經(jīng)過點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0返回到點(diǎn)SKIPIF1<0，共用時(shí)SKIPIF1<0秒，他的同桌小陳在固定點(diǎn)SKIPIF1<0位置觀察小李跑步的過程，設(shè)小李跑步的時(shí)間為SKIPIF1<0（單位：秒），他與同桌小陳間的距離為SKIPIF1<0（單位：米），若SKIPIF1<0，則SKIPIF1<0的圖象大致為（

）

A．

B．

C．

D．

【答案】D【解析】由題圖知，小李從點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的過程中，SKIPIF1<0的值先增后減，從點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的過程中，SKIPIF1<0的值先減后增，從點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的過程中，SKIPIF1<0的值先增后減，從點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的過程中，SKIPIF1<0的值先減后增，所以，在整個(gè)運(yùn)動過程中，小李和小陳之間的距離（即SKIPIF1<0的值）的增減性為：增、減、增、減、增，D選項(xiàng)合乎題意，故選：D.14．某天0時(shí)，小鵬同學(xué)生病了，體溫上升，吃過藥后感覺好多了，中午時(shí)他的體溫基本正常（正常體溫為37℃），但是下午他的體溫又開始上升，直到半夜才感覺身上不那么發(fā)燙了.下面能大致反映出小鵬這一天（0時(shí)至24時(shí)）體溫變化情況的圖像是（）A．

B．

C．

D．

【答案】C【解析】選項(xiàng)A反映，體溫逐漸降低，不符合題意；選項(xiàng)B不能反映下午體溫又開始上升的過程；選項(xiàng)D不能反映下午他的體溫又開始上升，直到半夜才感覺身上不那么發(fā)燙這一過程.故選：C15．如圖，點(diǎn)SKIPIF1<0在邊長為1的正方形SKIPIF1<0上運(yùn)動，設(shè)點(diǎn)SKIPIF1<0為SKIPIF1<0的中點(diǎn)，當(dāng)點(diǎn)SKIPIF1<0沿SKIPIF1<0運(yùn)動時(shí)，點(diǎn)SKIPIF1<0經(jīng)過的路程設(shè)為SKIPIF1<0，SKIPIF1<0面積設(shè)為SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象只可能是下圖中的（

）

A．

B．

C．

D．

【答案】A【解析】當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)：SKIPIF1<0；當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)：SKIPIF1<0SKIPIF1<0；當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)：SKIPIF1<0，所以SKIPIF1<0，由函數(shù)解析式可知，有三段線段，又當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)是減函數(shù)，故符合題意的為A.故選：A16．“龜兔賽跑”講述了這樣的故事：領(lǐng)先的兔子看著慢慢爬行的烏龜，驕傲起來，睡了一覺，當(dāng)它醒來時(shí)，發(fā)現(xiàn)烏龜快到終點(diǎn)了，于是急忙追趕，但為時(shí)已晚，烏龜還是先到達(dá)了終點(diǎn).用s1，s2分別表示烏龜和兔子經(jīng)過的路程，t為時(shí)間，則與故事情節(jié)相吻合的是（

）A． B．C． D．【答案】B【解析】由題意可得SKIPIF1<0始終是勻速增長,開始時(shí),SKIPIF1<0的增長比較快,但中間有一段時(shí)間SKIPIF1<0停止增長，在最后一段時(shí)間里,SKIPIF1<0的增長又較快,但SKIPIF1<0的值沒有超過SKIPIF1<0的值,結(jié)合所給的圖象可知，B選項(xiàng)正確;故選：B.題型五：函數(shù)圖象的變換17．函數(shù)SKIPIF1<0的圖象向右平移1個(gè)單位長度，所得圖象與SKIPIF1<0關(guān)于y軸對稱，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0關(guān)于SKIPIF1<0軸對稱的解析式為SKIPIF1<0，把SKIPIF1<0的圖象向左平移1個(gè)單位長度得出SKIPIF1<0，SKIPIF1<0，故選：D．18．若函數(shù)SKIPIF1<0的圖象如下圖所示，函數(shù)SKIPIF1<0的圖象為（

）

A．

B．

C．

D．

【答案】C【解析】函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0對稱可得函數(shù)SKIPIF1<0的圖象，再向右平移2個(gè)單位得函數(shù)SKIPIF1<0，即SKIPIF1<0的圖象.故選：C.19．把函數(shù)SKIPIF1<0的圖象按向量SKIPIF1<0平移，得到SKIPIF1<0的圖象，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】把函數(shù)SKIPIF1<0的圖象按向量SKIPIF1<0平移，即向右平移2個(gè)單位，再向上平移3個(gè)單位，平移后得到SKIPIF1<0的圖象，所以SKIPIF1<0，故選：SKIPIF1<0.20．將函數(shù)SKIPIF1<0向左、向下分別平移2個(gè)、3個(gè)單位長度，所得圖像為（

）A．

B．

C．

D．

【答案】C【解析】因?yàn)镾KIPIF1<0，可得函數(shù)的大致圖像如圖所示，將其向左、向下分別平移2個(gè)、3個(gè)單位長度，所得函數(shù)圖像為C選項(xiàng)中的圖像.故選：C21．要得到函數(shù)SKIPIF1<0的圖象，只需將指數(shù)函數(shù)SKIPIF1<0的圖象（

）A．向左平移SKIPIF1<0個(gè)單位 B．向右平移SKIPIF1<0個(gè)單位C．向左平移SKIPIF1<0個(gè)單位 D．向右平移SKIPIF1<0個(gè)單位【答案】D【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以，為了得到函數(shù)SKIPIF1<0的圖象，只需將指數(shù)函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位，故選：D.題型六：利用函數(shù)的圖像研究函數(shù)的性質(zhì)、最值22．記實(shí)數(shù)SKIPIF1<0，SKIPIF1<0中的最小值為SKIPIF1<0，例如SKIPIF1<0，當(dāng)SKIPIF1<0取任意實(shí)數(shù)時(shí)，則SKIPIF1<0的最大值為（

）A．5 B．4 C．3 D．2【答案】C【解析】畫出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象如圖：由圖可知：SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，可得當(dāng)SKIPIF1<0時(shí)，函數(shù)有最大值，最大值為3.故選：C.23．定義SKIPIF1<0為SKIPIF1<0中的最小值，設(shè)SKIPIF1<0，則SKIPIF1<0的最大值是．【答案】2【解析】將三個(gè)解析式的圖像作在同一坐標(biāo)系下，則SKIPIF1<0為三段函數(shù)圖像中靠下的部分，從而通過數(shù)形結(jié)合可得SKIPIF1<0的最大值點(diǎn)為SKIPIF1<0與SKIPIF1<0在第一象限的交點(diǎn)，即SKIPIF1<0，所以SKIPIF1<0．故答案為：2.24．定義一種運(yùn)算SKIPIF1<0，設(shè)SKIPIF1<0（t為常數(shù)），且SKIPIF1<0，則使函數(shù)SKIPIF1<0最大值為4的t值是．【答案】SKIPIF1<0【解析】若SKIPIF1<0在SKIPIF1<0上的最大值為4，所以由SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以要使函數(shù)SKIPIF1<0最大值為4，則根據(jù)新定義，結(jié)合SKIPIF1<0與SKIPIF1<0圖像可知，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)解得SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)解得SKIPIF1<0，故SKIPIF1<0或4，故答案為：SKIPIF1<0或4.25．已知函數(shù)SKIPIF1<0，SKIPIF1<0，對SKIPIF1<0，用SKIPIF1<0表示SKIPIF1<0，SKIPIF1<0中的較大者，記為SKIPIF1<0，則SKIPIF1<0的最小值為.【答案】SKIPIF1<0【解析】如圖，在同一直角坐標(biāo)系中分別作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，因?yàn)閷KIPIF1<0，SKIPIF1<0，故函數(shù)SKIPIF1<0的圖象如圖所示：由圖可知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最小值SKIPIF1<0.故答案為：SKIPIF1<0.題型七：利用函數(shù)的圖像解不等式26．如圖為函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，則不等式SKIPIF1<0的解集為（）

A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由圖象可得當(dāng)SKIPIF1<0，此時(shí)需滿足SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0，此時(shí)需滿足SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0.綜上所述，SKIPIF1<0.故選：D.27．（2024·北京平谷·模擬預(yù)測）已知函數(shù)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】B【解析】不等式SKIPIF1<0，分別畫出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，由圖象可知SKIPIF1<0和SKIPIF1<0有兩個(gè)交點(diǎn)，分別是SKIPIF1<0和SKIPIF1<0，由圖象可知SKIPIF1<0的解集是SKIPIF1<0即不等式SKIPIF1<0的解集是SKIPIF1<0.故選：B28．已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0有個(gè)零點(diǎn)；不等式SKIPIF1<0的解集為【答案】2SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0與SKIPIF1<0交點(diǎn)個(gè)數(shù)，即為SKIPIF1<0零點(diǎn)個(gè)數(shù)，由SKIPIF1<0在定義域上均遞增，且都過SKIPIF1<0，圖象如圖所示，所以兩函數(shù)有且僅有2個(gè)交點(diǎn)，故SKIPIF1<0有2個(gè)零點(diǎn)，由SKIPIF1<0，得SKIPIF1<0，由上圖知SKIPIF1<0．故答案為：SKIPIF1<0；SKIPIF1<0.題型八：利用函數(shù)的圖像求恒成立問題29．當(dāng)SKIPIF1<0,不等式SKIPIF1<0成立,則實(shí)數(shù)k的取值范圍是.【答案】SKIPIF1<0【解析】設(shè)SKIPIF1<0,畫出這兩個(gè)函數(shù)圖象,如圖所示,觀察圖象可知,當(dāng)直線SKIPIF1<0經(jīng)過函數(shù)SKIPIF1<0的最高點(diǎn)(1,1)和最低點(diǎn)(0,0)時(shí),k取得最大值,所以SKIPIF1<0.30．已知函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則非零實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】在同一坐標(biāo)系內(nèi)作出SKIPIF1<0與SKIPIF1<0的圖象，當(dāng)射線SKIPIF1<0與曲線SKIPIF1<0相切時(shí)，即方程SKIPIF1<0時(shí)，由SKIPIF1<0，解得SKIPIF1<0，結(jié)合圖象可得SKIPIF1<0時(shí)，SKIPIF1<0，所以a的的取值范圍是SKIPIF1<0，故選：B31．定義在SKIPIF1<0上函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則使得SKIPIF1<0在SKIPIF1<0上恒成立的SKIPIF1<0的最小值是．【答案】SKIPIF1<0【解析】由題設(shè)知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，同理：在SKIPIF1<0上，SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.函數(shù)SKIPIF1<0的圖象，如下圖示：在SKIPIF1<0上，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.由圖象知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故答案為：SKIPIF1<0.題型九：利用函數(shù)的圖像判斷零點(diǎn)的個(gè)數(shù)32．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】令SKIPIF1<0，故SKIPIF1<0，畫出SKIPIF1<0與SKIPIF1<0的圖象，函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，即SKIPIF1<0與SKIPIF1<0圖象有3個(gè)不同的交點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0.故選：D33．已知函數(shù)SKIPIF1<0，若存在SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0兩兩不相等，則SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】畫出函數(shù)SKIPIF1<0的圖象，如圖所示：設(shè)SKIPIF1<0，則方程SKIPIF1<0有3個(gè)根，根據(jù)圖可得SKIPIF1<0，不妨設(shè)SKIPIF1<0與SKIPIF1<0的兩個(gè)交點(diǎn)的橫坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0與SKIPIF1<0交點(diǎn)的橫坐標(biāo)為SKIPIF1<0則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最大，由SKIPIF1<0，解得SKIPIF1<0當(dāng)m接近SKIPIF1<0時(shí)，SKIPIF1<0接近最小，由SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0SKIPIF1<0的取值范圍是SKIPIF1<0故選：C.34．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．35．已知函數(shù)SKIPIF1<0，SKIPIF1<0．若SKIPIF1<0有2個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，令SKIPIF1<0可得SKIPIF1<0，作出函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象如圖所示：由上圖可知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有2個(gè)交點(diǎn)，此時(shí)，函數(shù)SKIPIF1<0有2個(gè)零點(diǎn)．因此，實(shí)數(shù)a的取值范圍是SKIPIF1<0．故選：D．1．（2024·內(nèi)蒙古呼和浩特·二模）函數(shù)SKIPIF1<0的部分圖象大致如圖所示，則SKIPIF1<0的解析式可能為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由圖可知，SKIPIF1<0的圖象關(guān)于原點(diǎn)對稱，則SKIPIF1<0為奇函數(shù)，且SKIPIF1<0，在SKIPIF1<0上先增后減.A：SKIPIF1<0，函數(shù)的定義域?yàn)镽，SKIPIF1<0，故A符合題意；B：SKIPIF1<0，函數(shù)的定義域?yàn)镽，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故B不符合題意；C：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)顯然沒有意義，故C不符合題意；D：SKIPIF1<0，函數(shù)的定義域?yàn)镽，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故D不符合題意.故選：A2．（2024·浙江溫州·三模）已知函數(shù)SKIPIF1<0，則關(guān)于SKIPIF1<0方程SKIPIF1<0的根個(gè)數(shù)不可能是（

）A．0個(gè) B．1個(gè) C．2個(gè) D．3個(gè)【答案】C【解析】作出函數(shù)SKIPIF1<0的圖象，如圖所示：將原問題轉(zhuǎn)化為直線SKIPIF1<0（過定點(diǎn)SKIPIF1<0）與函數(shù)SKIPIF1<0的圖象交點(diǎn)的個(gè)數(shù)，由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象只有一個(gè)交點(diǎn)；當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象沒有交點(diǎn)；當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有三個(gè)交點(diǎn)；所以直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象不可能有兩個(gè)交點(diǎn)．故選：C．3．（2024·全國·模擬預(yù)測）函數(shù)SKIPIF1<0的大致圖象是（

）A．

B．

C．

D．

【答案】A【解析】由題意得SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0；又SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，其圖象關(guān)于原點(diǎn)對稱，排除B，C；又SKIPIF1<0，所以排除D．故選：A．4．（2024·湖南邵陽·模擬預(yù)測）函數(shù)SKIPIF1<0的大致圖象為（

）A． B．C． D．【答案】A【解析】依題意，SKIPIF1<0，SKIPIF1<0恒成立，即函數(shù)SKIPIF1<0的定義域?yàn)镽，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，BC不滿足；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，則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，即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，SKIPIF1<0，D不滿足，A滿足.故選：A5．（2024·四川成都·三模）函數(shù)SKIPIF1<0的圖象大致是（

）A． B．C． D．【答案】A【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0是奇函數(shù)，圖象關(guān)于原點(diǎn)對稱，BD不滿足；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，C不滿足，A滿足.故選：A6．（2024·四川成都·模擬預(yù)測）華羅庚是享譽(yù)世界的數(shù)學(xué)大師，國際上以華氏命名的數(shù)學(xué)科研成果有“華氏定理”“華氏不等式”“華氏算子”“華—王方法”等，其斐然成績早為世人所推崇．他曾說：“數(shù)缺形時(shí)少直觀，形缺數(shù)時(shí)難入微”，告知我們把“數(shù)”與“形”，“式”與“圖”結(jié)合起來是解決數(shù)學(xué)問題的有效途徑．在數(shù)學(xué)的學(xué)習(xí)和研究中，常用函數(shù)的圖象來研究函數(shù)的性質(zhì)，也常用函數(shù)的解析式來分析函數(shù)圖象的特征．已知函數(shù)SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的解析式可能是（

）

A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由函數(shù)圖象可知，SKIPIF1<0的圖象不關(guān)SKIPIF1<0軸對稱,而SKIPIF1<0，SKIPIF1<0，即這兩個(gè)函數(shù)均關(guān)于SKIPIF1<0軸對稱，則排除選項(xiàng)SKIPIF1<0、SKIPIF1<0；由指數(shù)函數(shù)的性質(zhì)可知SKIPIF1<0為單調(diào)遞增函數(shù)，SKIPIF1<0為單調(diào)遞減函數(shù)，由SKIPIF1<0的圖象可知存在一個(gè)極小的值SKIPIF1<0，使得SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，由復(fù)合函數(shù)的單調(diào)性可知，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，由圖象可知SKIPIF1<0符合題意，故選：SKIPIF1<0.7．（2024·廣東·一模）如圖所示，設(shè)點(diǎn)SKIPIF1<0是單位圓上的一定點(diǎn)，動點(diǎn)SKIPIF1<0從點(diǎn)SKIPIF1<0出發(fā)在圓上按逆時(shí)針方向旋轉(zhuǎn)一周，點(diǎn)SKIPIF1<0所旋轉(zhuǎn)過的SKIPIF1<0的長為SKIPIF1<0，弦SKIPIF1<0的長為SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象大致是（

）A． B．C． D．【答案】C【解析】取SKIPIF1<0的中點(diǎn)為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，根據(jù)正弦函數(shù)的圖象知，C中的圖象符合解析式.故選：C.8．（2024·全國·模擬預(yù)測）若方程SKIPIF1<0在區(qū)間SKIPIF1<0上有解，SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)榉匠蘏KIPIF1<0，即SKIPIF1<0在區(qū)間SKIPIF1<0上有解，設(shè)函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0在區(qū)間SKIPIF1<0上有交點(diǎn)．因?yàn)镾KIPIF1<0，所以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，在SKIPIF1<0上單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，在區(qū)間SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0．則SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，綜上，實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．故選：A．9．（多選題）（2024·江蘇連云港·模擬預(yù)測）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有兩個(gè)不同的實(shí)數(shù)解，則下列選項(xiàng)中可以作為實(shí)數(shù)SKIPIF1<0取值范圍的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【解析】因?yàn)殛P(guān)于SKIPIF1<0的方程SKIPIF1<0恰有兩個(gè)不同的實(shí)數(shù)解，所以函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，作出函數(shù)圖象，如下圖所示，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，所以實(shí)數(shù)m的取值范圍是SKIPIF1<0．四個(gè)選項(xiàng)中只要是SKIPIF1<0的子集就滿足要求.故選：BCD.10．（多選題）（2024·高三·山東濱州·期末）在平面直角坐標(biāo)系中，如圖放置的邊長為SKIPIF1<0的正方形SKIPIF1<0沿SKIPIF1<0軸滾動(無滑動滾動)，點(diǎn)SKIPIF1<0恰好經(jīng)過坐標(biāo)原點(diǎn)，設(shè)頂點(diǎn)SKIPIF1<0的軌跡方程是SKIPIF1<0，則對函數(shù)SKIPIF1<0的判斷正確的是（

）.

A．函數(shù)SKIPIF1<0是奇函數(shù)B．對任意SKIPIF1<0，都有SKIPIF1<0C．函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0D．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增【答案】BCD【解析】由題意得，當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0為圓心，SKIPIF1<0為半徑的SKIPIF1<0圓；當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0的軌跡是以原點(diǎn)為圓心，SKIPIF1<0為半徑的SKIPIF1<0圓；當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0為圓心，SKIPIF1<0為半徑的SKIPIF1<0圓，如圖所示：此后依次重復(fù)，所以函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)，由圖象可知，函數(shù)SKIPIF1<0為偶函數(shù)，故A錯(cuò)誤；因?yàn)镾KIPIF1<0以SKIPIF1<0為周期，所以SKIPIF1<0，即SKIPIF1<0，故B正確；由圖象可知，SKIPIF1<0的值域?yàn)镾KIPIF1<0，故C正確；由圖象可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0以SKIPIF1<0為周期，所以SKIPIF1<0在SKIPIF1<0上的圖象和在SKIPIF1<0上的圖象相同，即單調(diào)遞增，故D正確．故選：BCD.11．（多選題）（2024·全國·模擬預(yù)測）已知SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】ABC【解析】令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0為偶函數(shù).當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0為偶函數(shù)，且其圖象過點(diǎn)SKIPIF1<0，顯然四個(gè)選項(xiàng)都不滿足.當(dāng)SKIPIF1<0為偶數(shù)且SKIPIF1<0時(shí)，易知函數(shù)SKIPIF1<0為偶函數(shù)，所以函數(shù)SKIPIF1<0為偶函數(shù)，其圖象關(guān)于SKIPIF1<0軸對稱，則選項(xiàng)SKIPIF1<0，SKIPIF1<0符合；若SKIPIF1<0為正偶數(shù)，因?yàn)镾KIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因?yàn)楹瘮?shù)SKIPIF1<0為偶函數(shù)，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，選項(xiàng)SKIPIF1<0符合；若SKIPIF1<0為負(fù)偶數(shù)，易知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，排除選項(xiàng)SKIPIF1<0.當(dāng)SKIPIF1<0為奇數(shù)時(shí)，易知函數(shù)SKIPIF1<0為奇函數(shù)，所以函數(shù)SKIPIF1<0為奇函數(shù)，其圖象關(guān)于坐標(biāo)原點(diǎn)對稱，則選項(xiàng)SKIPIF1<0符合，若SKIPIF1<0為正奇數(shù)，因?yàn)镾KIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，選項(xiàng)SKIPIF1<0符合；若SKIPIF1<0為負(fù)奇數(shù)，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，不妨取SKIPIF1<0，則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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0趨向于正無窮時(shí)，因?yàn)橹笖?shù)函數(shù)的增長速率比冪函數(shù)的快，所以SKIPIF1<0趨向于正無窮；所以SKIPIF1<0內(nèi)SKIPIF1<0先減后增，故選項(xiàng)SKIPIF1<0符合.故選：SKIPIF1<0.12．（2024·上海寶山·一模）設(shè)SKIPIF1<0為常數(shù)，若SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象必定不經(jīng)過第象限【答案】二【解析】已知SKIPIF1<0,則指數(shù)函數(shù)SKIPIF1<0單調(diào)遞增，過定點(diǎn)SKIPIF1<0，且SKIPIF1<0，函數(shù)SKIPIF1<0的圖象是由函數(shù)函數(shù)SKIPIF1<0向下平移SKIPIF1<0個(gè)單位，作出函數(shù)SKIPIF1<0的圖象，可知圖象必定不經(jīng)過第二象限.故答案為：二.13．（2024·貴州黔東南·模擬預(yù)測）設(shè)函數(shù)SKIPIF1<0，則滿足條件“方程SKIPIF1<0有三個(gè)實(shí)數(shù)解”的實(shí)數(shù)a的一個(gè)值為.【答案】3（答案不唯一，只要滿足SKIPIF1<0均可）．【解析】由于函數(shù)SKIPIF1<0為對勾函數(shù)，且SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0取等號，函數(shù)SKIPIF1<0為單調(diào)遞增函數(shù)，且SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示，由圖象可知，要使方程SKIPIF1<0有三個(gè)實(shí)數(shù)解，則需SKIPIF1<0，則符合題意的一個(gè)SKIPIF1<0的值為3．故答案為：3（答案不唯一，只要滿足SKIPIF1<0均可）．14．（2024·北京西城·二模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0．①若函數(shù)SKIPIF1<0無零點(diǎn)，則SKIPIF1<0的一個(gè)取值為；②若函數(shù)SKIPIF1<0有4個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0SKIPIF1<0【解析】畫函數(shù)SKIPIF1<0的圖象如下：①函數(shù)SKIPIF1<0無零點(diǎn)，即SKIPIF1<0無解，即SKIPIF1<0與SKIPIF1<0的圖象無交點(diǎn)，所以SKIPIF1<0，可取SKIPIF1<0；②函數(shù)SKIPIF1<0有4個(gè)零點(diǎn)，即SKIPIF1<0有4個(gè)根，即SKIPIF1<0與SKIPIF1<0的圖象有4個(gè)交點(diǎn)，由SKIPIF1<0關(guān)于SKIPIF1<0對稱，所以SKIPIF1<0，