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答案第=page11頁,共=sectionpages22頁專題06函數(shù)的圖像、方程與零點(diǎn)【練基礎(chǔ)】一、單選題1.(2022·安徽·六安市裕安區(qū)新安中學(xué)高三階段練習(xí))函數(shù)SKIPIF1<0的圖象大致為(
)A. B.C. D.【答案】D【分析】首先求出函數(shù)的定義域,再判斷函數(shù)的奇偶性,最后根據(jù)函數(shù)值的情況判斷即可.【詳解】解:因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0是偶函數(shù),函數(shù)圖象關(guān)于SKIPIF1<0軸對(duì)稱,排除A,B;當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,排除C.故選:D.2.(2022·安徽·安慶一中高三階段練習(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】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),結(jié)合圖像分析.【詳解】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.3.(2022·甘肅·高臺(tái)縣第一中學(xué)高三階段練習(xí)(文))如圖,SKIPIF1<0是邊長(zhǎng)為2的正三角形,記SKIPIF1<0位于直線SKIPIF1<0左側(cè)的圖形的面積為SKIPIF1<0,則SKIPIF1<0的函數(shù)圖象是(
).A.B.C. D.【答案】A【分析】根據(jù)題意,求出函數(shù)解析式,據(jù)此分析選項(xiàng),即可得答案【詳解】解:根據(jù)題意,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以只有A選項(xiàng)符合,故選:A4.(2023·全國(guó)·高三專題練習(xí))函數(shù)SKIPIF1<0的圖像與函數(shù)SKIPIF1<0的圖像的交點(diǎn)個(gè)數(shù)為(
)A.2 B.3 C.4 D.0【答案】C【分析】作出兩個(gè)函數(shù)的圖像,由圖像可得交點(diǎn)個(gè)數(shù).【詳解】SKIPIF1<0在SKIPIF1<0上是增函數(shù),SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上是減函數(shù),在SKIPIF1<0和SKIPIF1<0上是增函數(shù),SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,作出函數(shù)SKIPIF1<0SKIPIF1<0的圖像,如圖,由圖像可知它們有4個(gè)交點(diǎn).故選:C.5.(2021·云南省楚雄天人中學(xué)高三階段練習(xí))已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù),在區(qū)間SKIPIF1<0上單調(diào)遞減,且SKIPIF1<0,則不等式SKIPIF1<0的解集為(
)A.SKIPIF1<0或SKIPIF1<0 B.SKIPIF1<0或SKIPIF1<0C.SKIPIF1<0或SKIPIF1<0 D.SKIPIF1<0或SKIPIF1<0【答案】D【分析】先根據(jù)題意畫出函數(shù)SKIPIF1<0的簡(jiǎn)圖,再分SKIPIF1<0,SKIPIF1<0兩種情況討論,結(jié)合圖像解不等式即可【詳解】由題意,函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù),在區(qū)間SKIPIF1<0上單調(diào)遞減,且SKIPIF1<0,可畫出函數(shù)簡(jiǎn)圖如下圖所示:當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,解得SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,解得SKIPIF1<0;綜上不等式SKIPIF1<0的解集為:SKIPIF1<0或SKIPIF1<0故選:D6.(2021·甘肅省民樂縣第一中學(xué)高三階段練習(xí)(理))函數(shù)SKIPIF1<0在SKIPIF1<0的零點(diǎn)個(gè)數(shù)為()A.2 B.3 C.4 D.5【答案】B【解析】令SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,再根據(jù)x的取值范圍可求得零點(diǎn).【詳解】由SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.SKIPIF1<0在SKIPIF1<0的零點(diǎn)個(gè)數(shù)是3,故選B.【點(diǎn)睛】本題考查在一定范圍內(nèi)的函數(shù)的零點(diǎn)個(gè)數(shù),滲透了直觀想象和數(shù)學(xué)運(yùn)算素養(yǎng).采取特殊值法,利用數(shù)形結(jié)合和方程思想解題.7.(2022·全國(guó)·高三專題練習(xí))函數(shù)SKIPIF1<0的所有零點(diǎn)之和為(
)A.0 B.2 C.4 D.6【答案】B【分析】結(jié)合函數(shù)的對(duì)稱性求得正確答案.【詳解】令SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0圖象關(guān)于SKIPIF1<0對(duì)稱,在SKIPIF1<0上遞減.SKIPIF1<0,令SKIPIF1<0,所以SKIPIF1<0是奇函數(shù),圖象關(guān)于原點(diǎn)對(duì)稱,所以SKIPIF1<0圖象關(guān)于SKIPIF1<0對(duì)稱,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上遞增,所以SKIPIF1<0與SKIPIF1<0有兩個(gè)交點(diǎn),兩個(gè)交點(diǎn)關(guān)于SKIPIF1<0對(duì)稱,所以函數(shù)SKIPIF1<0的所有零點(diǎn)之和為SKIPIF1<0.故選:B8.(2022·河南·高三階段練習(xí)(理))已知函數(shù)SKIPIF1<0,若函數(shù)SKIPIF1<0有4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】轉(zhuǎn)化為兩個(gè)函數(shù)交點(diǎn)問題分析【詳解】SKIPIF1<0即SKIPIF1<0分別畫出SKIPIF1<0和SKIPIF1<0的函數(shù)圖像,則兩圖像有4個(gè)交點(diǎn)所以SKIPIF1<0,即SKIPIF1<0故選:C二、多選題9.(2021·重慶市第十一中學(xué)校高三階段練習(xí))關(guān)于函數(shù)SKIPIF1<0,正確的說法是(
)A.SKIPIF1<0有且僅有一個(gè)零點(diǎn)B.SKIPIF1<0在定義域內(nèi)單調(diào)遞減C.SKIPIF1<0的定義域?yàn)镾KIPIF1<0D.SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】ACD【分析】將函數(shù)SKIPIF1<0分離系數(shù)可得SKIPIF1<0,數(shù)形結(jié)合,逐一分析即可;【詳解】解:SKIPIF1<0,作出函數(shù)SKIPIF1<0圖象如圖:由圖象可知,函數(shù)只有一個(gè)零點(diǎn),定義域?yàn)镾KIPIF1<0,在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減,圖象關(guān)于SKIPIF1<0對(duì)稱,故B錯(cuò)誤,故選:ACD.10.(2023·全國(guó)·高三專題練習(xí))已知函數(shù)SKIPIF1<0,則下列說法正確的是(
)A.SKIPIF1<0為奇函數(shù) B.SKIPIF1<0為減函數(shù)C.SKIPIF1<0有且只有一個(gè)零點(diǎn) D.SKIPIF1<0的值域?yàn)镾KIPIF1<0【答案】AC【分析】化簡(jiǎn)函數(shù)解析式,分析函數(shù)的奇偶性,單調(diào)性,值域,零點(diǎn)即可求解.【詳解】SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故SKIPIF1<0為奇函數(shù),又SKIPIF1<0,SKIPIF1<0在R上單調(diào)遞增,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,即函數(shù)值域?yàn)镾KIPIF1<0令SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,故函數(shù)有且只有一個(gè)零點(diǎn)0.綜上可知,AC正確,BD錯(cuò)誤.故選:AC11.(2022·湖南省祁東縣育賢中學(xué)高三階段練習(xí))如圖是函數(shù)SKIPIF1<0的部分圖像,則(
)A.SKIPIF1<0的最小正周期為SKIPIF1<0B.將函數(shù)SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位后,得到的函數(shù)為奇函數(shù)C.SKIPIF1<0是函數(shù)SKIPIF1<0的一條對(duì)稱軸D.若函數(shù)SKIPIF1<0在SKIPIF1<0上有且僅有兩個(gè)零點(diǎn),則SKIPIF1<0【答案】AD【分析】先根據(jù)圖像可得SKIPIF1<0,即可判斷A,接下來求得SKIPIF1<0,即可得到SKIPIF1<0的解析式,根據(jù)圖像平移判斷B,令SKIPIF1<0解出SKIPIF1<0即可判斷C,令SKIPIF1<0,解出函數(shù)零點(diǎn),然后根據(jù)在SKIPIF1<0上有且僅有兩個(gè)零點(diǎn)列出不等式解SKIPIF1<0即可判斷D【詳解】由圖像可知,SKIPIF1<0SKIPIF1<0,即SKIPIF1<0,故A正確SKIPIF1<0
此時(shí)SKIPIF1<0又SKIPIF1<0在圖像上,SKIPIF1<0,解得SKIPIF1<0SKIPIF1<0將SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位后得到的圖像對(duì)應(yīng)的解析式為SKIPIF1<0不為奇函數(shù),故B錯(cuò)誤SKIPIF1<0,SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0是函數(shù)SKIPIF1<0的一條對(duì)稱軸時(shí),此時(shí)SKIPIF1<0不符合題意,故C錯(cuò)誤令SKIPIF1<0,解得SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,不合題意SKIPIF1<0時(shí),SKIPIF1<0;SKIPIF1<0時(shí),SKIPIF1<0;SKIPIF1<0時(shí),SKIPIF1<0又因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上有且僅有兩個(gè)零點(diǎn)SKIPIF1<0,解得SKIPIF1<0,故D正確故選:AD12.(2021·福建·福清西山學(xué)校高三階段練習(xí))已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0恰有2個(gè)零點(diǎn),則實(shí)數(shù)m可以是(
)A.SKIPIF1<0 B.0 C.1 D.2【答案】ABC【分析】轉(zhuǎn)化為函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0恰有兩個(gè)交點(diǎn),畫出函數(shù)SKIPIF1<0的圖象,根據(jù)圖象可得解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0恰有2個(gè)零點(diǎn),所以函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0恰有兩個(gè)交點(diǎn),畫出函數(shù)SKIPIF1<0的圖象如圖:由圖可知,SKIPIF1<0或SKIPIF1<0,結(jié)合選項(xiàng),因此SKIPIF1<0可以為-1,0,1.故選:ABC.【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.三、填空題13.(2020·廣東·北京師范大學(xué)珠海分校附屬外國(guó)語學(xué)校高三階段練習(xí))若函數(shù)f(x)=x2-ax-b的兩個(gè)零點(diǎn)分別是是2和3,則函數(shù)g(x)=bx2-ax-1的零點(diǎn)為____________.【答案】SKIPIF1<0【詳解】主要考查二次函數(shù)零點(diǎn)的性質(zhì)及零點(diǎn)的確定方法.首先將2,3分別代入方程SKIPIF1<0-ax-b=0,求得a,b,然后解方程bSKIPIF1<0-ax-1=0,得到函數(shù)g(x)零點(diǎn).14.(2022·全國(guó)·高三專題練習(xí)(理))若函數(shù)SKIPIF1<0的一個(gè)零點(diǎn)為SKIPIF1<0,則常數(shù)SKIPIF1<0的一個(gè)取值為___________.【答案】SKIPIF1<0【分析】根據(jù)零點(diǎn)的概念及特殊角的三角函數(shù)值即可求解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的一個(gè)零點(diǎn)為SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0時(shí),滿足條件,SKIPIF1<0是常數(shù)SKIPIF1<0的一個(gè)取值.故答案為:SKIPIF1<015.(2021·福建省南平市高級(jí)中學(xué)高三階段練習(xí))若方程SKIPIF1<0的實(shí)根在區(qū)間SKIPIF1<0上,則SKIPIF1<0_______.【答案】-2或1【分析】依題意可得SKIPIF1<0,在同一平面直角坐標(biāo)系中作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象,結(jié)合函數(shù)圖象即可判斷方程的根所在區(qū)間,即可得解;【詳解】解:由于方程SKIPIF1<0,顯然SKIPIF1<0,所以SKIPIF1<0,在同一平面直角坐標(biāo)系中作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象,由圖象上可得出:方程SKIPIF1<0在區(qū)間SKIPIF1<0和SKIPIF1<0內(nèi)各有一個(gè)實(shí)根.所以SKIPIF1<0或SKIPIF1<0故答案為:SKIPIF1<0或SKIPIF1<0.16.(2022·北京·北師大實(shí)驗(yàn)中學(xué)高三階段練習(xí))若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn),則實(shí)數(shù)SKIPIF1<0的取值范圍是_____.【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn),和的圖象有兩個(gè)交點(diǎn),畫出和的圖象,如圖,要有兩個(gè)交點(diǎn),那么四、解答題17.(2021·寧夏·青銅峽市寧朔中學(xué)高三階段練習(xí)(文))若函數(shù)SKIPIF1<0.(1)在所給的坐標(biāo)系內(nèi)畫出函數(shù)SKIPIF1<0圖像;(2)求方程SKIPIF1<0恰有三個(gè)不同實(shí)根時(shí)的實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)圖象見解析;(2)SKIPIF1<0.【分析】(1)結(jié)合二次函數(shù)的圖象與性質(zhì),對(duì)數(shù)函數(shù)的圖象與性質(zhì)利用描點(diǎn)法作函數(shù)的圖象,(2)觀察SKIPIF1<0圖象,根據(jù)SKIPIF1<0的圖象與SKIPIF1<0的圖象有三個(gè)交點(diǎn)確定m的范圍.【詳解】(1)作圖如下:(2)方程SKIPIF1<0有3個(gè)解等價(jià)于函數(shù)SKIPIF1<0的圖象與SKIPIF1<0的圖象有三個(gè)交點(diǎn),觀察圖象可得SKIPIF1<0.18.(2022·河南·模擬預(yù)測(cè)(文))已知函數(shù)SKIPIF1<0,SKIPIF1<0.(1)在給出的平面直角坐標(biāo)系中畫出SKIPIF1<0和SKIPIF1<0的圖象;(2)若關(guān)于x的不等式SKIPIF1<0恒成立,求實(shí)數(shù)a的取值范圍.【答案】(1)詳見解析;(2)SKIPIF1<0.【分析】(1)根據(jù)絕對(duì)值函數(shù)分區(qū)間去絕對(duì)值后變成分段函數(shù),然后作圖;(2)由題可得SKIPIF1<0,然后利用數(shù)形結(jié)合可得參數(shù)取值范圍.【詳解】(1)由題意得:SKIPIF1<0,SKIPIF1<0,畫出SKIPIF1<0和SKIPIF1<0的圖象如圖所示.(2)∵SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0或SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0,要使SKIPIF1<0恒成立,則SKIPIF1<0,解得SKIPIF1<0,所以實(shí)數(shù)a的取值范圍為SKIPIF1<0.19.(2020·內(nèi)蒙古·巴彥淖爾市臨河區(qū)第三中學(xué)高三階段練習(xí)(理))已知函數(shù)SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0的解析式.(2)若方程SKIPIF1<0有實(shí)數(shù)根,求實(shí)數(shù)a的取值范圍.【答案】(1)SKIPIF1<0,SKIPIF1<0;(2)SKIPIF1<0.【解析】(1)考查了函數(shù)解析式的求解,需要采用換元法,設(shè)SKIPIF1<0,表示出SKIPIF1<0,再寫出SKIPIF1<0,最后換元成SKIPIF1<0即可;(2)SKIPIF1<0有實(shí)根,轉(zhuǎn)化為SKIPIF1<0SKIPIF1<0,所以需要求函數(shù)SKIPIF1<0的值域,再解不等式.【詳解】解:(1)設(shè)SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0;且SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0;(2)設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí)函數(shù)有最小值SKIPIF1<0,而SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.【點(diǎn)睛】本題主要考查的是換元法求函數(shù)的解析式,利用函數(shù)值域求參數(shù)范圍的問題,需要注意:(1)采用換元法求解函數(shù)解析式時(shí),注意換元必?fù)Q域,不要漏掉SKIPIF1<0的范圍;(2)求解參數(shù)范圍時(shí)需要轉(zhuǎn)化為求解函數(shù)的最值問題,即求函數(shù)的值域,再利用SKIPIF1<0的范圍解不等式即可,需要注意定義域的限制.20.(2022·山東省青島第九中學(xué)高三階段練習(xí))已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間,(2)若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn),求實(shí)數(shù)m的取值范圍.【答案】(1)單調(diào)遞減區(qū)間是SKIPIF1<0,單調(diào)遞增區(qū)間是SKIPIF1<0;(2)SKIPIF1<0【分析】(1)根據(jù)題意,列出方程組求得SKIPIF1<0,得到SKIPIF1<0,進(jìn)而求得函數(shù)的單調(diào)區(qū)間;(2)由題意得到SKIPIF1<0,結(jié)合條件列出不等式組,即得.【詳解】(1)由題可得SKIPIF1<0,由題意得SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0,單調(diào)遞增區(qū)間是SKIPIF1<0;(2)因?yàn)镾KIPIF1<0,由(1)可知,SKIPIF1<0在SKIPIF1<0處取得極大值,在SKIPIF1<0處取得極小值,SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0,單調(diào)遞增區(qū)間是SKIPIF1<0,依題意,要使SKIPIF1<0有三個(gè)零點(diǎn),則SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,經(jīng)檢驗(yàn),SKIPIF1<0,根據(jù)零點(diǎn)存在定理,可以確定函數(shù)有三個(gè)零點(diǎn),所以m的取值范圍為SKIPIF1<0.21.(2021·貴州·遵義一中高三階段練習(xí)(理))已知函數(shù)SKIPIF1<0.(1)若函數(shù)SKIPIF1<0在范圍SKIPIF1<0上存在零點(diǎn),求SKIPIF1<0的取值范圍;(2)當(dāng)SKIPIF1<0時(shí),求函數(shù)SKIPIF1<0的最小值SKIPIF1<0.【答案】(1)SKIPIF1<0
(2)SKIPIF1<0【分析】(1)參變分離轉(zhuǎn)化為存在SKIPIF1<0,使得SKIPIF1<0成立,求導(dǎo)分析SKIPIF1<0的單調(diào)性和取值范圍,即得解;(2)函數(shù)SKIPIF1<0對(duì)稱軸為SKIPIF1<0,分SKIPIF1<0,SKIPIF1<0,SKIPIF1<0三種情況討論,即得解【詳解】(1)由題意,函數(shù)SKIPIF1<0在范圍SKIPIF1<0上存在零點(diǎn)即存在SKIPIF1<0,使得SKIPIF1<0成立令SKIPIF1<0,則SKIPIF1<0令SKIPIF1<0(舍)所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0即SKIPIF1<0在SKIPIF1<0單調(diào)遞增,在SKIPIF1<0單調(diào)遞減,又SKIPIF1<0SKIPIF1<0即SKIPIF1<0的取值范圍是SKIPIF1<0(2)SKIPIF1<0,對(duì)稱軸為SKIPIF1<0當(dāng)SKIPIF1<0時(shí),即SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),即SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),即SKIPIF1<0時(shí),SKIPIF1<0;綜上:SKIPIF1<022.(2020·江蘇省盱眙中學(xué)高三階段練習(xí))已知SKIPIF1<0是偶函數(shù).(1)求SKIPIF1<0的值;(2)若函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有公共點(diǎn),求a的取值范圍.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0.【解析】(1)由偶函數(shù)的定義結(jié)合對(duì)數(shù)的運(yùn)算性質(zhì)可求出實(shí)數(shù)SKIPIF1<0的值;(2)利用參變量分離法得出關(guān)于SKIPIF1<0的方程SKIPIF1<0有解,然后利用指數(shù)函數(shù)和對(duì)數(shù)的函數(shù)的基本性質(zhì)求出SKIPIF1<0的取值范圍,即可得出實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】(1)SKIPIF1<0是偶函數(shù),SKIPIF1<0,SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0對(duì)任意的SKIPIF1<0都成立,SKIPIF1<0;(2)由題意知,方程SKIPIF1<0有解,亦即SKIPIF1<0,即SKIPIF1<0有解,SKIPIF1<0有解,由SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,即SKIPIF1<0的取值范圍是SKIPIF1<0.【點(diǎn)睛】本題考查利用函數(shù)的奇偶性求參數(shù),同時(shí)也考查了利用函數(shù)的零點(diǎn)個(gè)數(shù)求參數(shù),涉及對(duì)數(shù)運(yùn)算性質(zhì)的應(yīng)用,靈活利用參變量分離法能簡(jiǎn)化計(jì)算,考查運(yùn)算求解能力,屬于中等題.【提能力】一、單選題1.(2020·全國(guó)·高三專題練習(xí)(文))函數(shù)SKIPIF1<0的圖像大致為()A. B.C. D.【答案】B【詳解】分析:通過研究函數(shù)奇偶性以及單調(diào)性,確定函數(shù)圖像.詳解:SKIPIF1<0為奇函數(shù),舍去A,SKIPIF1<0舍去D;SKIPIF1<0,所以舍去C;因此選B.點(diǎn)睛:有關(guān)函數(shù)圖象識(shí)別問題的常見題型及解題思路(1)由函數(shù)的定義域,判斷圖象左右的位置,由函數(shù)的值域,判斷圖象的上下位置;②由函數(shù)的單調(diào)性,判斷圖象的變化趨勢(shì);③由函數(shù)的奇偶性,判斷圖象的對(duì)稱性;④由函數(shù)的周期性,判斷圖象的循環(huán)往復(fù).2.(2019·全國(guó)·高三專題練習(xí))如圖所示,設(shè)點(diǎn)SKIPIF1<0是單位圓上的一定點(diǎn),動(dòng)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(zhǎng)為SKIPIF1<0,弦SKIPIF1<0的長(zhǎng)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,在直角三角形求出SKIPIF1<0的表達(dá)式,根據(jù)弧長(zhǎng)公式求出SKIPIF1<0的表達(dá)式,再用SKIPIF1<0表示SKIPIF1<0,再根據(jù)解析式得答案.【詳解】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.【點(diǎn)睛】本題考查正弦函數(shù)的圖象,考查弧長(zhǎng)公式,其中表示出弦長(zhǎng)SKIPIF1<0和弧長(zhǎng)SKIPIF1<0的解析式是解題的關(guān)鍵,屬于基礎(chǔ)題.3.(2008·四川·高考真題(理))直線SKIPIF1<0繞原點(diǎn)逆時(shí)針旋轉(zhuǎn)SKIPIF1<0,再向右平移1個(gè)單位,所得到的直線為()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【詳解】∵直線SKIPIF1<0繞原點(diǎn)逆時(shí)針旋轉(zhuǎn)SKIPIF1<0的直線為SKIPIF1<0,從而淘汰(C),(D)又∵將SKIPIF1<0向右平移1個(gè)單位得SKIPIF1<0,即SKIPIF1<0故選A;【點(diǎn)評(píng)】此題重點(diǎn)考察互相垂直的直線關(guān)系,以及直線平移問題;【突破】熟悉互相垂直的直線斜率互為負(fù)倒數(shù),過原點(diǎn)的直線無常數(shù)項(xiàng);重視平移方法:“左加右減”;4.(2022·全國(guó)·高三專題練習(xí))定義在R上的偶函數(shù)SKIPIF1<0滿足SKIPIF1<0,且當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,若關(guān)于x的方程SKIPIF1<0至少有8個(gè)實(shí)數(shù)解,則實(shí)數(shù)m的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】根據(jù)條件可得出函數(shù)SKIPIF1<0是以4為周期的周期函數(shù),作出SKIPIF1<0,SKIPIF1<0的圖象,根據(jù)函數(shù)為偶函數(shù),原問題可轉(zhuǎn)化為當(dāng)SKIPIF1<0時(shí)兩函數(shù)圖象至少有4個(gè)交點(diǎn),根據(jù)數(shù)形結(jié)合求解即可.【詳解】因?yàn)镾KIPIF1<0,且SKIPIF1<0為偶函數(shù)所以SKIPIF1<0,即SKIPIF1<0,所以函數(shù)SKIPIF1<0是以4為周期的周期函數(shù),作出SKIPIF1<0,SKIPIF1<0在同一坐標(biāo)系的圖象,如圖,因?yàn)榉匠蘏KIPIF1<0至少有8個(gè)實(shí)數(shù)解,所以SKIPIF1<0,SKIPIF1<0圖象至少有8個(gè)交點(diǎn),根據(jù)SKIPIF1<0,SKIPIF1<0的圖象都為偶函數(shù)可知,圖象在y軸右側(cè)至少有4個(gè)交點(diǎn),由圖可知,當(dāng)SKIPIF1<0時(shí),只需SKIPIF1<0,即SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),只需SKIPIF1<0,即SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),由圖可知顯然成立,綜上可知,SKIPIF1<0.故選:B5.(2021·全國(guó)·高三專題練習(xí))如圖,函數(shù)SKIPIF1<0的圖象由一條射線和拋物線的一部分構(gòu)成,SKIPIF1<0的零點(diǎn)為SKIPIF1<0,若不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立,則a的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】由條件可知,SKIPIF1<0的圖象是由SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到,再利用數(shù)形結(jié)合,分析圖象的臨界條件,得到SKIPIF1<0的取值范圍.【詳解】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,圖象過點(diǎn)SKIPIF1<0和SKIPIF1<0,即SKIPIF1<0,解得:SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),設(shè)拋物線SKIPIF1<0,代入點(diǎn)SKIPIF1<0得,SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0的圖象是由SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到,因?yàn)镾KIPIF1<0,對(duì)SKIPIF1<0恒成立,所以SKIPIF1<0的圖象恒在SKIPIF1<0的上方,當(dāng)兩圖象如圖所示,相切時(shí),拋物線SKIPIF1<0SKIPIF1<0,SKIPIF1<0,與直線SKIPIF1<0相切,即SKIPIF1<0,解得:SKIPIF1<0,SKIPIF1<0,切點(diǎn)SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,解得:SKIPIF1<0或SKIPIF1<0.故選:A【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛:本題考查根據(jù)不等式恒成立,求參數(shù)的取值范圍,本題的關(guān)鍵是數(shù)形結(jié)合,分析臨界條件,利用直線與拋物線相切,求參數(shù)的取值范圍.6.(2023·全國(guó)·高三專題練習(xí))正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0之間的大小關(guān)系為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】由SKIPIF1<0,得SKIPIF1<0,而SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn),從而可得SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0,令SKIPIF1<0,然后利用零點(diǎn)存在性定理可求得SKIPIF1<0,同理可求出SKIPIF1<0的范圍,從而可比較出SKIPIF1<0的大小【詳解】SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn),則SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0;SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,由SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn),則SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0;SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,由SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0只有一個(gè)交點(diǎn),則SKIPIF1<0在SKIPIF1<0只有一個(gè)根SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0;SKIPIF1<0故選:A.7.(2022·全國(guó)·高三專題練習(xí))設(shè)函數(shù)SKIPIF1<0,則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為(
)A.1個(gè) B.2個(gè) C.3個(gè) D.4個(gè)【答案】C【分析】畫出函數(shù)SKIPIF1<0的草圖,分析函數(shù)的值域及SKIPIF1<0的解,由SKIPIF1<0解的個(gè)數(shù),可得答案【詳解】函數(shù)SKIPIF1<0的圖象如圖所示,由SKIPIF1<0,得SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,由圖象可知方程有兩個(gè)實(shí)根,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,由圖象可知,方程有1個(gè)實(shí)根,綜上,方程SKIPIF1<0有3個(gè)實(shí)根,所以函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為3,故選:C8.(2020·全國(guó)·高三專題練習(xí)(理))已知定義在SKIPIF1<0上的偶函數(shù)SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0時(shí),SKIPIF1<0,則函數(shù)SKIPIF1<0在SKIPIF1<0上的所有零點(diǎn)之和為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】把函數(shù)g(x)SKIPIF1<0f(x)﹣cosπx的零點(diǎn)轉(zhuǎn)化為兩函數(shù)y=f(x)與y=cosπx圖象交點(diǎn)的橫坐標(biāo),再由已知可得函數(shù)f(x)的對(duì)稱軸與周期,作出函數(shù)y=f(x)與y=cosπx的圖象,數(shù)形結(jié)合得答案.【詳解】函數(shù)g(x)SKIPIF1<0f(x)﹣cosπx的零點(diǎn),即方程f(x)﹣cosπx=0的根,也就是兩函數(shù)y=f(x)與y=cosπx圖象交點(diǎn)的橫坐標(biāo).由f(x)是定義在R上的偶函數(shù),且SKIPIF1<0可得函數(shù)周期為2.又當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,作出函數(shù)y=f(x)與y=cosπx的圖象如圖:由圖可知,函數(shù)g(x)SKIPIF1<0f(x)﹣cosπx在區(qū)間[﹣2,4]上的所有零點(diǎn)之和為﹣SKIPIF1<02+SKIPIF1<02+SKIPIF1<02=6.故選:C.【點(diǎn)睛】本題考查函數(shù)零點(diǎn)的判定,考查數(shù)形結(jié)合的解題思想方法和數(shù)學(xué)轉(zhuǎn)化思想方法,是中檔題.二、多選題9.(2023·全國(guó)·高三專題練習(xí))對(duì)任意兩個(gè)實(shí)數(shù)SKIPIF1<0,定義SKIPIF1<0若SKIPIF1<0,SKIPIF1<0,下列關(guān)于函數(shù)SKIPIF1<0的說法正確的是(
)A.函數(shù)SKIPIF1<0是偶函數(shù)B.方程SKIPIF1<0有三個(gè)解C.函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D.函數(shù)SKIPIF1<0有4個(gè)單調(diào)區(qū)間【答案】ABD【分析】結(jié)合題意作出函數(shù)SKIPIF1<0的圖象,進(jìn)而數(shù)形結(jié)合求解即可.【詳解】解:根據(jù)函數(shù)SKIPIF1<0與SKIPIF1<0,,畫出函數(shù)SKIPIF1<0的圖象,如圖.由圖象可知,函數(shù)SKIPIF1<0關(guān)于y軸對(duì)稱,所以A項(xiàng)正確;函數(shù)SKIPIF1<0的圖象與x軸有三個(gè)交點(diǎn),所以方程SKIPIF1<0有三個(gè)解,所以B項(xiàng)正確;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,在SKIPIF1<0上單調(diào)遞減,所以C項(xiàng)錯(cuò)誤,D項(xiàng)正確.故選:ABD10.(2022·全國(guó)·高三專題練習(xí))已知函數(shù)SKIPIF1<0,若方程SKIPIF1<0有四個(gè)不同的實(shí)根SKIPIF1<0,滿足SKIPIF1<0,則下列說法正確的是(
)A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0【答案】BCD【分析】由解析式可得函數(shù)圖象,由方程有四個(gè)不等實(shí)根可得到SKIPIF1<0與SKIPIF1<0有四個(gè)不同的交點(diǎn),從而確定四個(gè)根的范圍和SKIPIF1<0的取值范圍;由SKIPIF1<0可化簡(jiǎn)知A錯(cuò)誤;由SKIPIF1<0與SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱知B正確;根據(jù)SKIPIF1<0與SKIPIF1<0是方程SKIPIF1<0的根,結(jié)合韋達(dá)定理和SKIPIF1<0的取值范圍可知C正確;由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0,由此可確定零點(diǎn)知D正確.【詳解】由解析式可得SKIPIF1<0圖象如下圖所示:若SKIPIF1<0有四個(gè)不同的實(shí)數(shù)根,則SKIPIF1<0與SKIPIF1<0有四個(gè)不同的交點(diǎn),由圖象可知:SKIPIF1<0,SKIPIF1<0;對(duì)于A,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,整理可得:SKIPIF1<0,A錯(cuò)誤;對(duì)于B,SKIPIF1<0,SKIPIF1<0與SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱,SKIPIF1<0,B正確;對(duì)于C,SKIPIF1<0與SKIPIF1<0是方程SKIPIF1<0的兩根,SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,C正確;對(duì)于D,SKIPIF1<0,由SKIPIF1<0得:SKIPIF1<0或SKIPIF1<0,SKIPIF1<0SKIPIF1<0的根為SKIPIF1<0;SKIPIF1<0的根為SKIPIF1<0,SKIPIF1<0的零點(diǎn)為SKIPIF1<0,D正確.故選:BCD.11.(2022·山東·日照國(guó)開中學(xué)高三階段練習(xí))已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù),SKIPIF1<0,且當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則下列說法正確的是(
)A.SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)B.SKIPIF1<0C.函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象有且僅有SKIPIF1<0個(gè)交點(diǎn)D.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0【答案】ACD【分析】推導(dǎo)出函數(shù)SKIPIF1<0的周期,可判斷A選項(xiàng)的正誤;求出SKIPIF1<0、SKIPIF1<0的值,可判斷B選項(xiàng)的正誤;數(shù)形結(jié)合可判斷C選項(xiàng)的正誤;求出函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的解析式,可判斷D選項(xiàng)的正誤.【詳解】對(duì)于A選項(xiàng),由已知條件可得SKIPIF1<0,所以,函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù),A選項(xiàng)正確;對(duì)于B選項(xiàng),SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,B選項(xiàng)錯(cuò)誤;對(duì)于C選項(xiàng),作出函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象如下圖所示:當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,結(jié)合圖象可知,SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象無交點(diǎn),由圖可知,函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有SKIPIF1<0個(gè)交點(diǎn),C選項(xiàng)正確;對(duì)于D選項(xiàng),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,所以,SKIPIF1<0,D選項(xiàng)正確.故選:ACD.【點(diǎn)睛】方法點(diǎn)睛:判定函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)的常用方法:(1)直接法:直接求解函數(shù)對(duì)應(yīng)方程的根,得到方程的根,即可得出結(jié)果;(2)數(shù)形結(jié)合法:先令SKIPIF1<0,將函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù),轉(zhuǎn)化為對(duì)應(yīng)方程的根,進(jìn)而轉(zhuǎn)化為兩個(gè)函數(shù)圖象的交點(diǎn)個(gè)數(shù),結(jié)合圖象,即可得出結(jié)果.12.(2020·全國(guó)·高三專題練習(xí))已知函數(shù)f(x)=x,g(x)=x-4,則下列結(jié)論正確的是(
)A.若h(x)=f(x)g(x),則函數(shù)h(x)的最小值為4B.若h(x)=f(x)|g(x)|,則函數(shù)h(x)的值域?yàn)镽C.若h(x)=|f(x)|-|g(x)|,則函數(shù)h(x)有且僅有一個(gè)零點(diǎn)D.若h(x)=|f(x)|-|g(x)|,則|h(x)|≤4恒成立【答案】BCD【解析】對(duì)選項(xiàng)逐一分析,由此確定結(jié)論正確的選項(xiàng).【詳解】對(duì)于A選項(xiàng),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0的最小值為SKIPIF1<0,所以A選項(xiàng)錯(cuò)誤.對(duì)于B選項(xiàng),SKIPIF1<0,畫出SKIPIF1<0圖像如下圖所示,由圖可知,SKIPIF1<0的值域?yàn)镾KIPIF1<0,故B選項(xiàng)正確.對(duì)于C選項(xiàng),SKIPIF1<0,畫出SKIPIF1<0圖像如下圖所示,由圖可知,SKIPIF1<0有唯一零點(diǎn)SKIPIF1<0,故C選項(xiàng)正確.對(duì)于D選項(xiàng),由C選項(xiàng)的分析,結(jié)合SKIPIF1<0圖像可知SKIPIF1<0恒成立,故D選項(xiàng)正確.故選:BCD【點(diǎn)睛】本小題主要考查函數(shù)的最值、值域和零點(diǎn),考查分段函數(shù),考查數(shù)形結(jié)合的思想方法,屬于基礎(chǔ)題.三、填空題13.(2022·全國(guó)·高三專題練習(xí))已知偶函數(shù)SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,若函數(shù)SKIPIF1<0恰有4個(gè)不同的零點(diǎn),則實(shí)數(shù)SKIPIF1<0的取值范圍為__________【答案】SKIPIF1<0【分析】作出函數(shù)SKIPIF1<0的圖象,將問題轉(zhuǎn)化為函數(shù)SKIPIF1<0與SKIPIF1<0有4個(gè)不同的交點(diǎn),由圖示可得答案.【詳解】解:作出函數(shù)SKIPIF1<0的圖象如下圖所示,令SKIPIF1<0,則SKIPIF1<0,若函數(shù)SKIPIF1<0恰有4個(gè)不同的零點(diǎn),則需函數(shù)SKIPIF1<0與SKIPIF1<0有4個(gè)不同的交點(diǎn),所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0,故答案為:SKIPIF1<0.14.(2020·上?!じ呷龑n}練習(xí))已知函數(shù)f(x)=logax+x-b(a>0,且a≠1).當(dāng)2<a<3<b<4時(shí),函數(shù)f(x)的零點(diǎn)為x0∈(n,n+1),n∈N*,則n=.【答案】2【分析】把要求零點(diǎn)的函數(shù),變成兩個(gè)基本初等函數(shù),根據(jù)所給的a,b的值,可以判斷兩個(gè)函數(shù)的交點(diǎn)的所在的位置,同所給的區(qū)間進(jìn)行比較,得到n的值.【詳解】設(shè)函數(shù)y=logax,m=﹣x+b根據(jù)2<a<3<b<4,對(duì)于函數(shù)y=logax在x=2時(shí),一定得到一個(gè)值小于1,而b-2>1,x=3時(shí),對(duì)數(shù)值在1和2之間,b-3<1在同一坐標(biāo)系中畫出兩個(gè)函數(shù)的圖象,判斷兩個(gè)函數(shù)的圖形的交點(diǎn)在(2,3)之間,∴函數(shù)f(x)的零點(diǎn)x0∈(n,n+1)時(shí),n=2.故答案為2.考點(diǎn):二分法求方程的近似解;對(duì)數(shù)函數(shù)的圖象與性質(zhì).15.(2023·全國(guó)·高三專題練習(xí))已知函數(shù)SKIPIF1<0SKIPIF1<0.若SKIPIF1<0存在2個(gè)零點(diǎn),則SKIPIF1<0的取值范圍是__________【答案】SKIPIF1<0【分析】由SKIPIF1
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