




版權(quán)說(shuō)明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡(jiǎn)介
專(zhuān)題05分段函數(shù)真題再現(xiàn)1.(2023·北京·統(tǒng)考高考真題)設(shè)SKIPIF1<0,函數(shù)SKIPIF1<0,給出下列四個(gè)結(jié)論:①SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減;②當(dāng)SKIPIF1<0時(shí),SKIPIF1<0存在最大值;③設(shè)SKIPIF1<0,則SKIPIF1<0;④設(shè)SKIPIF1<0.若SKIPIF1<0存在最小值,則a的取值范圍是SKIPIF1<0.其中所有正確結(jié)論的序號(hào)是____________.2.(2022·浙江·統(tǒng)考高考真題)已知函數(shù)SKIPIF1<0則SKIPIF1<0________;若當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0的最大值是_________.3.(2022·北京·統(tǒng)考高考真題)設(shè)函數(shù)SKIPIF1<0若SKIPIF1<0存在最小值,則a的一個(gè)取值為_(kāi)_______;a的最大值為_(kāi)__________.4.(2021·浙江·統(tǒng)考高考真題)已知SKIPIF1<0,函數(shù)SKIPIF1<0若SKIPIF1<0,則SKIPIF1<0___________.考點(diǎn)一分段函數(shù)函數(shù)值(解析式)一、單選題1.已知函數(shù)SKIPIF1<0,則SKIPIF1<0(
)A.5B.3 C.2 D.12.已知函數(shù)SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.1 C.-1 D.23.已知函數(shù)SKIPIF1<0,則SKIPIF1<0(
)A.4 B.8 C.16 D.324.已知函數(shù)SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0().A.SKIPIF1<01 B.SKIPIF1<02 C.SKIPIF1<03 D.SKIPIF1<045.已知函數(shù)SKIPIF1<0,則SKIPIF1<0的值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<06.已知函數(shù)SKIPIF1<0,則SKIPIF1<0(
)A.4 B.5 C.6 D.77.已知函數(shù)SKIPIF1<0,則SKIPIF1<0(
)A.1 B.e C.SKIPIF1<0 D.SKIPIF1<08.若函數(shù)SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、多選題9.函數(shù)f(x)的圖象如圖所示,則f(x)的解析式是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0三、填空題10.已知函數(shù)SKIPIF1<0,則SKIPIF1<0___________.11.已知函數(shù)SKIPIF1<0,則SKIPIF1<0___________.12.已知函數(shù)SKIPIF1<0,則SKIPIF1<0_________.13.設(shè)函數(shù)SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的解析式為_(kāi)___________.14.設(shè)SKIPIF1<0定義在SKIPIF1<0上且SKIPIF1<0,則SKIPIF1<0______.考點(diǎn)二分段函數(shù)定義域和值域一、單選題1.設(shè)函數(shù)SKIPIF1<0,SKIPIF1<0則SKIPIF1<0的值域是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<02.若定義運(yùn)算SKIPIF1<0,則函數(shù)SKIPIF1<0的值域是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、多選題3.已知函數(shù)SKIPIF1<0,關(guān)于函數(shù)SKIPIF1<0的結(jié)論正確的是(
)A.SKIPIF1<0的定義域是R B.SKIPIF1<0的值域是SKIPIF1<0C.若SKIPIF1<0,則x的值為SKIPIF1<0 D.SKIPIF1<0三、填空題4.函數(shù)SKIPIF1<0的定義域是________.5.已知SKIPIF1<0,則SKIPIF1<0的值域是______;6.函數(shù)SKIPIF1<0的值域是______.7.已知函數(shù)SKIPIF1<0SKIPIF1<0的最大值為m,SKIPIF1<0的最小值為n,則SKIPIF1<0______.8.已知函數(shù)SKIPIF1<0,則SKIPIF1<0的最小值為_(kāi)________.四、雙空題9.函數(shù)y=SKIPIF1<0的定義域?yàn)開(kāi)_______,值域?yàn)開(kāi)_______.10.如圖為一分段函數(shù)的圖象,則該函數(shù)的定義域?yàn)開(kāi)_______,值域?yàn)開(kāi)_______.五、解答題11.已知函數(shù)SKIPIF1<0的圖象如圖所示,其中SKIPIF1<0軸的左側(cè)為一條線段,右側(cè)為某拋物線的一段.(1)寫(xiě)出函數(shù)SKIPIF1<0的定義域和值域;(2)求SKIPIF1<0的值.考點(diǎn)三分段函數(shù)單調(diào)性一、單選題1.定義運(yùn)算:SKIPIF1<0,例如:SKIPIF1<0,SKIPIF1<0,則函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是(
)A.SKIPIF1<0 B.SKIPIF1<0和SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0和SKIPIF1<03.若函數(shù)SKIPIF1<0是SKIPIF1<0上的單調(diào)函數(shù),則實(shí)數(shù)SKIPIF1<0的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0二、多選題4.已知函數(shù)SKIPIF1<0在R上單調(diào)遞增,則實(shí)數(shù)a的取值可以是(
)A.0 B.1 C.2 D.3三、填空題5.設(shè)函數(shù)SKIPIF1<0,SKIPIF1<0,則函數(shù)SKIPIF1<0的遞增區(qū)間為_(kāi)_______.6.函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為_(kāi)_________.7.己知函數(shù)SKIPIF1<0滿足對(duì)任意SKIPIF1<0,且SKIPIF1<0,都有SKIPIF1<0成立,則實(shí)數(shù)a的取值范圍是__________.8.已知函數(shù)SKIPIF1<0,設(shè)SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0的取值范圍是____________.四、雙空題9.函數(shù)SKIPIF1<0的單調(diào)性為_(kāi)_____;奇偶性為_(kāi)_____.五、解答題10.已知函數(shù)SKIPIF1<0(1)求SKIPIF1<0,SKIPIF1<0的值;(2)若SKIPIF1<0,求實(shí)數(shù)a的值;(3)直接寫(xiě)出SKIPIF1<0的單調(diào)區(qū)間.考點(diǎn)四分段函數(shù)求參一、單選題1.已知函數(shù)SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.0 C.1 D.22.設(shè)SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.0 C.1 D.23.已知函數(shù)SKIPIF1<0SKIPIF1<0的值域?yàn)镾KIPIF1<0,則實(shí)數(shù)SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.已知函數(shù)SKIPIF1<0的最大值為0,則實(shí)數(shù)a的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<05.若函數(shù)SKIPIF1<0,在R上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<06.已知函數(shù)SKIPIF1<0,滿足對(duì)任意的實(shí)數(shù)SKIPIF1<0,SKIPIF1<0且SKIPIF1<0,都有SKIPIF1<0,則實(shí)數(shù)a的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<07.已知函數(shù)SKIPIF1<0在SKIPIF1<0是減函數(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<08.函數(shù)SKIPIF1<0,若SKIPIF1<0互不相等,且SKIPIF1<0,則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、填空題9.已知SKIPIF1<0,函數(shù)SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0________.10.已知函數(shù)SKIPIF1<0.若SKIPIF1<0,則實(shí)數(shù)m的取值范圍是______.11.設(shè)函數(shù)SKIPIF1<0,若SKIPIF1<0,則實(shí)數(shù)a=_____.12.已知函數(shù)SKIPIF1<0且SKIPIF1<0,則正數(shù)SKIPIF1<0的值為_(kāi)_____.13.設(shè)函數(shù)SKIPIF1<0存在最小值,則SKIPIF1<0的取值范圍是________.14.已知SKIPIF1<0,若存在三個(gè)不同實(shí)數(shù)SKIPIF1<0使得SKIPIF1<0,則SKIPIF1<0的取值范圍是______.15.設(shè)SKIPIF1<0且SKIPIF1<0,已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0是遞增數(shù)列,則a的取值范圍是____.16.已知SKIPIF1<0,其中SKIPIF1<0且SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0___________.考點(diǎn)五解分段函數(shù)不等式一、單選題1.已知SKIPIF1<0,則使SKIPIF1<0成立的SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<02.已知函數(shù)SKIPIF1<0,則滿足不等式SKIPIF1<0的x的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.已知函數(shù)SKIPIF1<0,若SKIPIF1<0,則實(shí)數(shù)a的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<04.已知函數(shù)SKIPIF1<0,則不等式SKIPIF1<0的解集是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<05.若SKIPIF1<0,且SKIPIF1<0的解集為SKIPIF1<0,則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<06.已知函數(shù)SKIPIF1<0,若SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0二、多選題7.已知SKIPIF1<0是定義在SKIPIF1<0的奇函數(shù),且SKIPIF1<0時(shí),SKIPIF1<0,則下列結(jié)論正確的是(
)A.SKIPIF1<0時(shí),SKIPIF1
溫馨提示
- 1. 本站所有資源如無(wú)特殊說(shuō)明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒(méi)有圖紙預(yù)覽就沒(méi)有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫(kù)網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。
最新文檔
- 學(xué)校教室裝修項(xiàng)目的施工合同
- 新建自建房購(gòu)買(mǎi)合同樣本
- 全新夫妻離婚前財(cái)產(chǎn)分割合同
- 建設(shè)工程合同管理規(guī)范
- 度渠道拓展合作合同
- 餐飲服務(wù)合同模板與消防相關(guān)
- 音樂(lè)藝人經(jīng)紀(jì)合同范本
- 化工產(chǎn)品出口代理合同書(shū)
- 簡(jiǎn)易彩鋼瓦合同范本
- Module 6 Unit 3 language in use 教學(xué)設(shè)計(jì) 2024-2025學(xué)年外研版八年級(jí)英語(yǔ)上冊(cè)
- 安全環(huán)保法律法規(guī)
- 2025年湖南環(huán)境生物職業(yè)技術(shù)學(xué)院高職單招職業(yè)適應(yīng)性測(cè)試近5年??及鎱⒖碱}庫(kù)含答案解析
- 建設(shè)工程質(zhì)量安全監(jiān)督人員考試題庫(kù)含答案
- 電氣控制技術(shù)項(xiàng)目化教程 第2版 課件 項(xiàng)目1、2 低壓電器的選用與維修、電動(dòng)機(jī)直接控制電路
- 2025年上半年山東人才發(fā)展集團(tuán)限公司社會(huì)招聘易考易錯(cuò)模擬試題(共500題)試卷后附參考答案
- 2025年度文化創(chuàng)意產(chǎn)業(yè)園區(qū)入駐及合作協(xié)議3篇
- 【MOOC期末】《大學(xué)體育射箭》(東南大學(xué))中國(guó)大學(xué)慕課答案
- 2024年山東理工職業(yè)學(xué)院高職單招語(yǔ)文歷年參考題庫(kù)含答案解析
- 《中華人民共和國(guó)學(xué)前教育法》專(zhuān)題培訓(xùn)
- 2023屆高考復(fù)習(xí)之文學(xué)類(lèi)文本閱讀訓(xùn)練
- 國(guó)家基礎(chǔ)教育實(shí)驗(yàn)中心外語(yǔ)教育研究中心
評(píng)論
0/150
提交評(píng)論