新高考數(shù)學(xué)一輪復(fù)習(xí) 專項(xiàng)分層精練第07課 冪函數(shù)與二次函數(shù)-2024年新高考數(shù)學(xué)一輪復(fù)習(xí)分層精練（新高考專用）（解析版）

試卷第=page22頁(yè)，共=sectionpages22頁(yè)第07課冪函數(shù)與二次函數(shù)（分層精練）【一層練基礎(chǔ)】一、單選題1．（2022·全國(guó)·高一專題練習(xí)）若函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022春·河南新鄉(xiāng)·高二校考階段練習(xí)）下列函數(shù)中，在SKIPIF1<0上單調(diào)遞減的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022春·陜西寶雞·高一寶雞市渭濱中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，則（

）A．最大值為2，最小值為1B．最大值為SKIPIF1<0，最小值為1C．最大值為SKIPIF1<0，最小值為1D．最大值為SKIPIF1<0，最小值為SKIPIF1<04．（2023·重慶酉陽(yáng)·重慶市酉陽(yáng)第一中學(xué)校?？家荒＃┤艉瘮?shù)SKIPIF1<0在SKIPIF1<0上是單調(diào)減函數(shù)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2023·江蘇鎮(zhèn)江·揚(yáng)中市第二高級(jí)中學(xué)校考模擬預(yù)測(cè)）已知冪函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0的值為（

）A．3 B．SKIPIF1<0 C．1 D．SKIPIF1<06．（2023·全國(guó)·高三專題練習(xí)）已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2023·河南信陽(yáng)·信陽(yáng)高中?？寄M預(yù)測(cè)）已知冪函數(shù)SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08．（2021·河北衡水·河北衡水中學(xué)?？既＃┮阎猄KIPIF1<0，SKIPIF1<0，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<09．（2022秋·江蘇連云港·高一期末）已知冪函數(shù)SKIPIF1<0的圖象過(guò)函數(shù)SKIPIF1<0的圖象所經(jīng)過(guò)的定點(diǎn)，則SKIPIF1<0的值等于（）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．SKIPIF1<010．（2023·全國(guó)·高三專題練習(xí)）若冪函數(shù)SKIPIF1<0滿足SKIPIF1<0，則下列關(guān)于函數(shù)SKIPIF1<0的判斷正確的是（

）A．SKIPIF1<0是周期函數(shù) B．SKIPIF1<0是單調(diào)函數(shù)C．SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱 D．SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱11．（2023·四川南充·閬中中學(xué)?？级＃┫铝泻瘮?shù)中，在SKIPIF1<0上是增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．（2021春·陜西延安·高二子長(zhǎng)市中學(xué)?？计谀﹥绾瘮?shù)SKIPIF1<0在SKIPIF1<0為增函數(shù)，則SKIPIF1<0的值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<013．（2022·全國(guó)·高三專題練習(xí)）若SKIPIF1<0則滿足SKIPIF1<0的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<014．（2023·廣西·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．8 D．9二、多選題15．（2022秋·高一單元測(cè)試）在下列四個(gè)圖形中，二次函數(shù)SKIPIF1<0與指數(shù)函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．16．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的大致圖象可能是（

）A． B．C． D．三、填空題17．（2012·江蘇·高考真題）已知函數(shù)的值域?yàn)镾KIPIF1<0，若關(guān)于x的不等式SKIPIF1<0的解集為SKIPIF1<0，則實(shí)數(shù)c的值為．18．（2020秋·廣東陽(yáng)江·高一陽(yáng)江市第一中學(xué)校考階段練習(xí)）如果二次函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)，那么SKIPIF1<0的取值范圍是.19．（2022秋·河南信陽(yáng)·高一統(tǒng)考期中）函數(shù)SKIPIF1<0是冪函數(shù)，且在SKIPIF1<0上是減函數(shù)，則實(shí)數(shù)SKIPIF1<0.20．（2020秋·全國(guó)·高一專題練習(xí)）已知冪函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，且在區(qū)間SKIPIF1<0上為減函數(shù)，則SKIPIF1<0的值為．【二層練綜合】一、單選題1．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<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<02．（2019·全國(guó)·高三專題練習(xí)）定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2023·全國(guó)·高一專題練習(xí)）函數(shù)SKIPIF1<0，且SKIPIF1<0與函數(shù)SKIPIF1<0在同一坐標(biāo)系內(nèi)的圖象不可能的是（

）A． B．C． D．4．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0滿足對(duì)任意的實(shí)數(shù)SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2023·全國(guó)·高三專題練習(xí)）已知二次函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022秋·寧夏中衛(wèi)·高三中寧一中校考階段練習(xí)）“冪函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)”是“函數(shù)SKIPIF1<0為奇函數(shù)”的（

）條件A．充分不必要 B．必要不充分C．充分必要 D．既不充分也不必要7．（2023·廣東東莞·?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象恒過(guò)定點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在冪函數(shù)SKIPIF1<0的圖象上，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（2022·全國(guó)·高三專題練習(xí)）已知定義在SKIPIF1<0上的冪函數(shù)SKIPIF1<0（SKIPIF1<0為實(shí)數(shù)）過(guò)點(diǎn)SKIPIF1<0，記SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2010·上海徐匯·統(tǒng)考高考模擬）下列函數(shù)中，與冪函數(shù)SKIPIF1<0有相同定義域的是（

）A．SKIPIF1<0； B．SKIPIF1<0； C．SKIPIF1<0； D．SKIPIF1<0．10．（2022秋·山西晉城·高三晉城市第一中學(xué)校校考階段練習(xí)）若集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<011．（2023·四川·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0與SKIPIF1<0都是定義在SKIPIF1<0上的函數(shù)，SKIPIF1<0是奇函數(shù)，SKIPIF1<0是偶函數(shù)，且SKIPIF1<0，SKIPIF1<0都不是常數(shù)函數(shù)，現(xiàn)有下列三個(gè)結(jié)論：①SKIPIF1<0；②SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱；③SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的單調(diào)性可能相同SKIPIF1<0其中正確結(jié)論的個(gè)數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．（2023春·四川成都·高一校聯(lián)考期末）冪函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則下列說(shuō)法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0是減函數(shù)C．SKIPIF1<0是奇函數(shù) D．SKIPIF1<0是偶函數(shù)二、多選題13．（2020秋·安徽安慶·高一桐城市第八中學(xué)?？茧A段練習(xí)）關(guān)于SKIPIF1<0的方程SKIPIF1<0，下列命題正確的有（

）A．存在實(shí)數(shù)SKIPIF1<0，使得方程無(wú)實(shí)根B．存在實(shí)數(shù)SKIPIF1<0，使得方程恰有2個(gè)不同的實(shí)根C．存在實(shí)數(shù)SKIPIF1<0，使得方程恰有3個(gè)不同的實(shí)根D．存在實(shí)數(shù)SKIPIF1<0，使得方程恰有4個(gè)不同的實(shí)根14．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．15．（2023·江蘇鎮(zhèn)江·揚(yáng)中市第二高級(jí)中學(xué)?？寄M預(yù)測(cè)）下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0B．函數(shù)SKIPIF1<0為奇函數(shù)C．冪函數(shù)SKIPIF1<0是減函數(shù)D．SKIPIF1<0圖像關(guān)于點(diǎn)SKIPIF1<0成中心對(duì)稱16．（2023·全國(guó)·高三專題練習(xí)）若a＞b＞0＞c，則(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0三、填空題17．（2023·遼寧葫蘆島·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，則關(guān)于x的不等式SKIPIF1<0的解集為．18．（2022秋·湖南郴州·高一安仁縣第一中學(xué)?？茧A段練習(xí)）若冪函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)則SKIPIF1<0．19．（2022·河南·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0，SKIPIF1<0，若對(duì)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是.20．（2023·陜西榆林·?？寄M預(yù)測(cè)）直線SKIPIF1<0：SKIPIF1<0與SKIPIF1<0，SKIPIF1<0軸的交點(diǎn)分別是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0與函數(shù)SKIPIF1<0，SKIPIF1<0的圖像的交點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0是線段SKIPIF1<0的三等分點(diǎn)，則SKIPIF1<0的值為.21．（2017·四川綿陽(yáng)·統(tǒng)考一模）SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0時(shí)，SKIPIF1<0，若對(duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是.22．（2022·高一課時(shí)練習(xí)）已知SKIPIF1<0.若函數(shù)SKIPIF1<0在SKIPIF1<0上遞減且為偶函數(shù)，則SKIPIF1<0.【三層練能力】一、單選題1．（2015·陜西·高考真題）對(duì)二次函數(shù)SKIPIF1<0（SKIPIF1<0為非零整數(shù)），四位同學(xué)分別給出下列結(jié)論，其中有且僅有一個(gè)結(jié)論是錯(cuò)誤的，則錯(cuò)誤的結(jié)論是A．SKIPIF1<0是SKIPIF1<0的零點(diǎn) B．1是SKIPIF1<0的極值點(diǎn)C．3是SKIPIF1<0的極值 D．點(diǎn)SKIPIF1<0在曲線SKIPIF1<0上2．（2022·全國(guó)·高三專題練習(xí)）已知冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，函數(shù)SKIPIF1<0時(shí)，總存在SKIPIF1<0使得SKIPIF1<0，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題3．（2022秋·重慶·高三校聯(lián)考階段練習(xí)）在三角函數(shù)部分，我們研究過(guò)二倍角公式SKIPIF1<0，實(shí)際上類似的還有三倍角公式，則下列說(shuō)法中正確的有（

）A．SKIPIF1<0B．存在SKIPIF1<0時(shí)，使得SKIPIF1<0C．給定正整數(shù)SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0D．設(shè)方程SKIPIF1<0的三個(gè)實(shí)數(shù)根為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，并且SKIPIF1<0，則SKIPIF1<0三、填空題4．（2023·四川成都·?？家荒＃┮阎瘮?shù)SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的取值范圍是．5．（2023·全國(guó)·高三專題練習(xí)）關(guān)于x的不等式SKIPIF1<0，解集為【一層練基礎(chǔ)】參考答案1．D【分析】先利用配湊法求出SKIPIF1<0的解析式，則可求出SKIPIF1<0的解析式，從而可求出函數(shù)的最小值【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.從而SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值，且最小值為SKIPIF1<0.故選：D2．D【分析】根據(jù)函數(shù)單調(diào)性的性質(zhì)可判斷每個(gè)選項(xiàng)中函數(shù)在SKIPIF1<0的單調(diào)性.【詳解】對(duì)于A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，故A錯(cuò)誤；對(duì)于B，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，故B錯(cuò)誤；對(duì)于C，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C錯(cuò)誤；對(duì)于D，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D正確故選：D.【點(diǎn)睛】本題主要考查對(duì)函數(shù)單調(diào)性的判斷，根據(jù)基本初等函數(shù)的復(fù)合函數(shù)單調(diào)性進(jìn)行判斷即可，屬于基礎(chǔ)題.3．B【分析】利用SKIPIF1<0化簡(jiǎn)f(x)解析式，根據(jù)二次函數(shù)的性質(zhì)即可求f(x)最值.【詳解】SKIPIF1<0，SKIPIF1<0時(shí)，sinx∈[SKIPIF1<0，1]，∴當(dāng)sinx＝SKIPIF1<0時(shí)，f(x)最大值為SKIPIF1<0；當(dāng)sinx＝1時(shí)，f(x)最小值為1.故選：B.4．A【分析】由求導(dǎo)公式和法則求出SKIPIF1<0，由導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系，列出不等式進(jìn)行分離常數(shù)，再構(gòu)造函數(shù)后，利用整體思想和二次函數(shù)的性質(zhì)求出函數(shù)的最值，可得a的取值范圍．【詳解】由題意得，SKIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0在[1，+∞）上是單調(diào)減函數(shù)，所以SKIPIF1<0≤0在[1，+∞）上恒成立，當(dāng)SKIPIF1<0≤0時(shí)，則SKIPIF1<0在[1，+∞）上恒成立，即aSKIPIF1<0，設(shè)g（x）SKIPIF1<0，因?yàn)閤∈[1，+∞），所以SKIPIF1<0∈（0，1]，當(dāng)SKIPIF1<0時(shí)，g（x）取到最大值是：SKIPIF1<0，所以aSKIPIF1<0，所以數(shù)a的取值范圍是（﹣∞，SKIPIF1<0]故選：A【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：根據(jù)求導(dǎo)公式和法則，導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系，將問(wèn)題轉(zhuǎn)化為恒成立問(wèn)題，利用分離常數(shù)法，求函數(shù)值域，屬于中檔題．5．C【分析】先根據(jù)SKIPIF1<0是冪函數(shù)，由SKIPIF1<0求得SKIPIF1<0，再根據(jù)函數(shù)在SKIPIF1<0上是減函數(shù)，確定SKIPIF1<0的值求解.【詳解】由函數(shù)SKIPIF1<0為冪函數(shù)知，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.∵SKIPIF1<0在SKIPIF1<0上是減函數(shù)，而當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在SKIPIF1<0是增函數(shù)，不符合題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，符合題意，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0.故選：C.6．B【分析】設(shè)冪函數(shù)SKIPIF1<0，依次將點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0坐標(biāo)代入，可得SKIPIF1<0，結(jié)合指數(shù)函數(shù)和對(duì)數(shù)函數(shù)性質(zhì)即可得到答案.【詳解】設(shè)冪函數(shù)SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0在SKIPIF1<0的圖象上，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又點(diǎn)SKIPIF1<0在SKIPIF1<0的圖象上，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故選：B7．D【分析】根據(jù)冪函數(shù)的定義求出函數(shù)SKIPIF1<0解析式，再利用冪函數(shù)的單調(diào)性比較大小而得解.【詳解】因冪函數(shù)SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，于是得SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0是R上的增函數(shù)，而SKIPIF1<0，則有SKIPIF1<0，所以SKIPIF1<0.故選：D8．A【分析】根據(jù)指對(duì)冪不等式，結(jié)合指對(duì)冪函數(shù)的性質(zhì)分別求參數(shù)a的范圍，再取交集即可.【詳解】由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，∴當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0同時(shí)成立時(shí)，取交集得SKIPIF1<0，故選：A.9．B【分析】先根據(jù)冪函數(shù)定義得SKIPIF1<0，再確定SKIPIF1<0的圖像所經(jīng)過(guò)的定點(diǎn)為SKIPIF1<0，代入SKIPIF1<0解得SKIPIF1<0的值.【詳解】由于SKIPIF1<0為冪函數(shù)，則SKIPIF1<0，解得：SKIPIF1<0，則SKIPIF1<0；函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0的圖像所經(jīng)過(guò)的定點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0,故選：B.10．C【分析】由題意得SKIPIF1<0，利用導(dǎo)數(shù)求出方程的根，進(jìn)而可求出結(jié)果.【詳解】由題意得SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；所以SKIPIF1<0，因此方程SKIPIF1<0有唯一解，解為SKIPIF1<0，因此SKIPIF1<0，所以不是周期函數(shù)，不是單調(diào)函數(shù)，關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，故選：C.11．C【解析】對(duì)AB：直接判斷其單調(diào)性；對(duì)C：把SKIPIF1<0化為SKIPIF1<0，判斷其單調(diào)性；對(duì)D：利用SKIPIF1<0判斷SKIPIF1<0的單調(diào)性.【詳解】本題考查函數(shù)的單調(diào)性.A項(xiàng)中，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故A錯(cuò)誤；B項(xiàng)中，二次函數(shù)SKIPIF1<0的圖像開(kāi)口向下，對(duì)稱軸方程為SKIPIF1<0，故該函數(shù)在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故B錯(cuò)誤；C項(xiàng)中，函數(shù)SKIPIF1<0，在SKIPIF1<0和SKIPIF1<0上分別單調(diào)遞增，故C正確；D項(xiàng)中，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D錯(cuò)誤.故選：C【點(diǎn)睛】方法點(diǎn)睛：四個(gè)選項(xiàng)互不相關(guān)的選擇題，需要對(duì)各個(gè)選項(xiàng)一一驗(yàn)證.12．B【分析】由冪函數(shù)解析式的形式可構(gòu)造方程求得SKIPIF1<0或SKIPIF1<0，分別驗(yàn)證兩種情況下SKIPIF1<0在SKIPIF1<0上的單調(diào)性即可得到結(jié)果.【詳解】SKIPIF1<0為冪函數(shù)，SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上為減函數(shù)，不合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上為增函數(shù)，符合題意；綜上所述：SKIPIF1<0.故選：B.13．B【分析】按SKIPIF1<0或0，SKIPIF1<0，SKIPIF1<0和SKIPIF1<0四種情況，分別化簡(jiǎn)解出不等式，可得x的取值范圍．【詳解】①當(dāng)SKIPIF1<0或0時(shí)，SKIPIF1<0成立；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可有SKIPIF1<0，解得SKIPIF1<0；③當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0所以SKIPIF1<0則原不等式的解為SKIPIF1<0，故選：B14．C【分析】利用誘導(dǎo)公式化簡(jiǎn)函數(shù)的表達(dá)式，利用三角函數(shù)和特殊冪函數(shù)的奇偶性進(jìn)行分析，可得到SKIPIF1<0，進(jìn)而計(jì)算得到答案.【詳解】由SKIPIF1<0，有SKIPIF1<0，可得SKIPIF1<0.故選：C15．ABD【分析】根據(jù)SKIPIF1<0的關(guān)系與各圖形一個(gè)個(gè)檢驗(yàn)即可判斷．【詳解】當(dāng)SKIPIF1<0時(shí)，A正確；當(dāng)SKIPIF1<0時(shí)，B正確；當(dāng)SKIPIF1<0時(shí)，D正確；當(dāng)SKIPIF1<0時(shí)，無(wú)此選項(xiàng)．故選：ABD．16．ABD【分析】先根據(jù)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，排除C，再舉出適當(dāng)?shù)腟KIPIF1<0的值，分別得到ABD三個(gè)圖象.【詳解】由題意知SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的大致圖象不可能為C，而當(dāng)SKIPIF1<0為其他值時(shí)，A，B，D均有可能出現(xiàn)，不妨設(shè)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，此時(shí)A選項(xiàng)符合要求；當(dāng)SKIPIF1<0時(shí)，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，故函數(shù)SKIPIF1<0為奇函數(shù)，所以B選項(xiàng)符合要求，當(dāng)SKIPIF1<0時(shí)，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，故函數(shù)SKIPIF1<0為偶函數(shù)，所以D選項(xiàng)符合要求.故選：ABD17．9．【詳解】∵f(x)＝x2＋ax＋b的值域?yàn)閇0，＋∞)，∴Δ＝0，∴b－SKIPIF1<0＝0，∴f(x)＝x2＋ax＋SKIPIF1<0a2＝SKIPIF1<02.又∵f(x)＜c的解集為(m，m＋6)，∴m，m＋6是方程x2＋ax＋SKIPIF1<0－c＝0的兩根．由一元二次方程根與系數(shù)的關(guān)系得SKIPIF1<0解得c＝9.18．SKIPIF1<0【詳解】SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)，則SKIPIF1<0，所以SKIPIF1<0.19．2【分析】根據(jù)函數(shù)為冪函數(shù)求參數(shù)m，討論所求得的m判斷函數(shù)是否在SKIPIF1<0上是減函數(shù)，即可確定m值.【詳解】由題設(shè)，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)在SKIPIF1<0上遞增，不合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)在SKIPIF1<0上遞減，符合題設(shè).綜上，SKIPIF1<0.故答案為：220．2【分析】由冪指數(shù)為偶數(shù)且小于可得.【詳解】SKIPIF1<0為偶數(shù)，且小于0，即SKIPIF1<0，解得SKIPIF1<0，驗(yàn)證得SKIPIF1<0．【點(diǎn)睛】?jī)绾瘮?shù)SKIPIF1<0中，當(dāng)SKIPIF1<0為奇數(shù)時(shí)，函數(shù)為奇函數(shù)，當(dāng)SKIPIF1<0為偶數(shù)時(shí)，函數(shù)為偶函數(shù)；當(dāng)SKIPIF1<0時(shí)，在第一象限內(nèi)函數(shù)為增函數(shù)，當(dāng)SKIPIF1<0時(shí)，在第一象限內(nèi)函數(shù)為減函數(shù).【二層練綜合】參考答案1．D【分析】求出函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)值的集合，函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)值的集合，再由已知并借助集合包含關(guān)系即可作答.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上值的集合是SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上值的集合為SKIPIF1<0，因函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，于是得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D2．A【分析】求出函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式，利用二次函數(shù)的基本性質(zhì)可求得函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由題意可得SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故選：A.3．D【分析】利用對(duì)數(shù)函數(shù)及二次函數(shù)的性質(zhì)逐項(xiàng)分析即得.【詳解】對(duì)于A，由對(duì)數(shù)函數(shù)圖象可知SKIPIF1<0，又函數(shù)SKIPIF1<0，對(duì)稱軸為SKIPIF1<0<1，對(duì)應(yīng)方程的兩個(gè)根為0，SKIPIF1<0，由圖知SKIPIF1<0，從而SKIPIF1<0，選項(xiàng)A可能；對(duì)于B，由對(duì)數(shù)函數(shù)圖象可知SKIPIF1<0，又函數(shù)SKIPIF1<0，對(duì)稱軸為SKIPIF1<0<1，對(duì)應(yīng)方程的兩個(gè)根為0，SKIPIF1<0，由圖知SKIPIF1<0，從而SKIPIF1<0，選項(xiàng)B可能；對(duì)于C，由對(duì)數(shù)函數(shù)圖象可知SKIPIF1<0，又函數(shù)SKIPIF1<0，對(duì)稱軸為SKIPIF1<0>1，對(duì)應(yīng)方程的兩個(gè)根為0，SKIPIF1<0，由圖知SKIPIF1<0，從而SKIPIF1<0，選項(xiàng)B可能；對(duì)于D，由對(duì)數(shù)函數(shù)圖象可知SKIPIF1<0，又函數(shù)SKIPIF1<0，對(duì)稱軸為SKIPIF1<0<1，對(duì)應(yīng)方程的兩個(gè)根為0，SKIPIF1<0，由圖知SKIPIF1<0，從而SKIPIF1<0，選項(xiàng)D不可能.故選：D.4．D【分析】根據(jù)題意，令SKIPIF1<0，SKIPIF1<0，進(jìn)而將問(wèn)題轉(zhuǎn)化為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且函數(shù)SKIPIF1<0為減函數(shù)，進(jìn)而得SKIPIF1<0，再解不等式組即可得答案.【詳解】解：因?yàn)閷?duì)任意的實(shí)數(shù)SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0成立，所以，對(duì)任意的實(shí)數(shù)SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，即函數(shù)SKIPIF1<0是SKIPIF1<0上的減函數(shù)．因?yàn)镾KIPIF1<0，令SKIPIF1<0，SKIPIF1<0，要使SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．另一方面，函數(shù)SKIPIF1<0為減函數(shù)，所以，SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)a的取值范圍是SKIPIF1<0．故選：D．5．B【分析】由二次函數(shù)的值域可得出SKIPIF1<0，可得出SKIPIF1<0，則有SKIPIF1<0，利用基本不等式可求得結(jié)果.【詳解】若SKIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，不合乎題意，因?yàn)槎魏瘮?shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故選：B.6．A【分析】要使函數(shù)SKIPIF1<0是冪函數(shù)，且在SKIPIF1<0上為增函數(shù)，求出SKIPIF1<0，可得函數(shù)SKIPIF1<0為奇函數(shù)，即充分性成立；函數(shù)SKIPIF1<0為奇函數(shù)，求出SKIPIF1<0，故必要性不成立，可得答案.【詳解】要使函數(shù)SKIPIF1<0是冪函數(shù)，且在SKIPIF1<0上為增函數(shù)，則SKIPIF1<0，解得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0為奇函數(shù)，即充分性成立；“函數(shù)SKIPIF1<0為奇函數(shù)”，則SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，故必要性不成立，故選：A．7．B【分析】令對(duì)數(shù)的真數(shù)等于0，求得x、y的值，可得圖象經(jīng)過(guò)的定點(diǎn)坐標(biāo)．再根據(jù)在冪函數(shù)y＝f（x）的圖象上，求出函數(shù)f（x）的解析式，從而求出SKIPIF1<0的值．【詳解】∵已知a＞0且a≠1，對(duì)于函數(shù)SKIPIF1<0，令x﹣1＝1，求得x＝2，ySKIPIF1<0，可得它的圖象恒過(guò)定點(diǎn)P（2，4），∵點(diǎn)P在冪函數(shù)y＝f（x）＝xn的圖象上，∴2nSKIPIF1<0，∴nSKIPIF1<0，∴f（x）SKIPIF1<0則f（2）SKIPIF1<0,故SKIPIF1<0SKIPIF1<0故選B．【點(diǎn)睛】本題主要考查對(duì)數(shù)函數(shù)的圖象經(jīng)過(guò)定點(diǎn)問(wèn)題，求函數(shù)值，屬于基礎(chǔ)題．8．A【分析】首先求出SKIPIF1<0，得到函數(shù)的單調(diào)性，再利用對(duì)數(shù)函數(shù)的圖象性質(zhì)得到SKIPIF1<0，即得解.【詳解】由題得SKIPIF1<0.函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù).因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A【點(diǎn)睛】方法點(diǎn)睛：比較對(duì)數(shù)式的大小，一般先利用對(duì)數(shù)函數(shù)的圖象和性質(zhì)比較每個(gè)式子和零的大小分成正負(fù)兩個(gè)集合，再利用對(duì)數(shù)函數(shù)的圖象和性質(zhì)比較同類數(shù)的大小.9．A【分析】由題知冪函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，再依次討論各選項(xiàng)即可得答案.【詳解】解：冪函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，對(duì)于A選項(xiàng)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，故正確；對(duì)于B選項(xiàng)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，故錯(cuò)誤；對(duì)于C選項(xiàng)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，故錯(cuò)誤；對(duì)于D選項(xiàng)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，故錯(cuò)誤；故選：A10．A【分析】先解出集合A、B,再求SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故選：A．11．D【分析】根據(jù)奇函數(shù)的性質(zhì)及賦值法得到SKIPIF1<0，從而判斷①正確；根據(jù)偶函數(shù)的性質(zhì)得到SKIPIF1<0，從而判斷②正確；取SKIPIF1<0，判斷兩者的單調(diào)性，從而判斷③正確.【詳解】對(duì)于①：由SKIPIF1<0是奇函數(shù)，即SKIPIF1<0，取SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，正確；對(duì)于②：由SKIPIF1<0是偶函數(shù)，得SKIPIF1<0，則SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，正確；對(duì)于③：取SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上都單調(diào)遞增，正確SKIPIF1<0故選：D.12．C【分析】根據(jù)冪函數(shù)的定義及單調(diào)性可判斷AB，再由奇函數(shù)的定義判斷CD.【詳解】函數(shù)SKIPIF1<0為冪函數(shù)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，不滿足條件，排除A；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，滿足題意.函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，但不是減函數(shù)，排除B；因?yàn)楹瘮?shù)定義域關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，不是偶函數(shù)，故C正確，D錯(cuò)誤.故選：C.13．AB【解析】通過(guò)換元法，設(shè)SKIPIF1<0，方程化為關(guān)于SKIPIF1<0的二次方程SKIPIF1<0的根的情況進(jìn)行分類討論.【詳解】設(shè)SKIPIF1<0，方程化為關(guān)于SKIPIF1<0的二次方程SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0無(wú)實(shí)根，故原方程無(wú)實(shí)根.當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，則SKIPIF1<0，原方程有兩個(gè)相等的實(shí)根SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有兩個(gè)實(shí)根SKIPIF1<0，由SKIPIF1<0可知，SKIPIF1<0，SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0無(wú)實(shí)根，SKIPIF1<0有兩個(gè)不同的實(shí)根.綜上可知：A，B項(xiàng)正確，C，D項(xiàng)錯(cuò)誤.故選：AB【點(diǎn)睛】此題考查方程的根的問(wèn)題，利用換元法討論二次方程的根的分布，涉及分類討論思想.14．ABC【分析】令SKIPIF1<0，先分析函數(shù)SKIPIF1<0的奇偶性，再分情況討論SKIPIF1<0的奇偶性，然后逐項(xiàng)分析四個(gè)選項(xiàng)即可求解.【詳解】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(guò)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(duì)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(duì)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(wú)窮時(shí)，因?yàn)橹笖?shù)函數(shù)的增長(zhǎng)速率比冪函數(shù)的快，所以SKIPIF1<0趨向于正無(wú)窮；所以SKIPIF1<0內(nèi)SKIPIF1<0先減后增，故選項(xiàng)SKIPIF1<0符合.故選：SKIPIF1<0.15．ABD【分析】利用函數(shù)性質(zhì)相關(guān)的定義以及復(fù)合函數(shù)的同增異減性質(zhì)逐項(xiàng)分析.【詳解】對(duì)于A，SKIPIF1<0，SKIPIF1<0是減函數(shù)，SKIPIF1<0在SKIPIF1<0是減函數(shù)，在SKIPIF1<0是增函數(shù)，根據(jù)復(fù)合函數(shù)同增異減的性質(zhì)，在SKIPIF1<0時(shí)是增函數(shù)，正確；對(duì)于B，SKIPIF1<0，是奇函數(shù)，正確；對(duì)于C，SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0并且是減函數(shù)，所以SKIPIF1<0是增函數(shù)，錯(cuò)誤；對(duì)于D，SKIPIF1<0，相當(dāng)于函數(shù)SKIPIF1<0先向左平移2個(gè)單位，再向上平移2個(gè)單位，而SKIPIF1<0是關(guān)于原點(diǎn)對(duì)稱的，所以SKIPIF1<0是關(guān)于SKIPIF1<0對(duì)稱的，正確；故選：ABD.16．ABD【分析】利用作差法可判斷AB，根據(jù)冪函數(shù)單調(diào)性可判斷C，根據(jù)基本不等式可判斷D.【詳解】A：SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故A正確；B：SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，故B正確；C：SKIPIF1<0時(shí)，SKIPIF1<0