新高考數(shù)學(xué)一輪復(fù)習(xí) 專(zhuān)項(xiàng)分層精練第05課 函數(shù)的單調(diào)性與最值（解析版）

第02課函數(shù)的單調(diào)性與最值（分層專(zhuān)項(xiàng)精練）【一層練基礎(chǔ)】一、單選題1．（2012·天津·高考真題）下列函數(shù)中，既是偶函數(shù)，又在區(qū)間（1,2）內(nèi)是增函數(shù)的為A．SKIPIF1<0，xSKIPIF1<0RB．SKIPIF1<0，xSKIPIF1<0R且x≠0C．SKIPIF1<0,xSKIPIF1<0RD．SKIPIF1<0，xSKIPIF1<0R2．（2022秋·浙江杭州·高一?？计谥校┖瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)SKIPIF1<0，則a，b，c的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2020春·山西太原·高二山西大附中?？茧A段練習(xí)）若關(guān)于x的不等式SKIPIF1<0在SKIPIF1<0區(qū)間上有解，則k的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·吉林白城·校考模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0存在平行于SKIPIF1<0軸的切線，則實(shí)數(shù)SKIPIF1<0取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題6．（2021秋·甘肅蘭州·高一蘭州一中校考期中）已知函數(shù)SKIPIF1<0，則下列結(jié)論正確的是（

）A．函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)B．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱C．函數(shù)SKIPIF1<0的圖象上存在兩點(diǎn)SKIPIF1<0，SKIPIF1<0，使得直線SKIPIF1<0軸D．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱7．（2023春·重慶九龍坡·高一四川外國(guó)語(yǔ)大學(xué)附屬外國(guó)語(yǔ)學(xué)校?？茧A段練習(xí)）設(shè)函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的一個(gè)周期為SKIPIF1<0 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0在SKIPIF1<0上有最大值SKIPIF1<0 D．SKIPIF1<0圖象的一條對(duì)稱軸為直線SKIPIF1<08．（2021秋·福建三明·高三?？计谥校┫铝泻瘮?shù)中是偶函數(shù)，且在區(qū)間SKIPIF1<0上單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<09．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的定義域?yàn)镾KIPIF1<0 B．SKIPIF1<0是偶函數(shù)C．函數(shù)SKIPIF1<0的零點(diǎn)為0 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最大值為SKIPIF1<0三、填空題10．（2022秋·陜西西安·高一西安市第八十三中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為．11．（2023春·安徽亳州·高二亳州二中?？计谀┮阎x域?yàn)镾KIPIF1<0的減函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為.12．（2016·北京·高考真題）函數(shù)SKIPIF1<0的最大值為.13．（2022秋·天津西青·高一天津市西青區(qū)楊柳青第一中學(xué)?？计谥校┤魞蓚€(gè)正實(shí)數(shù)x，y滿足SKIPIF1<0，且不等式SKIPIF1<0恒成立，則實(shí)數(shù)m的取值范圍是．14．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知實(shí)數(shù)a，b滿足SKIPIF1<0，則SKIPIF1<0的最小值是.【二層練綜合】一、單選題1．（2022秋·高一單元測(cè)試）已知函數(shù)SKIPIF1<0滿足SKIPIF1<0，且對(duì)任意的SKIPIF1<0，都有SKIPIF1<0SKIPIF1<0，則滿足不等式SKIPIF1<0的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·全國(guó)·高一專(zhuān)題練習(xí)）函數(shù)y＝SKIPIF1<0，x∈(m，n]的最小值為0，則m的取值范圍是（

）A．(1，2) B．(－1，2) C．[1，2) D．[－1，2)3．（2023·四川·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0且SKIPIF1<0在定義域上是單調(diào)函數(shù)，則實(shí)數(shù)t的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·安徽滁州·高二校考學(xué)業(yè)考試）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，滿足：①對(duì)任意SKIPIF1<0，都有SKIPIF1<0，②對(duì)任意SKIPIF1<0且SKIPIF1<0，都有SKIPIF1<0，則函數(shù)SKIPIF1<0叫“成功函數(shù)”，下列函數(shù)是“成功函數(shù)”的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．若函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為3，則實(shí)數(shù)a的值為（

）A．1 B．2 C．3 D．4二、多選題6．（2022秋·山東濰坊·高三?？计谥校┫铝兴膫€(gè)函數(shù)中，以SKIPIF1<0為周期且在SKIPIF1<0上單調(diào)遞增的偶函數(shù)有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<07．（2023春·湖北恩施·高一校聯(lián)考期中）已知函數(shù)SKIPIF1<0，SKIPIF1<0，則（

）A．函數(shù)SKIPIF1<0為偶函數(shù)B．函數(shù)SKIPIF1<0為奇函數(shù)C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值之和為0D．設(shè)SKIPIF1<0，則SKIPIF1<0的解集為SKIPIF1<08．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<09．（2022·全國(guó)·高三專(zhuān)題練習(xí)）高斯是德國(guó)著名數(shù)學(xué)家，近代數(shù)學(xué)奠基者之一，享有“數(shù)學(xué)王子”的稱號(hào)，他和阿基米德，牛頓并列為世界三大數(shù)學(xué)家，用SKIPIF1<0表示不超過(guò)x的最大整數(shù)，則SKIPIF1<0稱為高斯函數(shù)，例如SKIPIF1<0，SKIPIF1<0.則下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0（SKIPIF1<0）上單調(diào)遞增B．若函數(shù)SKIPIF1<0，則SKIPIF1<0的值域?yàn)镾KIPIF1<0C．若函數(shù)SKIPIF1<0，則SKIPIF1<0的值域?yàn)镾KIPIF1<0D．SKIPIF1<0，SKIPIF1<0三、填空題10．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為1，則SKIPIF1<0的值為.11．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0（x>0），若SKIPIF1<0的最大值為SKIPIF1<0，則正實(shí)數(shù)a=.12．（2022·全國(guó)·高三專(zhuān)題練習(xí)）若SKIPIF1<0，SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為.13．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若SKIPIF1<0，使SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是．14．（2021秋·湖北荊州·高一荊州市沙市第五中學(xué)?？茧A段練習(xí)）同學(xué)們，你們是否注意到：自然下垂的鐵鏈；空曠的田野上，兩根電線桿之間的電線；峽谷的上空，橫跨深澗的觀光索道的鋼索．這些現(xiàn)象中都有相似的曲線形態(tài)．事實(shí)上，這些曲線在數(shù)學(xué)上常常被稱為懸鏈線．懸鏈線的相關(guān)理論在工程、航海、光學(xué)等方面有廣泛的應(yīng)用．在恰當(dāng)?shù)淖鴺?biāo)系中，這類(lèi)函數(shù)的表達(dá)式可以為SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0是非零常數(shù)，無(wú)理數(shù)SKIPIF1<0…），對(duì)于函數(shù)SKIPIF1<0以下結(jié)論正確的是．①如果SKIPIF1<0，那么函數(shù)SKIPIF1<0為奇函數(shù)；②如果SKIPIF1<0，那么SKIPIF1<0為單調(diào)函數(shù)；③如果SKIPIF1<0，那么函數(shù)SKIPIF1<0沒(méi)有零點(diǎn)；④如果SKIPIF1<0那么函數(shù)SKIPIF1<0的最小值為2．【三層練能力】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知a，b，c滿足SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<02．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0有且僅有兩個(gè)正整數(shù)解（其中SKIPIF1<0為自然對(duì)數(shù)的底數(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<0二、多選題4．（2023·全國(guó)·校聯(lián)考一模）曲線的曲率就是針對(duì)曲線上?個(gè)克的切線方向角對(duì)弧長(zhǎng)的轉(zhuǎn)動(dòng)率，表明曲線偏離直線的程度，曲率越大，表示曲線的彎曲程度越大．曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的曲率SKIPIF1<0SKIPIF1<0，其中SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)．（

）A．若函數(shù)f（x）=SKIPIF1<0．則曲線y=f（x）在點(diǎn)（-SKIPIF1<0，-SKIPIF1<0）與點(diǎn)（SKIPIF1<0，SKIPIF1<0）處的彎曲程度相同B．若SKIPIF1<0是二次函數(shù)．則曲線SKIPIF1<0的曲率在頂點(diǎn)處取得最小值C．若函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0D．若函數(shù)SKIPIF1<0，則曲線SKIPIF1<0上任意一點(diǎn)的曲率的最大值為SKIPIF1<05．（2023春·重慶江北·高二字水中學(xué)?？茧A段練習(xí)）定義：在區(qū)間SKIPIF1<0上，若函數(shù)SKIPIF1<0是減函數(shù)，且SKIPIF1<0是增函數(shù)，則稱SKIPIF1<0在區(qū)間SKIPIF1<0上是“弱減函數(shù)”.根據(jù)定義可得（

）A．SKIPIF1<0在SKIPIF1<0上是“弱減函數(shù)”B．SKIPIF1<0在SKIPIF1<0上是“弱減函數(shù)”C．若SKIPIF1<0在SKIPIF1<0上是“弱減函數(shù)”，則SKIPIF1<0D．若SKIPIF1<0在SKIPIF1<0上是“弱減函數(shù)”，則SKIPIF1<06．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)周期B．SKIPIF1<0是函數(shù)SKIPIF1<0的一條對(duì)稱軸C．函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，最小值為SKIPIF1<0D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增【一層練基礎(chǔ)】參考答案1．B【詳解】首先判斷奇偶性：A,B為偶函數(shù)，C為奇函數(shù)，D既不是奇函數(shù)也不是偶函數(shù),所以排除C、D，對(duì)于先減后增，排除A，故選B.考點(diǎn)：函數(shù)的奇偶性、單調(diào)性.2．B【分析】由分段函數(shù)單調(diào)性列不等式組求解【詳解】SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，由題意得SKIPIF1<0解得SKIPIF1<0，故選：B3．D【分析】根據(jù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的單調(diào)性結(jié)合中間量法即可求解.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選：SKIPIF1<0.4．D【分析】用分離參數(shù)法得出不等式k＞SKIPIF1<0﹣x在x∈[1，2]上成立，根據(jù)函數(shù)f（x）=SKIPIF1<0﹣x在x∈[1，2]上的單調(diào)性，即可求出k的取值范圍．【詳解】關(guān)于x的不等式x2+kx﹣1＞0在區(qū)間[1，2]上有解，∴kx＞1﹣x2在x∈[1，2]上有解，即k＞SKIPIF1<0﹣x在x∈[1，2]上成立；設(shè)函數(shù)f（x）=SKIPIF1<0﹣x，x∈[1，2]，∴f′（x）=﹣SKIPIF1<0﹣1＜0恒成立，∴f（x）在x∈[1，2]上是單調(diào)減函數(shù)，且f（x）的值域?yàn)閇﹣SKIPIF1<0，0]，要k＞SKIPIF1<0﹣x在x∈[1，2]上有解，則k＞﹣SKIPIF1<0，即實(shí)數(shù)k的取值范圍為（﹣SKIPIF1<0，+∞）．故答案為：D【點(diǎn)睛】（1）本題主要考查了不等式的有解問(wèn)題，考查利用導(dǎo)數(shù)求函數(shù)的值域，意在考查學(xué)生對(duì)這些知識(shí)的掌握水平和分析推理能力.(2)處理參數(shù)的問(wèn)題常用的有分離參數(shù)法和分類(lèi)討論法，本題利用的是分離參數(shù)法，解題效率比分類(lèi)討論法解題效率高.5．C【分析】由題意，SKIPIF1<0在SKIPIF1<0上有解，分離參數(shù)SKIPIF1<0轉(zhuǎn)化為值域問(wèn)題即可求解.【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0存在平行于SKIPIF1<0軸的切線，所以SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0在SKIPIF1<0上有解，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故選：C.6．AC【分析】SKIPIF1<0，然后畫(huà)出其圖象可得答案.【詳解】SKIPIF1<0，其大致圖象如下，結(jié)合函數(shù)圖象可得AC正確，BD錯(cuò)誤.故選：AC7．BD【分析】利用誘導(dǎo)公式化簡(jiǎn)可得SKIPIF1<0，可判斷選項(xiàng)A；利用換元法和函數(shù)的單調(diào)性，可判斷選項(xiàng)B和C；利用誘導(dǎo)公式化簡(jiǎn)可得SKIPIF1<0，可判斷選項(xiàng)D．【詳解】對(duì)A：SKIPIF1<0，故SKIPIF1<0不是SKIPIF1<0的周期，A錯(cuò)誤；對(duì)B：令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，又∵SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，B正確；對(duì)C：∵SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，又∵SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上最大值為SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上有最大值SKIPIF1<0，C錯(cuò)誤；對(duì)D：SKIPIF1<0，故SKIPIF1<0圖象的一條對(duì)稱軸為直線SKIPIF1<0，D正確.故選：BD.【點(diǎn)睛】結(jié)論點(diǎn)睛：若SKIPIF1<0，則SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，特別地SKIPIF1<0，則SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱；若SKIPIF1<0，則SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，特別地SKIPIF1<0，則SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱.8．AD【解析】利用函數(shù)的奇偶性的定義判斷奇偶性，根據(jù)函數(shù)解析式判斷單調(diào)性.【詳解】A，因?yàn)镾KIPIF1<0，SKIPIF1<0是偶函數(shù)，在區(qū)間SKIPIF1<0上為增函數(shù)，符合題意；B，因?yàn)镾KIPIF1<0，SKIPIF1<0是奇函數(shù)，且在區(qū)間SKIPIF1<0上為減函數(shù)，不符合題意；C，因?yàn)镾KIPIF1<0，SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，不符合題意；D，因?yàn)镾KIPIF1<0，SKIPIF1<0是偶函數(shù)，且在區(qū)間SKIPIF1<0上為增函數(shù)，符合題意.故選：AD9．AD【分析】根據(jù)函數(shù)的解析式，分別從定義域、奇偶性、零點(diǎn)、最值考察即可求解.【詳解】對(duì)A，由解析式可知SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故A正確；對(duì)B，因?yàn)镾KIPIF1<0，可知SKIPIF1<0是奇函數(shù)，故B不正確；對(duì)C,SKIPIF1<0，得SKIPIF1<0，故C不正確；對(duì)D,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故D正確.故選：AD10．SKIPIF1<0【分析】根據(jù)復(fù)合函數(shù)的單調(diào)性即得.【詳解】函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，在定義域內(nèi)函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是SKIPIF1<0，而函數(shù)SKIPIF1<0的單調(diào)增區(qū)間就是在定義域內(nèi)函數(shù)SKIPIF1<0的增區(qū)間，所以函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0.11．SKIPIF1<0【分析】根據(jù)題意可得SKIPIF1<0，SKIPIF1<0，進(jìn)而將原不等式轉(zhuǎn)換為不等式組，解之即可.【詳解】由題意知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0故答案為：SKIPIF1<0.12．2【分析】分離常量，由函數(shù)SKIPIF1<0可得函數(shù)單調(diào)遞減，然后求解函數(shù)的最大值即可.【詳解】[方法一]：分離常量法由函數(shù)SKIPIF1<0，得SKIPIF1<0在SKIPIF1<0單調(diào)遞減，即SKIPIF1<0SKIPIF1<0.故答案為：2.[方法二]：換元法由函數(shù)SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，可知SKIPIF1<0在SKIPIF1<0是單調(diào)遞減的，即SKIPIF1<0SKIPIF1<0.故答案為：2.[方法三]：反函數(shù)法由函數(shù)SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，從而有SKIPIF1<0，即SKIPIF1<0SKIPIF1<0.故答案為：2.[方法四]：函數(shù)圖像法由函數(shù)SKIPIF1<0SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0單調(diào)遞增，得SKIPIF1<0在SKIPIF1<0單調(diào)遞減，即SKIPIF1<0SKIPIF1<0.故答案為：2.13．SKIPIF1<0【分析】不等式SKIPIF1<0恒成立，即為SKIPIF1<0不大于xy的最小值，運(yùn)用基本不等式，計(jì)算即可得到所求最小值，解不等式可得m的范圍．【詳解】∵正實(shí)數(shù)x，y滿足SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，由SKIPIF1<0恒成立，可得SKIPIF1<0，解得SKIPIF1<0故答案為：SKIPIF1<014．SKIPIF1<0【分析】先判斷出SKIPIF1<0，且SKIPIF1<0.令SKIPIF1<0，利用判別式法求出SKIPIF1<0的最小值.【詳解】因?yàn)閷?shí)數(shù)a，b滿足SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，代入SKIPIF1<0，則有SKIPIF1<0，所以關(guān)于b的一元二次方程SKIPIF1<0有正根，只需SKIPIF1<0，解得：SKIPIF1<0.此時(shí)，關(guān)于b的一元二次方程SKIPIF1<0的兩根SKIPIF1<0，所以兩根同號(hào)，只需SKIPIF1<0，解得SKIPIF1<0.綜上所述：SKIPIF1<0.即SKIPIF1<0的最小值是SKIPIF1<0（此時(shí)SKIPIF1<0，解得：SKIPIF1<0）.故答案為：SKIPIF1<0.【二層練綜合】參考答案1．A【分析】SKIPIF1<0可化為SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，再結(jié)合奇偶性可知該函數(shù)在R上單調(diào)遞增，又將所求不等式變形，即可由單調(diào)性解該抽象不等式.【詳解】根據(jù)題意可知，SKIPIF1<0可轉(zhuǎn)化為SKIPIF1<0，所以SKIPIF1<0在[0，+∞)上是增函數(shù)，又SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0在R上為增函數(shù)，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即x的取值范圍是SKIPIF1<0.故選：A.【關(guān)鍵點(diǎn)點(diǎn)睛】本題的關(guān)鍵是將不等式SKIPIF1<0化為SKIPIF1<0，從而構(gòu)造函數(shù)SKIPIF1<0，再根據(jù)奇偶性和單調(diào)性解抽象不等式.2．D【分析】先將函數(shù)化簡(jiǎn)轉(zhuǎn)化，判斷函數(shù)的單調(diào)性，根據(jù)條件求出SKIPIF1<0的值，即可求出SKIPIF1<0的取值范圍.【詳解】函數(shù)SKIPIF1<0，可以判斷函數(shù)在區(qū)間(－1，＋∞)上是減函數(shù)，且f（2）＝0，所以n＝2，根據(jù)題意，x∈(m，n]時(shí)，ymin＝0，∴m的取值范圍是[－1，2).故選：D.【點(diǎn)睛】本題考查已知給定函數(shù)在未知區(qū)間的最值，求區(qū)間內(nèi)參數(shù)范圍，判斷函數(shù)的單調(diào)性是解決問(wèn)題的關(guān)鍵.3．A【分析】先判斷SKIPIF1<0的單調(diào)性，然后對(duì)SKIPIF1<0進(jìn)行分類(lèi)討論，由此求得SKIPIF1<0的取值范圍.【詳解】由于函數(shù)SKIPIF1<0在定義域上單調(diào)遞增，所以函數(shù)SKIPIF1<0在定義域上是單調(diào)遞增函數(shù)．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在定義域上不單調(diào)，不符合題意；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0圖象的對(duì)稱軸為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，不符合題意，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，要使函數(shù)SKIPIF1<0在定義域上單調(diào)遞增，則需SKIPIF1<0，解得SKIPIF1<0．故實(shí)數(shù)t的取值范圍為SKIPIF1<0．故選：A4．B【解析】根據(jù)已知可得判斷“成功函數(shù)”為定義域上單調(diào)遞增的奇函數(shù)，逐項(xiàng)判斷，即可得出結(jié)論.【詳解】由任意SKIPIF1<0，都有SKIPIF1<0知SKIPIF1<0是奇函數(shù)，由任意SKIPIF1<0且SKIPIF1<0，都有SKIPIF1<0，知SKIPIF1<0是增函數(shù)，因?yàn)镾KIPIF1<0在定義域上是奇函數(shù)，但在定義域上不是單增函數(shù)，故A錯(cuò)；因?yàn)镾KIPIF1<0是奇函數(shù)，SKIPIF1<0，所以在定義域上是增函數(shù)，故B正確；因?yàn)镾KIPIF1<0在定義域SKIPIF1<0是減函數(shù)，故C錯(cuò)；因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D錯(cuò).故選:B.【點(diǎn)睛】本題考查函數(shù)的奇偶性和單調(diào)性的判斷，熟練掌握初等函數(shù)單調(diào)性，以及應(yīng)用導(dǎo)數(shù)法判斷函數(shù)的單調(diào)性，屬于中檔題.5．D【分析】由已知結(jié)合奇函數(shù)定義先求出當(dāng)SKIPIF1<0時(shí)的函數(shù)解析式，然后利用導(dǎo)數(shù)對(duì)SKIPIF1<0進(jìn)行分類(lèi)討論，確定函數(shù)單調(diào)性，進(jìn)而可求．【詳解】因?yàn)镾KIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上恒成立，函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，函數(shù)取得最小值SKIPIF1<0，解得SKIPIF1<0（舍SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，函數(shù)單調(diào)遞減；SKIPIF1<0，SKIPIF1<0，函數(shù)單調(diào)遞增，SKIPIF1<0時(shí)，函數(shù)取得最小值SKIPIF1<0，解得SKIPIF1<0，綜上，SKIPIF1<0．故選：D．6．BD【分析】根據(jù)題意，結(jié)合三角函數(shù)的圖象性質(zhì)以及圖象的變換，一一判斷即可.【詳解】對(duì)于選項(xiàng)A，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0SKIPIF1<0上單調(diào)遞減，故A錯(cuò)；對(duì)于選項(xiàng)B，結(jié)合SKIPIF1<0的圖象性質(zhì)，易知SKIPIF1<0是以SKIPIF1<0為周期且在SKIPIF1<0上單調(diào)遞增的偶函數(shù)，故B正確；對(duì)于選項(xiàng)C，結(jié)合SKIPIF1<0的圖象性質(zhì)，易知SKIPIF1<0沒(méi)有周期性，故C錯(cuò)；對(duì)于選項(xiàng)D，令SKIPIF1<0，易知SKIPIF1<0是以SKIPIF1<0為周期且在SKIPIF1<0上單調(diào)遞增的偶函數(shù)，因SKIPIF1<0也是單調(diào)遞增的，所以SKIPIF1<0是以SKIPIF1<0為周期且在SKIPIF1<0上單調(diào)遞增的偶函數(shù)，故D正確.故選：BD.7．BCD【分析】根據(jù)題意，利用奇偶性，單調(diào)性，依次分析選項(xiàng)是否正確，即可得到答案【詳解】對(duì)于A：SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0為奇函數(shù)，故A錯(cuò)誤；對(duì)于B：SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0為奇函數(shù)，故B正確；對(duì)于C：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0都為奇函數(shù)，則SKIPIF1<0為奇函數(shù)，SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值互為相反數(shù)，必有SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值之和為0，故C正確；對(duì)于D：SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上為減函數(shù)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上為減函數(shù)，則SKIPIF1<0在SKIPIF1<0上為減函數(shù)，若SKIPIF1<0即SKIPIF1<0，則必有SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0的解集為SKIPIF1<0，故D正確；故選：BCD8．BD【分析】由奇偶函數(shù)的單調(diào)性的關(guān)系確定兩函數(shù)的單調(diào)性，再結(jié)合SKIPIF1<0，SKIPIF1<0逐項(xiàng)判斷即可.【詳解】因?yàn)镾KIPIF1<0是定義在R上的偶函數(shù)，SKIPIF1<0是定義在R上的奇函數(shù)，且兩函數(shù)在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以BD正確，C錯(cuò)誤；若SKIPIF1<0，則SKIPIF1<0，A錯(cuò)誤.故選：BD9．AC【分析】求出函數(shù)式確定單調(diào)性判斷A；舉特例說(shuō)明判斷B，D；變形函數(shù)式，分析計(jì)算判斷C作答.【詳解】對(duì)于A，SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，A正確；對(duì)于B，SKIPIF1<0，則SKIPIF1<0，B不正確；對(duì)于C，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0的值域?yàn)镾KIPIF1<0，C正確；對(duì)于D，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，D不正確.故選：AC10．1【分析】分SKIPIF1<0，SKIPIF1<0討論，利用函數(shù)的單調(diào)性求最值即得.【詳解】由題意得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，∴SKIPIF1<0的最小值為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0不成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0的最小值為SKIPIF1<0，符合題意.故SKIPIF1<0.故答案為：1.11．1【分析】依據(jù)題意列出關(guān)于a的方程即可求得正實(shí)數(shù)a的值.【詳解】令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0令SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0則SKIPIF1<0，即SKIPIF1<0的最大值為SKIPIF1<0則SKIPIF1<0，解之得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立）則SKIPIF1<0，即SKIPIF1<0的最大值為SKIPIF1<0則SKIPIF1<0，解之得SKIPIF1<0（舍）綜上，所求正實(shí)數(shù)SKIPIF1<0故答案為：112．SKIPIF1<0【分析】利用基本不等式SKIPIF1<0的最小值，由此可得出實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，由基本不等式可得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0，因此實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.13．SKIPIF1<0【分析】利用不等式的基本性質(zhì)分離參數(shù)，利用函數(shù)的單調(diào)性求相應(yīng)最值即可得到結(jié)論.【詳解】由SKIPIF1<0可得，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，根據(jù)題意，SKIPIF1<0即可，設(shè)SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<014．②③【分析】利用函數(shù)的奇偶性，單調(diào)性，零點(diǎn)和基本不等式等性質(zhì)對(duì)①②③④逐一分析即可得到結(jié)論.【詳解】對(duì)①：當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，此時(shí)SKIPIF1<0為偶函數(shù)，故①錯(cuò)誤.對(duì)②：當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，函數(shù)SKIPIF1<0在其定義域上為單調(diào)遞增函數(shù)，函數(shù)SKIPIF1<0在其定義域上也為單調(diào)遞增函數(shù)，故函數(shù)SKIPIF1<0在其定義域上為單調(diào)遞增函數(shù)；當(dāng)SKIPIF1<0，函數(shù)SKIPIF1<0在其定義域上為單調(diào)遞減函數(shù)，函數(shù)SKIPIF1<0在其定義域上也為單調(diào)遞減函數(shù)，故函數(shù)SKIPIF1<0在其定義域上為單調(diào)遞減函數(shù)；綜上：如果SKIPIF1<0，那么SKIPIF1<0為單調(diào)函數(shù)；故②正確.對(duì)③：當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0；綜上：如果SKIPIF1<0，那么函數(shù)SKIPIF1<0沒(méi)有零點(diǎn)；故③正確.對(duì)④：由SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0；故SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0沒(méi)有最小值；故④錯(cuò)誤.故答案為：②③【點(diǎn)睛】本題考查了函數(shù)的奇偶性和最值的應(yīng)用，考查基本不等式，考查指數(shù)函數(shù)的性質(zhì)，屬于中檔題.【三層練能力】參考答案1．B【分析】構(gòu)造函數(shù)SKIPIF1<0，利用其單調(diào)性，分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0討論即可.【詳解】由題意得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，根據(jù)減函數(shù)加減函數(shù)為減函數(shù)的結(jié)論知：SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，SKIPIF1<0，兩邊同取以5為底的對(duì)數(shù)得SKIPIF1<0，對(duì)SKIPIF1<0通過(guò)移項(xiàng)得SKIPIF1<0，兩邊同取以3為底的對(duì)數(shù)得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，故此時(shí)，SKIPIF1<0，故C,D選項(xiàng)錯(cuò)誤，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，故A錯(cuò)誤,下面嚴(yán)格證明當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0根據(jù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，下面證明：SKIPIF1<0，SKIPIF1<0要證：SKIPIF1<0，即證：SKIPIF1<0，等價(jià)于證明SKIPIF1<0，即證：SKIPIF1<0，此式開(kāi)頭已證明，對(duì)SKIPIF1<0，左邊同除分子分母同除SKIPIF1<0，右邊分子分母同除SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，SKIPIF1<0，兩邊同取以5為底的對(duì)數(shù)得SKIPIF1<0，對(duì)SKIPIF1<0通過(guò)移項(xiàng)得SKIPIF1<0，兩邊同取以3為底的對(duì)數(shù)得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，故SKIPIF1<0，故此時(shí)，SKIPIF1<0，下面嚴(yán)格證明當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，根據(jù)函數(shù)SKIPIF1<0，且其在SKIPIF1<0上單調(diào)遞減，可知SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，根據(jù)函數(shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，下面證明：SKIPIF1<0，要證：SKIPIF1<0即證：SKIPIF1<0，等價(jià)于證SKIPIF1<0，即證：SKIPIF1<0，此式已證明，對(duì)SKIPIF1<0，左邊同除分子分母同除SKIPIF1<0，右邊分子分母同除SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，綜上SKIPIF1<0，SKIPIF1<0，故選：B.【點(diǎn)睛】關(guān)鍵點(diǎn)睛：本題的關(guān)鍵在于構(gòu)造函數(shù)SKIPIF1<0，利用其單調(diào)性及SKIPIF1<0，從而得到SKIPIF1<0之間的大小關(guān)系，同時(shí)需要先求出SKIPIF1<0的范圍，然后再對(duì)SKIPIF1<0進(jìn)行分類(lèi)討論.2．D【分析】由SKIPIF1<0，可得SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，利用函數(shù)的導(dǎo)數(shù)與單調(diào)性的關(guān)系，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，進(jìn)而可得SKIPIF1<0，SKIPIF1<0，從而即可得答案.【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0；令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0；同理SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.綜上，SKIPIF1<0，故選：D.3．D【分析】問(wèn)題轉(zhuǎn)化為SKIPIF1<0（SKIPIF1<0）有且僅有兩個(gè)正整數(shù)解，討論SKIPIF1<0、SKIPIF1<0并構(gòu)造SKIPIF1<0、SKIPIF1<0，利用導(dǎo)數(shù)研究單調(diào)性，進(jìn)而數(shù)形結(jié)合列出不等式組求參數(shù)范圍.【詳解】當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，可得SKIPIF1<0（SKIPIF1<0），顯然當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0在SKIPIF1<0恒成立，不合題意；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，又SKIPIF1<0且SKIPIF1<0趨向正無(wú)窮時(shí)SKIPIF1<0趨向0，故SKIPIF1<0，綜上，SKIPIF1<0圖象如下：由圖知：要使SKIPIF1<0有兩個(gè)正整數(shù)解，則SKIPI