新高考數(shù)學(xué)二輪復(fù)習(xí)分層訓(xùn)練專題04 函數(shù)及其性質(zhì)（解析版）

答案第=page11頁(yè)，共=sectionpages22頁(yè)解密04講：函數(shù)及其性質(zhì)【練基礎(chǔ)】一、單選題1．（2018·全國(guó)·高三課時(shí)練習(xí)（文））已知集合SKIPIF1<0，SKIPIF1<0，若SKIPIF1<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<0【答案】A【詳解】由已知得SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故選A.2．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的值域是SKIPIF1<0

SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由SKIPIF1<0，知SKIPIF1<0，解得SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0.，即為SKIPIF1<0和SKIPIF1<0兩函數(shù)圖象有交點(diǎn)，作出函數(shù)圖象，如圖所示：由圖可知，當(dāng)直線和半圓相切時(shí)SKIPIF1<0最小，當(dāng)直線過(guò)點(diǎn)A(4,0)時(shí)，SKIPIF1<0最大.當(dāng)直線和半圓相切時(shí)，SKIPIF1<0，解得SKIPIF1<0，由圖可知SKIPIF1<0.當(dāng)直線過(guò)點(diǎn)A(4,0)時(shí)，SKIPIF1<0，解得SKIPIF1<0.所以SKIPIF1<0，即SKIPIF1<0.故選A.3．（2020·全國(guó)·高三課時(shí)練習(xí)（理））已知SKIPIF1<0是定義在R上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0(m為常數(shù))，則SKIPIF1<0的值為(

)A．4 B．6 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】先由函數(shù)在R上是奇函數(shù)求出參數(shù)m的值，求函數(shù)函數(shù)的解板式，再由奇函數(shù)的性質(zhì)得到f（﹣log35）=﹣f（log35）代入解析式即可求得所求的函數(shù)值.【詳解】由題意，f（x）是定義在R上的奇函數(shù)，當(dāng)x≥0時(shí)f（x）=3x+m（m為常數(shù)），∴f（0）=30+m=0，解得m=﹣1，故有x≥0時(shí)f（x）=3x﹣1∴f（﹣log35）=﹣f（log35）=﹣（SKIPIF1<0）=﹣4故選C．【點(diǎn)睛】本題考查函數(shù)奇偶性質(zhì)，解題的關(guān)鍵是利用f（0）=0求出參數(shù)m的值，再利用性質(zhì)轉(zhuǎn)化求值，本題考查了轉(zhuǎn)化的思想，方程的思想．4．（2021·全國(guó)·高三專題練習(xí)（理））設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若對(duì)任意SKIPIF1<0，都有SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】分別求得SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，對(duì)應(yīng)函數(shù)SKIPIF1<0的值域，根據(jù)二次函數(shù)圖像及性質(zhì)可知SKIPIF1<0時(shí)，令SKIPIF1<0，可解得SKIPIF1<0的最大值.【詳解】SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，值域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，值域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，值域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，對(duì)任意SKIPIF1<0都有SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0，故選：C【點(diǎn)睛】本題主要考查了函數(shù)與方程的應(yīng)用問(wèn)題，二次函數(shù)的圖象與性質(zhì)，也考查了運(yùn)算與求解能力，以及分類討論的解題思想，屬于中檔題.5．（2019·天津·高考真題（文））已知函數(shù)，若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有兩個(gè)互異的實(shí)數(shù)解，則SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】畫(huà)出SKIPIF1<0圖象及直線SKIPIF1<0，借助圖象分析．【詳解】如圖，當(dāng)直線SKIPIF1<0位于SKIPIF1<0點(diǎn)及其上方且位于SKIPIF1<0點(diǎn)及其下方，或者直線SKIPIF1<0與曲線SKIPIF1<0相切在第一象限時(shí)符合要求．即SKIPIF1<0，即SKIPIF1<0，或者SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0．故選D．【點(diǎn)睛】根據(jù)方程實(shí)根個(gè)數(shù)確定參數(shù)范圍，常把其轉(zhuǎn)化為曲線交點(diǎn)個(gè)數(shù)，特別是其中一條為直線時(shí)常用此法．6．（2022·四川省仁壽縣文宮中學(xué)高三開(kāi)學(xué)考試（理））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<0【答案】A【分析】由分段函數(shù)單調(diào)遞增的特性結(jié)合單調(diào)增函數(shù)的圖象特征列出不等式組求解即得.【詳解】因函數(shù)SKIPIF1<0在R上單調(diào)遞增，則有SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0在SKIPIF1<0上也遞增，根據(jù)增函數(shù)圖象特征知，點(diǎn)SKIPIF1<0不能在點(diǎn)SKIPIF1<0上方，于是得SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)a的取值范圍是SKIPIF1<0.故選：A7．（2019·全國(guó)·高考真題（理））函數(shù)SKIPIF1<0在SKIPIF1<0的圖像大致為（）A． B．C． D．【答案】B【分析】由分子、分母的奇偶性，易于確定函數(shù)為奇函數(shù)，由SKIPIF1<0的近似值即可得出結(jié)果．【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，圖象關(guān)于原點(diǎn)成中心對(duì)稱，排除選項(xiàng)C．又SKIPIF1<0排除選項(xiàng)D；SKIPIF1<0，排除選項(xiàng)A，故選B．【點(diǎn)睛】本題通過(guò)判斷函數(shù)的奇偶性，縮小考察范圍，通過(guò)計(jì)算特殊函數(shù)值，最后做出選擇．本題較易，注重了基礎(chǔ)知識(shí)、基本計(jì)算能力的考查．8．（2022·河南·南陽(yáng)中學(xué)模擬預(yù)測(cè)（文））已知定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則下列結(jié)論一定正確的是（

）A．SKIPIF1<0 B．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．函數(shù)SKIPIF1<0是奇函數(shù) D．SKIPIF1<0【答案】B【分析】推導(dǎo)出SKIPIF1<0可判斷A選項(xiàng)的正誤；推導(dǎo)出SKIPIF1<0可判斷B選項(xiàng)的正誤；分析得出SKIPIF1<0可判斷C選項(xiàng)的正誤；推導(dǎo)出SKIPIF1<0可判斷D選項(xiàng)的正誤.【詳解】對(duì)于A選項(xiàng)，因?yàn)镾KIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，A錯(cuò)；對(duì)于B選項(xiàng)，因?yàn)镾KIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，故函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，B對(duì)；對(duì)于C選項(xiàng)，因?yàn)镾KIPIF1<0，故函數(shù)SKIPIF1<0是偶函數(shù)，C錯(cuò)；對(duì)于D選項(xiàng)，因?yàn)镾KIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，D錯(cuò).故選：B.二、多選題9．（2022·全國(guó)·高三專題練習(xí)）已知偶函數(shù)SKIPIF1<0滿足：SKIPIF1<0，且當(dāng)0≤x≤2時(shí)，SKIPIF1<0，則下列說(shuō)法正確的是（

）A．－2≤x≤0時(shí)，SKIPIF1<0B．點(diǎn)（1，0）是f(x)圖象的一個(gè)對(duì)稱中心C．f(x)在區(qū)間[－10，10]上有10個(gè)零點(diǎn)D．對(duì)任意SKIPIF1<0，都有SKIPIF1<0【答案】AC【分析】由偶函數(shù)的定義得解析式，判斷A，由SKIPIF1<0上的解析式判斷B，已知條件得SKIPIF1<0是一條對(duì)稱軸，這樣函數(shù)SKIPIF1<0是周期函數(shù)，周期為4，利用周期性可判斷零點(diǎn)個(gè)數(shù)，判斷C，由最值判斷D．【詳解】因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0時(shí)，SKIPIF1<0，A正確；在SKIPIF1<0上，SKIPIF1<0不關(guān)于SKIPIF1<0對(duì)稱，因此SKIPIF1<0不是SKIPIF1<0的一個(gè)對(duì)稱中心，B錯(cuò)；由SKIPIF1<0得SKIPIF1<0，因此在SKIPIF1<0上，SKIPIF1<0有兩個(gè)零點(diǎn)，又SKIPIF1<0，所以SKIPIF1<0是函數(shù)圖象的一條對(duì)稱軸，SKIPIF1<0，所以SKIPIF1<0是周期函數(shù)，周期為4，因此SKIPIF1<0在SKIPIF1<0上各有2個(gè)零點(diǎn)，在SKIPIF1<0上共有10個(gè)零點(diǎn)，C正確；由周期性知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，D錯(cuò)．故選：AC．【點(diǎn)睛】思路點(diǎn)睛：本題考查函數(shù)的奇偶性、對(duì)稱性與周期性，解題關(guān)鍵是由兩個(gè)對(duì)稱性得出函數(shù)具有周期性，因此只要在一個(gè)周期內(nèi)確定函數(shù)的零點(diǎn)，從而可得函數(shù)的性質(zhì)可得整個(gè)定義域上函數(shù)的性質(zhì)．10．（2021·福建·寧化濱江實(shí)驗(yàn)中學(xué)高三期中）下列函數(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<0【答案】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：AD11．（2022·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0是定義在R上的偶函數(shù)，且對(duì)任意SKIPIF1<0，有SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則（

）A．SKIPIF1<0是以2為周期的周期函數(shù)B．點(diǎn)SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)對(duì)稱中心C．SKIPIF1<0D．函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)【答案】BD【分析】首先根據(jù)函數(shù)的對(duì)稱性求出SKIPIF1<0的周期和對(duì)稱中心，然后求得SKIPIF1<0.利用圖象法即可判斷D.【詳解】依題意，SKIPIF1<0為偶函數(shù)，且SKIPIF1<0，有SKIPIF1<0，即SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，則SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是周期為4的周期函數(shù)，故A錯(cuò)誤；因?yàn)镾KIPIF1<0的周期為4，SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，所以SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)對(duì)稱中心，故B正確；因?yàn)镾KIPIF1<0的周期為4，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故C錯(cuò)誤；作函數(shù)SKIPIF1<0和SKIPIF1<0的圖象如下圖所示，由圖可知，兩個(gè)函數(shù)圖象有3個(gè)交點(diǎn)，所以函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，故D正確.故選：BD.12．（2020·全國(guó)·模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則下列說(shuō)法正確的是（

）A．4是函數(shù)SKIPIF1<0的周期B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0C．函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱D．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】ACD【解析】選項(xiàng)A.由SKIPIF1<0奇函數(shù)結(jié)合條件可得SKIPIF1<0，可判斷;

選項(xiàng)B.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0可判斷;選項(xiàng)C.由條件可得SKIPIF1<0，可判斷;

選項(xiàng)D.由條件可得SKIPIF1<0，即SKIPIF1<0，從而可判斷.【詳解】由函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)及SKIPIF1<0可得SKIPIF1<0，所以4是函數(shù)SKIPIF1<0的周期，故A正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B錯(cuò)誤；由SKIPIF1<0及SKIPIF1<0為奇函數(shù)可得SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故C正確；易知SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，故D正確．故選：ACD三、填空題13．（2007·重慶·高考真題（理））若函數(shù)f(x)=SKIPIF1<0的定義域?yàn)镽，則SKIPIF1<0的取值范圍為_(kāi)______.【答案】SKIPIF1<0【詳解】SKIPIF1<0恒成立，SKIPIF1<0恒成立，SKIPIF1<014．（2021·廣東·高州一中高三階段練習(xí)）已知y=f(x)的圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，且對(duì)任意的x∈R，f(x+2)=f(-x)恒成立，當(dāng)SKIPIF1<0時(shí)，f(x)=2x，則f(2021)=_____________.【答案】SKIPIF1<0【分析】由已知條件推出函數(shù)SKIPIF1<0的周期，利用函數(shù)的周期和奇偶性求值即可．【詳解】y=f(x)的圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，則SKIPIF1<0又SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0的周期為SKIPIF1<0SKIPIF1<0故答案為：SKIPIF1<015．（2021·全國(guó)·高三專題練習(xí)）已知f（x）是定義域?yàn)镽的偶函數(shù)，且在區(qū)間（-∞，0]上單調(diào)遞增，則不等式，f（3x-1）＞f（2）的解集是________．【答案】SKIPIF1<0【解析】根據(jù)函數(shù)的奇偶性，可知函數(shù)在SKIPIF1<0上遞減，即可求解.【詳解】因?yàn)閒（x）是定義域?yàn)镽的偶函數(shù)，且在區(qū)間（-∞，0]上單調(diào)遞增，所以函數(shù)在SKIPIF1<0上遞減，因?yàn)閒（3x-1）＞f（2），所以SKIPIF1<0所以SKIPIF1<0即-2＜3x-1＜2，解得SKIPIF1<0．故答案為：SKIPIF1<016．（2022·全國(guó)·高三專題練習(xí)（文））若“SKIPIF1<0，使得SKIPIF1<0成立”是假命題，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)__________.【答案】SKIPIF1<0【分析】轉(zhuǎn)化為“SKIPIF1<0，使得SKIPIF1<0成立”是真命題，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】若“SKIPIF1<0，使得SKIPIF1<0成立”是假命題，則“SKIPIF1<0，使得SKIPIF1<0成立”是真命題，分離SKIPIF1<0，進(jìn)而SKIPIF1<0.【點(diǎn)睛】本題考查存在性命題的真假判定，涉及不等式的恒成立問(wèn)題，函數(shù)的單調(diào)性和最值問(wèn)題，轉(zhuǎn)化為“SKIPIF1<0，使得SKIPIF1<0成立”是真命題是關(guān)鍵步驟，分離參數(shù)法是本題的關(guān)鍵思想方法.四、解答題17．（2019·全國(guó)·高三專題練習(xí)（文））設(shè)SKIPIF1<0，且SKIPIF1<0.（1）求實(shí)數(shù)SKIPIF1<0的值及函數(shù)SKIPIF1<0的定義域；（2）求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值.【答案】(1)SKIPIF1<0;SKIPIF1<0;(2)1.【分析】(1)代入可求出參數(shù)SKIPIF1<0的值，根據(jù)對(duì)數(shù)函數(shù)的真數(shù)大于零，求出函數(shù)的定義域；（2）先化簡(jiǎn)函數(shù)SKIPIF1<0的解析式，再根據(jù)二次函數(shù)性質(zhì)求最值.【詳解】（1）∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.由SKIPIF1<0得SKIPIF1<0，∴函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.（2）SKIPIF1<0.∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是增函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是減函數(shù)，故函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值是SKIPIF1<0.【點(diǎn)睛】研究二次函數(shù)最值，一般通過(guò)研究對(duì)稱軸與定義區(qū)間位置關(guān)系得函數(shù)單調(diào)性，再根據(jù)單調(diào)性確定函數(shù)最值取法.18．（2022·全國(guó)·高三專題練習(xí)（理））已知函數(shù)SKIPIF1<0定義在SKIPIF1<0上有SKIPIF1<0恒成立，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.（1）求SKIPIF1<0的值及函數(shù)SKIPIF1<0的解析式；（2）求函數(shù)SKIPIF1<0的值域.【答案】（1）SKIPIF1<0，SKIPIF1<0（2）SKIPIF1<0【分析】（1）利用奇函數(shù)的性質(zhì)進(jìn)行計(jì)算．（2）利用換元法結(jié)合一元二次函數(shù)的性質(zhì)求出當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的取值范圍，再根據(jù)奇函數(shù)的性質(zhì)，即可求出函數(shù)的值域．【詳解】解：（1）因?yàn)楹瘮?shù)SKIPIF1<0定義在SKIPIF1<0上有SKIPIF1<0恒成立所以函數(shù)SKIPIF1<0為奇函數(shù)，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0．所以SKIPIF1<0，因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，所以SKIPIF1<0，即SKIPIF1<0．所以函數(shù)SKIPIF1<0的解析式為SKIPIF1<0．（2）令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0可寫(xiě)為SKIPIF1<0，所以SKIPIF1<0．由SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0．即函數(shù)的值域?yàn)镾KIPIF1<0．19．（2022·廣東·小欖中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0是奇函數(shù)．(1)求SKIPIF1<0的值；(2)已知SKIPIF1<0，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）由SKIPIF1<0求出參數(shù)值，再檢驗(yàn)即可；（2）先判斷函數(shù)SKIPIF1<0的單調(diào)性，然后根據(jù)單調(diào)性列出不等式求解即可．【詳解】（1）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，又因?yàn)镾KIPIF1<0是奇函數(shù)，則SKIPIF1<0，解得SKIPIF1<0；經(jīng)檢驗(yàn)SKIPIF1<0，故SKIPIF1<0成立；（2）因?yàn)镾KIPIF1<0對(duì)任意SKIPIF1<0，有SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增又SKIPIF1<0，所以SKIPIF1<0解得SKIPIF1<020．（2022·河南·濮陽(yáng)南樂(lè)一高高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0是偶函數(shù).當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.(1)求函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，求實(shí)數(shù)a的取值范圍；(3)已知SKIPIF1<0，試討論SKIPIF1<0的零點(diǎn)個(gè)數(shù)，并求對(duì)應(yīng)的m的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0(3)答案見(jiàn)解析【分析】（1）根據(jù)偶函數(shù)的定義求解即可.（2）根據(jù)（1）做出SKIPIF1<0圖像，數(shù)形結(jié)合.（3）根據(jù)（1）做出SKIPIF1<0圖像，數(shù)形結(jié)合.【詳解】（1）設(shè)SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0∵SKIPIF1<0為偶函數(shù)∴SKIPIF1<0綜上，有SKIPIF1<0（2）由（1）作出SKIPIF1<0的圖像如圖：因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上具有單調(diào)性，由圖可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0或SKIPIF1<0.（3）由（1）作出SKIPIF1<0的圖像如圖：由圖像可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有兩個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有四個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有六個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有三個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0沒(méi)有零點(diǎn).21．（2022·陜西·漢中市龍崗學(xué)校高三階段練習(xí)（文））設(shè)函數(shù)SKIPIF1<0．(1)求不等式SKIPIF1<0的解集；(2)若SKIPIF1<0的最大值為m，實(shí)數(shù)a，b滿足SKIPIF1<0，求SKIPIF1<0的最小值．【答案】(1)SKIPIF1<0；(2)1【分析】（1）將函數(shù)寫(xiě)成分段函數(shù)，再分類討論，即可求出不等式的解集；（2）首先得到SKIPIF1<0的函數(shù)圖象，即可得到SKIPIF1<0，則SKIPIF1<0，在根據(jù)SKIPIF1<0的幾何意義計(jì)算可得；【詳解】（1）解：SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0，即不等式的解集為SKIPIF1<0．（2）解：由（1）可得SKIPIF1<0的函數(shù)圖象如下所示：所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0的幾何意義為圓SKIPIF1<0上的點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0距離的平方，顯然SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0的最小值為SKIPIF1<0．22．（2022·福建省漳州市第八中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且它的圖象關(guān)于直線SKIPIF1<0對(duì)稱.(1)求證：SKIPIF1<0是周期為4的周期函數(shù)；(2)若SKIPIF1<0，求SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的解析式.【答案】(1)證明見(jiàn)解析(2)SKIPIF1<0【分析】(1)由函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，可得SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0是奇函數(shù)，所以SKIPIF1<0，從而得SKIPIF1<0，即可得周期為4；(2)先求得SKIPIF1<0時(shí)，SKIPIF1<0，再結(jié)合周期為4，即求得SKIPIF1<0在SKIPIF1<0上的解析式.【詳解】（1）解：證明：由函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，有SKIPIF1<0，即有SKIPIF1<0，又函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，有SKIPIF1<0，故SKIPIF1<0，從而SKIPIF1<0，即SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)；（2）解：由函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，有SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，從而，SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的解析式為SKIPIF1<0【提能力】一、單選題1．（2022·天津市建華中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】先求出SKIPIF1<0的定義域，再根據(jù)分母不為零和前者可求題設(shè)中函數(shù)的定義域.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故函數(shù)SKIPIF1<0中的SKIPIF1<0需滿足：SKIPIF1<0，故SKIPIF1<0，故函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故選：C2．（2017·河北定州中學(xué)高三階段練習(xí)）若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】令SKIPIF1<0，則SKIPIF1<0取遍SKIPIF1<0上的所有實(shí)數(shù)，就SKIPIF1<0結(jié)合對(duì)應(yīng)函數(shù)的圖象可得實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】由值域?yàn)镾KIPIF1<0，可知SKIPIF1<0取遍SKIPIF1<0上的所有實(shí)數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0能取遍SKIPIF1<0上的所有實(shí)數(shù)，只需定義域滿足SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，要保證SKIPIF1<0能取遍SKIPIF1<0上的所有實(shí)數(shù)，需SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故選：D.【點(diǎn)睛】本題考查函數(shù)的值域，要注意定義域是SKIPIF1<0、與值域是為SKIPIF1<0的兩個(gè)題型的區(qū)別，值域?yàn)镾KIPIF1<0，可知SKIPIF1<0取遍SKIPIF1<0上的所有實(shí)數(shù)，而定義域是SKIPIF1<0，是SKIPIF1<0SKIPIF1<0恒成立．3．（2022·湖北·南漳縣第二中學(xué)高三階段練習(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<0【答案】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：D4．（2022·天津市濱海新區(qū)塘沽第一中學(xué)高三階段練習(xí)）已知奇函數(shù)SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上是增函數(shù).若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則a，b，c的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0是奇函數(shù)，從而SKIPIF1<0是SKIPIF1<0上的偶函數(shù)，且在SKIPIF1<0上是增函數(shù)，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，所以即SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故選C．【考點(diǎn)】指數(shù)、對(duì)數(shù)、函數(shù)的單調(diào)性【名師點(diǎn)睛】比較大小是高考常見(jiàn)題，指數(shù)式、對(duì)數(shù)式的比較大小要結(jié)合指數(shù)函數(shù)、對(duì)數(shù)函數(shù)，借助指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的圖象，利用函數(shù)的單調(diào)性進(jìn)行比較大小，特別是靈活利用函數(shù)的奇偶性和單調(diào)性數(shù)形結(jié)合不僅能比較大小，還可以解不等式.5．（2022·天津市第四中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0的圖像為（

）A． B．C． D．【答案】D【分析】分析函數(shù)SKIPIF1<0的定義域、奇偶性、單調(diào)性及其在SKIPIF1<0上的函數(shù)值符號(hào)，結(jié)合排除法可得出合適的選項(xiàng).【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，函數(shù)SKIPIF1<0為奇函數(shù)，A選項(xiàng)錯(cuò)誤；又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，C選項(xiàng)錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0函數(shù)單調(diào)遞增，故B選項(xiàng)錯(cuò)誤；故選：D.6．（2020·山西省新絳中學(xué)校高三階段練習(xí)（文））已知定義在R上的函數(shù)SKIPIF1<0為偶函數(shù),記SKIPIF1<0SKIPIF1<0,則SKIPIF1<0,的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由SKIPIF1<0為偶函數(shù)得SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故選B.考點(diǎn)：本題主要考查函數(shù)奇偶性及對(duì)數(shù)運(yùn)算.7．（2022·河北·安新縣第二中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0對(duì)任意SKIPIF1<0，都有SKIPIF1<0的圖形關(guān)于SKIPIF1<0對(duì)稱，且SKIPIF1<0

則SKIPIF1<0（

）A．-1 B．1 C．0 D．2【答案】B【分析】根據(jù)題意得到函數(shù)周期為12，函數(shù)為奇函數(shù)，據(jù)此得到SKIPIF1<0，計(jì)算得到答案.【詳解】SKIPIF1<0函數(shù)周期為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖形關(guān)于SKIPIF1<0對(duì)稱，故SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，SKIPIF1<0.故SKIPIF1<0.故選：B.8．（2021·四川·眉山市彭山區(qū)第一中學(xué)高三階段練習(xí)（理））已知定義在R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0為偶函數(shù)，若SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞減，則下面結(jié)論正確的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由題意可判斷函數(shù)f（x）的周期為6，對(duì)稱軸為x＝3，所以有f（12.5）＝f（0.5），f（-4.5）＝f（1.5），f（3.5）＝f（2.5），因?yàn)?＜0.5＜1.5＜2.5＜3，且函數(shù)在（0，3）內(nèi)單調(diào)遞減，從而判斷大小【詳解】∵函數(shù)SKIPIF1<0滿足SKIPIF1<0，∴SKIPIF1<0=SKIPIF1<0，∴f（x）在R上是以6為周期的函數(shù)，∴f（12.5）＝f（12+0.5）＝f（0.5），SKIPIF1<0又SKIPIF1<0為偶函數(shù)，∴f（x）的對(duì)稱軸為x＝3，∴f（3.5）＝f（2.5），又∵0＜0.5＜1.5＜2.5＜3，且SKIPIF1<0在（0，3）內(nèi)單調(diào)遞減，∴f（2.5）＜f（1.5）＜f（0.5）即f（3.5）＜f（-4.5）＜f（12.5）故選B．【點(diǎn)睛】本題主要考查了函數(shù)周期性與對(duì)稱性的推導(dǎo)，考查了周期與單調(diào)性的綜合運(yùn)用，利用周期與對(duì)稱把所要比較的變量轉(zhuǎn)化到同一單調(diào)區(qū)間，利用函數(shù)的單調(diào)性比較函數(shù)值的大小，是解決此類問(wèn)題的常用方法，屬于中檔題．二、多選題9．（2022·山東·汶上圣澤中學(xué)高三階段練習(xí)）下列說(shuō)法正確的是（

）A．若SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則SKIPIF1<0的定義域?yàn)镾KIPIF1<0B．函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0C．函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0D．函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0【答案】AC【分析】根據(jù)抽象函數(shù)的定義域的求解判斷A；利用分離常數(shù)化簡(jiǎn)函數(shù)解析式，結(jié)合反比型函數(shù)的值域判斷B；利用換元法，結(jié)合二次函數(shù)的性質(zhì)求得其值域，判斷C；利用配方法，結(jié)合二次函數(shù)的性質(zhì)判斷D.【詳解】對(duì)于A，因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故A正確；對(duì)于B，SKIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，故B不正確；對(duì)于C，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，該函數(shù)取得最大值，最大值為SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，故C正確；對(duì)于D，SKIPIF1<0，其圖象的對(duì)稱軸為直線SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，故D不正確．故選：AC．10．（2022·山東省泰安英雄山中學(xué)高三階段練習(xí)）已知定義在R上的奇函數(shù)SKIPIF1<0對(duì)SKIPIF1<0都有SKIPIF1<0，則下列判斷正確的是（

）A．SKIPIF1<0是周期函數(shù)且周期為4 B．SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱 D．SKIPIF1<0在SKIPIF1<0上至少有5個(gè)零點(diǎn)【答案】ACD【分析】ABC選項(xiàng)根據(jù)函數(shù)的奇偶性和對(duì)稱性化簡(jiǎn)得出結(jié)論；D選項(xiàng)利用奇偶性得到SKIPIF1<0，以及周期性和對(duì)稱性得出結(jié)論.【詳解】A選項(xiàng)：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0周期為4，故A項(xiàng)正確；B選項(xiàng)：因?yàn)镾KIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故B項(xiàng)錯(cuò)誤；C選項(xiàng)：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，故C項(xiàng)正確；D選項(xiàng)：因?yàn)镾KIPIF1<0為定義在R上的奇函數(shù)，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故D項(xiàng)正確.故選：ACD．11．（2021·江蘇·無(wú)錫市第六高級(jí)中學(xué)高三階段練習(xí)）高斯是德國(guó)著名的數(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，用其名字命名的“高斯函數(shù)”為：設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過(guò)SKIPIF1<0的最大整數(shù)，則SKIPIF1<0稱為高斯函數(shù)，例如：SKIPIF1<0，SKIPIF1<0．已知函數(shù)SKIPIF1<0，則關(guān)于函數(shù)SKIPIF1<0的敘述中正確的是（

）A．SKIPIF1<0是偶函數(shù) B．SKIPIF1<0是奇函數(shù)C．SKIPIF1<0在SKIPIF1<0上是增函數(shù) D．SKIPIF1<0的值域是SKIPIF1<0【答案】BC【解析】計(jì)算SKIPIF1<0得出SKIPIF1<0判斷選項(xiàng)A不正確；用函數(shù)的奇偶性定義，可證SKIPIF1<0是奇函數(shù)，選項(xiàng)B正確；通過(guò)分離常數(shù)結(jié)合復(fù)合函數(shù)的單調(diào)性，可得出SKIPIF1<0在R上是增函數(shù)，判斷選項(xiàng)C正確；由SKIPIF1<0的范圍，利用不等式的關(guān)系，可求出SKIPIF1<0，選項(xiàng)D不正確，即可求得結(jié)果.【詳解】根據(jù)題意知，SKIPIF1<0.∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴函數(shù)SKIPIF1<0既不是奇函數(shù)也不是偶函數(shù)，A錯(cuò)誤；SKIPIF1<0，∴SKIPIF1<0是奇函數(shù)，B正確；SKIPIF1<0在R上是增函數(shù)，由復(fù)合函數(shù)的單調(diào)性知SKIPIF1<0在R上是增函數(shù)，C正確；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，D錯(cuò)誤.故選：BC.【點(diǎn)睛】關(guān)鍵點(diǎn)睛：本題是一道以數(shù)學(xué)文化為背景，判斷函數(shù)性質(zhì)的習(xí)題，屬于中檔題型，本題的關(guān)鍵是理解函數(shù)SKIPIF1<0，然后才會(huì)對(duì)函數(shù)SKIPIF1<0變形，并作出判斷.12．（2022·江蘇·南京市秦淮中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0對(duì)SKIPIF1<0恒成立，則SKIPIF1<0的取值可能是（

）A．0 B．2 C．1 D．3【答案】BD【分析】令SKIPIF1<0，將不等式SKIPIF1<0變成SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，分離常數(shù)可得SKIPIF1<0，令SKIPIF1<0，求出SKIPIF1<0的單調(diào)性即可得出答案.【詳解】令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，所以SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0在SKIPIF1<0上為減函數(shù)，在SKIPIF1<0上為增函數(shù)，且SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0的最大值為SKIPIF1<0，所以SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為增函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的取值范圍為SKIPIF1<0.故選：BD.三、填空題13．（2019·江蘇·南通一中高三階段練習(xí)）設(shè)奇函數(shù)f(x)在(0，＋∞)上為增函數(shù)，且f(1)＝0，則不等式SKIPIF1<0<0的解集為_(kāi)_______.【答案】(－1，0)∪(0，1)【分析】首先根據(jù)奇函數(shù)f(x)在(0，＋∞)上為增函數(shù)，且f(1)＝0，得到f(－1)＝0，且在(－∞，0)上也是增函數(shù)，從而將不等式轉(zhuǎn)化為SKIPIF1<0或SKIPIF1<0，進(jìn)而求得結(jié)果.【詳解】因?yàn)閒(x)為奇函數(shù)，且在(0，＋∞)上是增函數(shù)，f(1)＝0，所以f(－1)＝－f(1)＝0，且在(－∞，0)上也是增