新高考數(shù)學(xué)三輪沖刺 北京卷押題練習(xí) 第5題 函數(shù)的性質(zhì) （解析版）

函數(shù)的性質(zhì)核心考點(diǎn)考情統(tǒng)計(jì)考向預(yù)測(cè)備考策略定義域2020·北京卷T11可以預(yù)測(cè)2024年新高考命題方向?qū)⒗^續(xù)以函數(shù)的基本性質(zhì)等問(wèn)題展開(kāi)命題．函數(shù)的基本性質(zhì)單選題一般為中檔題，縱觀近幾年的新高考試題，分別考查函數(shù)的單調(diào)性、奇偶性、周期性及對(duì)稱性，考點(diǎn)綜合性強(qiáng)，思維難度較大，是高考沖刺的重點(diǎn)復(fù)習(xí)內(nèi)容。單調(diào)性2023·北京卷T4奇偶性2022·北京卷T41．（2023·北京卷T4）下列函數(shù)中，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】C【解析】對(duì)于A，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故A錯(cuò)誤；對(duì)于B，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故B錯(cuò)誤；對(duì)于C，因?yàn)镾KIPIF1<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在SKIPIF1<0上單調(diào)遞增，故C正確；對(duì)于D，因?yàn)镾KIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上不單調(diào)，D錯(cuò)誤.故選：C.2．（2022·北京卷T4））已知函數(shù)SKIPIF1<0，則對(duì)任意實(shí)數(shù)x，有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，故A錯(cuò)誤，C正確；SKIPIF1<0，不是常數(shù)，故BD錯(cuò)誤；故選：C．3．（2020·北京卷T11））函數(shù)SKIPIF1<0的定義域是．【答案】SKIPIF1<0【解析】由題意得SKIPIF1<0，SKIPIF1<01.求函數(shù)的定義域應(yīng)關(guān)注三點(diǎn)①要明確使各函數(shù)表達(dá)式有意義的條件是什么，函數(shù)有意義的準(zhǔn)則一般有：(ⅰ)分式的分母不為0；(ⅱ)偶次根式的被開(kāi)方數(shù)非負(fù)；(ⅲ)y＝x0要求x≠0.②不對(duì)解析式化簡(jiǎn)變形，以免定義域變化．③當(dāng)一個(gè)函數(shù)由兩個(gè)或兩個(gè)以上代數(shù)式的和、差、積、商的形式構(gòu)成時(shí)，定義域是使得各式子都有意義的公共部分的集合．2.函數(shù)單調(diào)性設(shè)SKIPIF1<0那么SKIPIF1<0SKIPIF1<0SKIPIF1<0上是增函數(shù)；SKIPIF1<0SKIPIF1<0上是減函數(shù).設(shè)函數(shù)SKIPIF1<0在某個(gè)區(qū)間內(nèi)可導(dǎo)，如果SKIPIF1<0，則SKIPIF1<0為增函數(shù)；如果SKIPIF1<0，則SKIPIF1<0為減函數(shù).3.奇偶性①具有奇偶性的函數(shù)定義域關(guān)于原點(diǎn)對(duì)稱（大前提）②奇偶性的定義：奇函數(shù)：SKIPIF1<0，圖象關(guān)于原點(diǎn)對(duì)稱偶函數(shù)：SKIPIF1<0，圖象關(guān)于SKIPIF1<0軸對(duì)稱4.周期性（差為常數(shù)有周期）SKIPIF1<05.對(duì)稱性（和為常數(shù)有對(duì)稱軸）SKIPIF1<0SKIPIF1<01．已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0=（

）A．2 B．3 C．4 D．5【答案】B【解析】由題意知SKIPIF1<0，且SKIPIF1<0，用SKIPIF1<0代換x，則SKIPIF1<0，即得SKIPIF1<0，故選B2．函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故選C.3．若函數(shù)SKIPIF1<0為奇函數(shù)，則實(shí)數(shù)SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】B【解析】由題意可得，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，整理可得，SKIPIF1<0對(duì)任意SKIPIF1<0都成立，SKIPIF1<0，SKIPIF1<0，故選B4．在下列函數(shù)中，即是偶函數(shù)又在SKIPIF1<0上單調(diào)遞增的函數(shù)的有（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】對(duì)于A，函數(shù)SKIPIF1<0是奇函數(shù)，在SKIPIF1<0上單調(diào)遞減，故A不符合；對(duì)于B，函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，又函數(shù)在SKIPIF1<0上單調(diào)遞減的函數(shù)，故B不符合；對(duì)于C，函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，故C不符合；對(duì)于D，函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0為偶函數(shù)，又SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)在SKIPIF1<0上單調(diào)遞增的函數(shù)，故D符合.故選：D．5．已知SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)SKIPIF1<0，顯然它定義域關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.故選：C.6．已知函數(shù)SKIPIF1<0在定義域SKIPIF1<0上是增函數(shù)，且SKIPIF1<0，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】C【解析】因?yàn)楹瘮?shù)SKIPIF1<0在定義域SKIPIF1<0上是增函數(shù)，且SKIPIF1<0，則有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C.7．已知函數(shù)SKIPIF1<0，則下列說(shuō)法中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0的圖像關(guān)于原點(diǎn)對(duì)稱C．SKIPIF1<0在定義域內(nèi)是增函數(shù) D．SKIPIF1<0存在最大值【答案】B【解析】對(duì)于選項(xiàng)A:因?yàn)镾KIPIF1<0，可得SKIPIF1<0，故選項(xiàng)A錯(cuò)誤；對(duì)于選項(xiàng)B:因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0，定義域關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，可得SKIPIF1<0為奇函數(shù)，故選項(xiàng)B正確；對(duì)于選項(xiàng)C:因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0為單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0為單調(diào)遞增，由于SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，所以SKIPIF1<0在SKIPIF1<0為單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0單調(diào)遞增，不滿足在定義域SKIPIF1<0單調(diào)遞增，（可取特殊值排除），故選項(xiàng)C錯(cuò)誤；對(duì)于選項(xiàng)D:SKIPIF1<0在SKIPIF1<0為單調(diào)遞增，故無(wú)最大值，故選項(xiàng)D錯(cuò)誤.故選：B.8．函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的增函數(shù)，則滿足SKIPIF1<0的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的增函數(shù)，則由SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，故選D9．下列函數(shù)中，既是偶函數(shù)又在區(qū)間SKIPIF1<0上單調(diào)遞增的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】對(duì)于A：SKIPIF1<0，則SKIPIF1<0，偶函數(shù)，另外當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減，A錯(cuò)誤；對(duì)于B：SKIPIF1<0，則SKIPIF1<0，偶函數(shù)，另外當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞增，B正確；對(duì)于C：SKIPIF1<0，則SKIPIF1<0，奇函數(shù)，C錯(cuò)誤；對(duì)于D：SKIPIF1<0，則SKIPIF1<0，偶函數(shù)，另外當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減，D錯(cuò)誤.故選：B.10．定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．3 D．9【答案】C【解析】由函數(shù)SKIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0的周期為3，SKIPIF1<0，故選：C.11．已知函數(shù)SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是增函數(shù)，而SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：A12．已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，對(duì)任意的SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，則（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)閷?duì)任意的SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，所以由函數(shù)單調(diào)性的定義可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，又SKIPIF1<0是偶函數(shù)，SKIPIF1<0，所以SKIPIF1<0，故選：A13．函數(shù)SKIPIF1<0的定義域是.【答案】SKIPIF1<0【解析】由SKIPIF1<0的解析式可得SKIPIF1<0，解得SKIPIF1<0；所以其定義域?yàn)镾KIPIF1<0.14．函數(shù)SKIPIF1<0的值域?yàn)?【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，即SKIPIF1<0，綜上SKIPIF1<0的值域?yàn)镾KIPIF1<0，15．函數(shù)SKIPIF1<0在SKIPIF1<0上是單調(diào)遞減函數(shù)，則SKIPIF1<0的單調(diào)遞增區(qū)間是【答案】SKIPIF1<0【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，即SKIPIF1<0的定義域?yàn)镾KIPIF1<0.由于SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故單調(diào)遞增區(qū)間是SKIPIF1<0.16．求函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為【答案】SKIPIF1<0和SKIPIF1<0【解析】SKIPIF1<0，畫(huà)出函數(shù)圖象（如圖所示）結(jié)合圖象得函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0和SKIPIF1<0.17．設(shè)奇函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0．若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖象如圖，則不等式SKIPIF1<0的解集是．【答案】SKIPIF1<0【解析】因?yàn)楹瘮?shù)SKIPIF1<0是奇函數(shù)，所以利用函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，可得SKIPIF1<0的解集為SKIPIF1<0．18．設(shè)函數(shù)f(x)＝SKIPIF1<0，則f(SKIPIF1<0)＋f(SKIPIF1<0)＋…＋f(SKIPIF1<0)＝．【答案】1012【解析】∵f(x)＝，∴f(1－x)＝＝，∴f(x