新高考數(shù)學(xué)一輪復(fù)習(xí)講練測(cè)專題3.3函數(shù)的奇偶性與周期性（練）解析版

專題3.3函數(shù)的奇偶性與周期性練基礎(chǔ)練基礎(chǔ)1．（2021·海南?？谑小じ呷渌M）已知函數(shù)SKIPIF1<0，則“SKIPIF1<0”是“函數(shù)SKIPIF1<0為奇函數(shù)”的（）A．充分而不必要條件 B．必要而不充分條件 C．充分必要條件 D．既不充分也不必要條件【答案】C【解析】化簡(jiǎn)“SKIPIF1<0”和“函數(shù)SKIPIF1<0為奇函數(shù)”，再利用充分必要條件的定義判斷得解.【詳解】SKIPIF1<0，所以SKIPIF1<0，函數(shù)SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0.所以“SKIPIF1<0”是“函數(shù)SKIPIF1<0為奇函數(shù)”的充分必要條件.故選：C2．（2021·福建高三三模）若函數(shù)SKIPIF1<0的大致圖象如圖所示，則SKIPIF1<0的解析式可能是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】利用排除法，取特殊值分析判斷即可得答案【詳解】解：由圖可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，取SKIPIF1<0，則對(duì)于B，SKIPIF1<0，所以排除B，對(duì)于D，SKIPIF1<0，所以排除D，當(dāng)SKIPIF1<0時(shí)，對(duì)于A，SKIPIF1<0，此函數(shù)是由SKIPIF1<0向右平移1個(gè)單位，再向上平移1個(gè)單位，所以SKIPIF1<0時(shí)，SKIPIF1<0恒成立，而圖中，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0可以小于1，所以排除A,故選：C3．（2021·廣東高三其他模擬）下列函數(shù)中，既是奇函數(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】C【解析】利用函數(shù)奇偶性的定義和函數(shù)的解析式判斷.【詳解】A.函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，所以函數(shù)是非奇非偶函數(shù)，故錯(cuò)誤；B.SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故錯(cuò)誤；C.因?yàn)镾KIPIF1<0，所以函數(shù)是奇函數(shù)，且在SKIPIF1<0上單調(diào)遞增，正確；D.因?yàn)镾KIPIF1<0，所以函數(shù)是偶函數(shù)，故錯(cuò)誤；故選：C．4．（2021·湖南高三月考）定義函數(shù)SKIPIF1<0則下列命題中正確的是（）A．SKIPIF1<0不是周期函數(shù) B．SKIPIF1<0是奇函數(shù)C．SKIPIF1<0的圖象存在對(duì)稱軸 D．SKIPIF1<0是周期函數(shù)，且有最小正周期【答案】C【解析】當(dāng)SKIPIF1<0為有理數(shù)時(shí)恒有SKIPIF1<0，所以SKIPIF1<0是周期函數(shù)，且無(wú)最小正周期，又因?yàn)闊o(wú)論SKIPIF1<0是有理數(shù)還是無(wú)理數(shù)總有SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，圖象關(guān)于SKIPIF1<0軸對(duì)稱．【詳解】當(dāng)SKIPIF1<0為有理數(shù)時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0任何一個(gè)有理數(shù)SKIPIF1<0都是SKIPIF1<0的周期，SKIPIF1<0是周期函數(shù)，且無(wú)最小正周期，SKIPIF1<0選項(xiàng)SKIPIF1<0，SKIPIF1<0錯(cuò)誤，若SKIPIF1<0為有理數(shù)，則SKIPIF1<0也為有理數(shù)，SKIPIF1<0，若SKIPIF1<0為無(wú)理數(shù)，則SKIPIF1<0也為無(wú)理數(shù)，SKIPIF1<0，綜上，總有SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0為偶函數(shù)，圖象關(guān)于SKIPIF1<0軸對(duì)稱，SKIPIF1<0選項(xiàng)B錯(cuò)誤，選項(xiàng)C正確，故選：C5．【多選題】（2021·淮北市樹人高級(jí)中學(xué)高一期末）對(duì)于定義在R上的函數(shù)SKIPIF1<0，下列說(shuō)法正確的是（）A．若SKIPIF1<0是奇函數(shù)，則SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱B．若對(duì)SKIPIF1<0，有SKIPIF1<0，則SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱C．若函數(shù)SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱，則SKIPIF1<0為偶函數(shù)D．若SKIPIF1<0，則SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】ACD【解析】四個(gè)選項(xiàng)都是對(duì)函數(shù)性質(zhì)的應(yīng)用，在給出的四個(gè)選項(xiàng)中靈活的把變量x加以代換，再結(jié)合函數(shù)的對(duì)稱性、周期性和奇偶性就可以得到正確答案.【詳解】對(duì)A，SKIPIF1<0是奇函數(shù)，故圖象關(guān)于原點(diǎn)對(duì)稱，將SKIPIF1<0的圖象向右平移1個(gè)單位得SKIPIF1<0的圖象，故SKIPIF1<0的圖象關(guān)于點(diǎn)（1，0）對(duì)稱，正確；對(duì)B，若對(duì)SKIPIF1<0，有SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0是一個(gè)周期為2的周期函數(shù)，不能說(shuō)明其圖象關(guān)于直線SKIPIF1<0對(duì)稱，錯(cuò)誤.；對(duì)C，若函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，則SKIPIF1<0的圖象關(guān)于y軸對(duì)稱，故為偶函數(shù)，正確；對(duì)D，由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖象關(guān)于（1，1）對(duì)稱，正確.故選：ACD.6．【多選題】（2020·江蘇南通市·金沙中學(xué)高一期中）已知偶函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù),則滿足SKIPIF1<0的SKIPIF1<0的取值是（）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】根據(jù)偶函數(shù)和單調(diào)性求得不等式的解，然后判斷各選項(xiàng)．．【詳解】由題意SKIPIF1<0，解得SKIPIF1<0，只有BC滿足．故選：BC．7．【多選題】（2021·廣東高三二模）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0與SKIPIF1<0都為奇函數(shù)，則下列說(shuō)法正確的是（）A．SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù) B．SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)C．SKIPIF1<0為奇函數(shù) D．SKIPIF1<0為奇函數(shù)【答案】BD【解析】AB選項(xiàng)，利用周期函數(shù)的定義判斷；CD選項(xiàng)，利用周期性結(jié)合SKIPIF1<0，SKIPIF1<0為奇函數(shù)判斷.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0與SKIPIF1<0都為奇函數(shù)，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故B正確A錯(cuò)誤；因?yàn)镾KIPIF1<0，且SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0為奇函數(shù)，故D正確；因?yàn)镾KIPIF1<0與SKIPIF1<0相差1，不是最小周期的整數(shù)倍，且SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0不為奇函數(shù)，故C錯(cuò)誤.故選：BD.8．（2021·吉林高三二模（文））寫出一個(gè)符合“對(duì)SKIPIF1<0，SKIPIF1<0”的函數(shù)SKIPIF1<0___________.【答案】SKIPIF1<0（答案不唯一）【解析】分析可知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且該函數(shù)為奇函數(shù)，由此可得結(jié)果.【詳解】由題意可知，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且該函數(shù)為奇函數(shù)，可取SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）.9．（2021·全國(guó)高三二模（理））已知SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，且其圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，若SKIPIF1<0，則SKIPIF1<0__________．【答案】1【解析】根據(jù)函數(shù)的對(duì)稱性及奇函數(shù)性質(zhì)求得函數(shù)周期為4，從而SKIPIF1<0.【詳解】函數(shù)關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，則SKIPIF1<0，又SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，則SKIPIF1<0，因此函數(shù)的周期為4，因此SKIPIF1<0.故答案為：1.10．（2021·上海高三二模）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，函數(shù)SKIPIF1<0是奇函數(shù)，且SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0___________．【答案】SKIPIF1<0【解析】通過(guò)計(jì)算SKIPIF1<0可得．【詳解】因?yàn)镾KIPIF1<0是奇函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0．練提升TIDHNEG練提升TIDHNEG1．（2021·安徽高三三模（文））若把定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0的圖象沿x軸左右平移后，可以得到關(guān)于原點(diǎn)對(duì)稱的圖象，也可以得到關(guān)于SKIPIF1<0軸對(duì)稱的圖象，則關(guān)于函數(shù)SKIPIF1<0的性質(zhì)敘述一定正確的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0是周期函數(shù) D．SKIPIF1<0存在單調(diào)遞增區(qū)間【答案】C【解析】通過(guò)舉例說(shuō)明選項(xiàng)ABD錯(cuò)誤；對(duì)于選項(xiàng)C可以證明判斷得解.【詳解】定義域?yàn)镽的函數(shù)SKIPIF1<0的圖象沿SKIPIF1<0軸左右平移后，可以得到關(guān)于原點(diǎn)對(duì)稱的圖象，也可以得到關(guān)于SKIPIF1<0軸對(duì)稱的圖象，∴SKIPIF1<0的圖象既有對(duì)稱中心又有對(duì)稱軸，但SKIPIF1<0不一定具有奇偶性，例如SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0為奇函數(shù)，故選項(xiàng)A錯(cuò)誤；由SKIPIF1<0，可得函數(shù)SKIPIF1<0圖象關(guān)于SKIPIF1<0對(duì)稱，故選項(xiàng)B錯(cuò)誤；由SKIPIF1<0時(shí)，SKIPIF1<0不存在單調(diào)遞增區(qū)間，故選項(xiàng)D錯(cuò)誤；由已知設(shè)SKIPIF1<0圖象的一條對(duì)稱抽為直線SKIPIF1<0，一個(gè)對(duì)稱中心為SKIPIF1<0，且SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0的一個(gè)周期SKIPIF1<0，故選項(xiàng)C正確.故選：C2．（2021·天津高三二模）已知函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，且滿足SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】根據(jù)對(duì)數(shù)運(yùn)算性質(zhì)和對(duì)數(shù)函數(shù)單調(diào)性可得SKIPIF1<0，根據(jù)指數(shù)函數(shù)單調(diào)性可知SKIPIF1<0；利用SKIPIF1<0為減函數(shù)可知SKIPIF1<0，結(jié)合SKIPIF1<0為奇函數(shù)可得大小關(guān)系.【詳解】SKIPIF1<0，SKIPIF1<0即：SKIPIF1<0又SKIPIF1<0是定義在SKIPIF1<0上的減函數(shù)SKIPIF1<0又SKIPIF1<0為奇函數(shù)SKIPIF1<0SKIPIF1<0，即：SKIPIF1<0.故選：B.3.（2021·陜西高三三模（理））已知函數(shù)f(x)為R上的奇函數(shù)，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則f(101)+f(105)的值為（）A．3 B．2 C．1 D．0【答案】A【解析】根據(jù)函數(shù)為奇函數(shù)可求得函數(shù)的解析式，再由SKIPIF1<0求得函數(shù)f(x)是周期為4的周期函數(shù)，由此可計(jì)算得選項(xiàng)．【詳解】解：根據(jù)題意，函數(shù)f(x)為R上的奇函數(shù)，則f(0)=0，又由x∈[0，1]時(shí)，SKIPIF1<0，則有f(0)=1+a=0，解可得：a=﹣1，則有SKIPIF1<0，又由f(﹣x)=f(2+x)，即f(x+2)=﹣f(x)，則有f(x+4)=﹣f(x+2)=f(x)，即函數(shù)f(x)是周期為4的周期函數(shù)，則SKIPIF1<0，故有f(101)+f(105)=3，故選：A．4．（2021·上海高三二模）若SKIPIF1<0是R上的奇函數(shù)，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則下列結(jié)論：①SKIPIF1<0是偶函數(shù)；②對(duì)任意的x∈R都有SKIPIF1<0；③SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；④反函數(shù)SKIPIF1<0存在且在SKIPIF1<0上單調(diào)遞增．其中正確結(jié)論的個(gè)數(shù)為（）A．1 B．2 C．3 D．4【答案】C【解析】根據(jù)奇函數(shù)定義以及單調(diào)性性質(zhì)，及反函數(shù)性質(zhì)逐一進(jìn)行判斷選擇.【詳解】對(duì)于①，由SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，得SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，故①正確；對(duì)于②，由SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，得SKIPIF1<0，而SKIPIF1<0不一定成立，所以對(duì)任意的SKIPIF1<0，不一定有SKIPIF1<0，故②錯(cuò)誤；對(duì)于③，因?yàn)镾KIPIF1<0是SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，因此SKIPIF1<0，利用復(fù)合函數(shù)的單調(diào)性，知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故③正確.對(duì)于④，由已知得SKIPIF1<0是SKIPIF1<0上的單調(diào)遞增函數(shù)，利用函數(shù)存在反函數(shù)的充要條件是，函數(shù)的定義域與值域是一一映射，且函數(shù)與其反函數(shù)在相應(yīng)區(qū)間內(nèi)單調(diào)性一致，故反函數(shù)SKIPIF1<0存在且在SKIPIF1<0上單調(diào)遞增，故④正確；故選：C5．【多選題】（2021·全國(guó)高三專題練習(xí)）已知函數(shù)SKIPIF1<0是偶函數(shù)，SKIPIF1<0是奇函數(shù)，并且當(dāng)SKIPIF1<0，SKIPIF1<0，則下列選項(xiàng)正確的是（）A．SKIPIF1<0在SKIPIF1<0上為減函數(shù) B．SKIPIF1<0在SKIPIF1<0上SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上為增函數(shù) D．SKIPIF1<0在SKIPIF1<0上SKIPIF1<0【答案】CD【解析】根據(jù)題意，分析可得SKIPIF1<0，結(jié)合函數(shù)的解析式可得當(dāng)SKIPIF1<0時(shí)函數(shù)的解析式，據(jù)此分析可得答案．【詳解】解：根據(jù)題意，函數(shù)SKIPIF1<0為奇函數(shù)，則有SKIPIF1<0，即SKIPIF1<0，又由SKIPIF1<0為偶函數(shù)，則SKIPIF1<0，則有SKIPIF1<0，即有SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，則SKIPIF1<0為增函數(shù)且SKIPIF1<0；故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，且SKIPIF1<0；故選：SKIPIF1<0．6．【多選題】（2021·全國(guó)高三專題練習(xí)）若函數(shù)SKIPIF1<0對(duì)任意SKIPIF1<0都有SKIPIF1<0成立，SKIPIF1<0，則下列的點(diǎn)一定在函數(shù)SKIPIF1<0圖象上的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABC【解析】根據(jù)任意SKIPIF1<0滿足SKIPIF1<0，得到SKIPIF1<0是奇函數(shù)判斷.【詳解】因?yàn)槿我釹KIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，又SKIPIF1<0，所以令SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0，所以點(diǎn)SKIPIF1<0，且點(diǎn)SKIPIF1<0與SKIPIF1<0也一定在SKIPIF1<0的圖象上，故選：ABC．7．【多選題】（2021·浙江高一期末）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則下列說(shuō)法正確的是（）A．函數(shù)SKIPIF1<0有2個(gè)零點(diǎn) B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0C．不等式SKIPIF1<0的解集是SKIPIF1<0 D．SKIPIF1<0，都有SKIPIF1<0【答案】BCD【解析】根據(jù)函數(shù)奇偶性定義和零點(diǎn)定義對(duì)選項(xiàng)一一判斷即可．【詳解】對(duì)A，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，又因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，所以SKIPIF1<0，故函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，則A錯(cuò)；對(duì)B，設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，則B對(duì)；對(duì)C，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0無(wú)解；則C對(duì)；對(duì)D，SKIPIF1<0，都有SKIPIF1<0，則D對(duì)．故選：BCD．8．【多選題】（2021·蘇州市第五中學(xué)校高一月考）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).設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過(guò)SKIPIF1<0的最大整數(shù)，SKIPIF1<0也被稱為“高斯函數(shù)”，例如：SKIPIF1<0，SKIPIF1<0.已知函數(shù)SKIPIF1<0，下列說(shuō)法中正確的是（）A．SKIPIF1<0是周期函數(shù) B．SKIPIF1<0的值域是SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上是減函數(shù) D．SKIPIF1<0，SKIPIF1<0【答案】AC【解析】根據(jù)SKIPIF1<0定義將函數(shù)SKIPIF1<0寫成分段函數(shù)的形式，再畫出函數(shù)的圖象，根據(jù)圖象判斷函數(shù)的性質(zhì).【詳解】由題意可知SKIPIF1<0，SKIPIF1<0，可畫出函數(shù)圖像，如圖：可得到函數(shù)SKIPIF1<0是周期為1的函數(shù)，且值域?yàn)镾KIPIF1<0，在SKIPIF1<0上單調(diào)遞減，故選項(xiàng)AC正確，B錯(cuò)誤；對(duì)于D，取SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，故D錯(cuò)誤.故選：AC．9．【多選題】（2021·湖南高三月考）函數(shù)SKIPIF1<0滿足以下條件：①SKIPIF1<0的定義域是SKIPIF1<0，且其圖象是一條連續(xù)不斷的曲線；②SKIPIF1<0是偶函數(shù)；③SKIPIF1<0在SKIPIF1<0上不是單調(diào)函數(shù)；④SKIPIF1<0恰有2個(gè)零點(diǎn).則函數(shù)SKIPIF1<0的解析式可以是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】CD【解析】利用函數(shù)圖象變換畫出選項(xiàng)A，B，C，D對(duì)應(yīng)的函數(shù)圖象，逐一分析即可求解.【詳解】解：顯然題設(shè)選項(xiàng)的四個(gè)函數(shù)均為偶函數(shù)，但SKIPIF1<0的定義域?yàn)镾KIPIF1<0，所以選項(xiàng)B錯(cuò)誤；函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，在SKIPIF1<0，SKIPIF1<0單調(diào)遞減，在SKIPIF1<0，SKIPIF1<0單調(diào)遞增，但SKIPIF1<0有3個(gè)零點(diǎn)，選項(xiàng)A錯(cuò)誤；函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖象對(duì)稱軸為SKIPIF1<0，其圖象是開(kāi)口向下的拋物線，故SKIPIF1<0在SKIPIF1<0，SKIPIF1<0單調(diào)遞增，在SKIPIF1<0，SKIPIF1<0單調(diào)遞減，由圖得SKIPIF1<0恰有2個(gè)零點(diǎn)，選項(xiàng)C正確；函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，在SKIPIF1<0，SKIPIF1<0單調(diào)遞減，在SKIPIF1<0，SKIPIF1<0單調(diào)遞增，且SKIPIF1<0有2個(gè)零點(diǎn)，選項(xiàng)D正確.故選：CD.10．（2021·黑龍江大慶市·高三二模（理））定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象與SKIPIF1<0的圖象的交點(diǎn)個(gè)數(shù)為_(kāi)__________.【答案】7【解析】由題設(shè)可知SKIPIF1<0的周期為2，結(jié)合已知區(qū)間的解析式及SKIPIF1<0，可得兩函數(shù)圖象，即知圖象交點(diǎn)個(gè)數(shù).【詳解】由題意知：SKIPIF1<0的周期為2，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0、SKIPIF1<0的圖象如下：即SKIPIF1<0與SKIPIF1<0共有7個(gè)交點(diǎn)，故答案為：7.【點(diǎn)睛】結(jié)論點(diǎn)睛：SKIPIF1<0有SKIPIF1<0的周期為SKIPIF1<0.練真題TIDHNEG練真題TIDHNEG1.（2020·天津高考真題）函數(shù)SKIPIF1<0的圖象大致為（）A． B．C． D．【答案】A【解析】【分析】由題意首先確定函數(shù)的奇偶性，然后考查函數(shù)在特殊點(diǎn)的函數(shù)值排除錯(cuò)誤選項(xiàng)即可確定函數(shù)的圖象.【詳解】由函數(shù)的解析式可得：SKIPIF1<0，則函數(shù)SKIPIF1<0為奇函數(shù)，其圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，選項(xiàng)CD錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，選項(xiàng)B錯(cuò)誤.故選：A.2.（2020·全國(guó)高考真題（理））設(shè)函數(shù)SKIPIF1<0，則f(x)（）A．是偶函數(shù)，且在SKIPIF1<0單調(diào)遞增 B．是奇函數(shù)，且在SKIPIF1<0單調(diào)遞減C．是偶函數(shù)，且在SKIPIF1<0單調(diào)遞增 D．是奇函數(shù)，且在SKIPIF1<0單調(diào)遞減【答案】D【解析】由SKIPIF1<0得SKIPIF1<0定義域?yàn)镾KIPIF1<0，關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，又SKIPIF1<0，SKIPIF1<0為定義域上