新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)專(zhuān)題02 函數(shù)的概念和性質(zhì)（練）（解析版）

第一篇熱點(diǎn)、難點(diǎn)突破篇專(zhuān)題02函數(shù)的概念和性質(zhì)（練）【對(duì)點(diǎn)演練】一、單選題1．（2022·山西太原·高三期中）已知集合SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】解不等式SKIPIF1<0和SKIPIF1<0，再求交集即可.【詳解】由SKIPIF1<0得：SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0得：SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：C2．（2022·海南昌茂花園學(xué)校高三階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（

）A．充要條件 B．必要不充分條件 C．充分不必要條件 D．既不充分也不必要條件【答案】C【分析】先化簡(jiǎn)題目中的不等式，然后根據(jù)充分性和必要性的定義進(jìn)行判斷即可【詳解】由SKIPIF1<0結(jié)合函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù)，可得SKIPIF1<0，由SKIPIF1<0結(jié)合函數(shù)SKIPIF1<0是SKIPIF1<0上的減函數(shù)，可得SKIPIF1<0，故“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件，故選：C3．（2022·河南·模擬預(yù)測(cè)（理））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則a，b，c的大小關(guān)系是（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】SKIPIF1<0，通過(guò)比較5和SKIPIF1<0，可得到SKIPIF1<0大小關(guān)系.通過(guò)比較SKIPIF1<0與SKIPIF1<0，可得到SKIPIF1<0大小關(guān)系.【詳解】SKIPIF1<0，因SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0.又SKIPIF1<0，SKIPIF1<0在在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0.故選：A二、多選題4．（2022·山東·青島超銀高級(jí)中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0是偶函數(shù)，SKIPIF1<0是奇函數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0是SKIPIF1<0的周期函數(shù)【答案】ACD【分析】根據(jù)給定條件，利用奇偶性判斷A，B，C；推理計(jì)算并結(jié)合周期的意義判斷D作答.【詳解】因函數(shù)SKIPIF1<0是偶函數(shù)，即SKIPIF1<0，于是得SKIPIF1<0，A正確；因函數(shù)SKIPIF1<0是奇函數(shù)，即SKIPIF1<0，B不正確；因函數(shù)SKIPIF1<0是奇函數(shù)，則SKIPIF1<0，C正確；由選項(xiàng)A，C知，SKIPIF1<0，即SKIPIF1<0，因此SKIPIF1<0，即SKIPIF1<0是SKIPIF1<0的周期函數(shù)，D正確.故選：ACD5．（2022·遼寧·丹東市教師進(jìn)修學(xué)院高三期中）已知定義域?yàn)镾KIPIF1<0的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0，則必有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)【答案】ACD【分析】根據(jù)函數(shù)SKIPIF1<0為奇函數(shù)可得SKIPIF1<0，又SKIPIF1<0則可得周期為3，從而可得SKIPIF1<0，再利用周期性與對(duì)稱(chēng)性逐項(xiàng)判斷即可.【詳解】解：已知定義域?yàn)镾KIPIF1<0的奇函數(shù)SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0又SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0是周期為3的函數(shù)所以SKIPIF1<0，故A正確；又由SKIPIF1<0可得SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，故SKIPIF1<0，故B錯(cuò)誤；由于SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，則SKIPIF1<0，故C正確；由SKIPIF1<0周期為3，SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，可得SKIPIF1<0圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，故D正確.故選：ACD.三、填空題6．（2022·江蘇·南京師大附中高三期中）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且函數(shù)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則滿足SKIPIF1<0的SKIPIF1<0取值范圍是______.【答案】SKIPIF1<0【分析】判斷出SKIPIF1<0是奇函數(shù)，結(jié)合函數(shù)的奇偶性、單調(diào)性化簡(jiǎn)不等式SKIPIF1<0，從而求得正確答案.【詳解】由于SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱(chēng)，SKIPIF1<0為奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為增函數(shù)，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0，所以滿足SKIPIF1<0的SKIPIF1<0取值范圍SKIPIF1<0.故答案為：SKIPIF1<07．（2022·天津市軍糧城中學(xué)高三期中）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是_________.【答案】SKIPIF1<0【分析】根據(jù)復(fù)合函數(shù)的單調(diào)性求解即可.【詳解】函數(shù)SKIPIF1<0定義域?yàn)镾KIPIF1<0，令SKIPIF1<0，則SKIPIF1<0為減函數(shù).當(dāng)SKIPIF1<0，SKIPIF1<0為減函數(shù)，則SKIPIF1<0為增函數(shù).故答案為：SKIPIF1<0.8．（2022·廣西北?！ひ荒＃ㄎ模┮阎婧瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,且SKIPIF1<0對(duì)任意SKIPIF1<0恒成立,若SKIPIF1<0,則SKIPIF1<0____________.【答案】2【分析】根據(jù)SKIPIF1<0的周期性和對(duì)稱(chēng)性,求出一個(gè)周期內(nèi)的整數(shù)點(diǎn)處的函數(shù)值及它們的和,再根據(jù)SKIPIF1<0,求出505個(gè)周期內(nèi)的和加上SKIPIF1<0即可.【詳解】解:由題知,SKIPIF1<0,所以SKIPIF1<0周期為4,因?yàn)槠婧瘮?shù)SKIPIF1<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,又SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0.故答案為:29．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0________，函數(shù)SKIPIF1<0的零點(diǎn)為_(kāi)_______．【答案】

SKIPIF1<0

SKIPIF1<0【分析】根據(jù)給定的分段函數(shù)求出函數(shù)值即可，再直接求出方程的解作答.【詳解】依題意，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0，無(wú)解，所以數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<010．（2022·北京市西城外國(guó)語(yǔ)學(xué)校高三階段練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)_____________________，單調(diào)遞增區(qū)間為_(kāi)__________．【答案】

SKIPIF1<0

SKIPIF1<0##SKIPIF1<0【分析】根據(jù)給定的函數(shù)，列出不等式，解不等式得定義域；結(jié)合對(duì)數(shù)函數(shù)、二次函數(shù)單調(diào)性求解單調(diào)增區(qū)間作答.【詳解】函數(shù)SKIPIF1<0有意義，則有SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0；因函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，而函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，因此函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0【沖刺提升】一、單選題1．（2022·河南·高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】求出SKIPIF1<0，不等式轉(zhuǎn)化為SKIPIF1<0，分SKIPIF1<0與SKIPIF1<0兩種情況進(jìn)行求解，得到不等式的解集.【詳解】∵SKIPIF1<0，∴不等式轉(zhuǎn)化為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0.綜上所述，不等式的解集為SKIPIF1<0.故選：A.2．（2022·黑龍江·哈爾濱七十三中高三階段練習(xí)）已知函數(shù)SKIPIF1<0，則“函數(shù)SKIPIF1<0為偶函數(shù)”是“SKIPIF1<0”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】B【分析】根據(jù)偶函數(shù)的定義求出當(dāng)函數(shù)SKIPIF1<0為偶函數(shù)時(shí)，實(shí)數(shù)SKIPIF1<0的值，再利用集合的包含關(guān)系判斷可得出結(jié)論.【詳解】若函數(shù)SKIPIF1<0為偶函數(shù)，則對(duì)任意的SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0，又因?yàn)镾KIPIF1<0SKIPIF1<0，因此，“函數(shù)SKIPIF1<0為偶函數(shù)”是“SKIPIF1<0”的必要不充分條件.故選：B.3．（2022·海南昌茂花園學(xué)校高三階段練習(xí)）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且在SKIPIF1<0上是單調(diào)遞增的，設(shè)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<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)偶函數(shù)的性質(zhì)以及函數(shù)在SKIPIF1<0上單調(diào)遞增，比較自變量絕對(duì)值的大小即可得解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，所以SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上是單調(diào)遞增的，故SKIPIF1<0在SKIPIF1<0上是單調(diào)遞減，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：B.4．（2022·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)SKIPIF1<0存在反函數(shù)，則常數(shù)a的取值范圍為（）A．（﹣∞，1] B．[1，2]C．[2，+∞） D．（﹣∞，1]∪[2，+∞）【答案】D【分析】依題意可得f（x）在[0，1]上單調(diào)，分兩種情況討論，參變分離，結(jié)合指數(shù)函數(shù)的性質(zhì)能求出常數(shù)a的取值范圍．【詳解】解：∵函數(shù)SKIPIF1<0存在反函數(shù)∴函數(shù)SKIPIF1<0在[0，1]上單調(diào)若單調(diào)遞增，即SKIPIF1<0，則SKIPIF1<0在x∈[0，1]上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立∵SKIPIF1<0在[0，1]上單調(diào)遞增∴SKIPIF1<0∴a≤1若單調(diào)遞減，即SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恒成立即SKIPIF1<0在SKIPIF1<0上恒成立∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增∴SKIPIF1<0∴SKIPIF1<0．綜上，常數(shù)a的取值范圍為SKIPIF1<0．故選：D．5．（2022·河南·高三階段練習(xí)（文））設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且滿足SKIPIF1<0是偶函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則下列說(shuō)法不正確的是（

）A．SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的取值范圍為SKIPIF1<0C．SKIPIF1<0為奇函數(shù)D．方程SKIPIF1<0僅有5個(gè)不同實(shí)數(shù)解【答案】D【分析】由已知條件可得函數(shù)的對(duì)稱(chēng)中心及對(duì)稱(chēng)軸，利用對(duì)稱(chēng)中心和對(duì)稱(chēng)軸將已知區(qū)間圖象進(jìn)行多次對(duì)稱(chēng)變換，可得函數(shù)SKIPIF1<0的圖象，依據(jù)圖象對(duì)各個(gè)選項(xiàng)進(jìn)行判斷即可.【詳解】∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的圖象如圖：∵SKIPIF1<0是偶函數(shù)，∴SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0在區(qū)間SKIPIF1<0的圖象如圖：∵SKIPIF1<0，∴將SKIPIF1<0中的SKIPIF1<0替換為SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0在區(qū)間SKIPIF1<0的圖象如圖：由函數(shù)圖象的對(duì)稱(chēng)軸直線SKIPIF1<0和對(duì)稱(chēng)中心SKIPIF1<0進(jìn)行多次對(duì)稱(chēng)變換，可得函數(shù)圖象如圖：由函數(shù)圖象可知，SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)，函數(shù)SKIPIF1<0的對(duì)稱(chēng)軸為直線SKIPIF1<0（SKIPIF1<0Z），對(duì)稱(chēng)中心為點(diǎn)SKIPIF1<0（SKIPIF1<0Z），另外，函數(shù)的周期性還可以通過(guò)以下方法進(jìn)行證明：將SKIPIF1<0中的SKIPIF1<0替換為SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，由已知有SKIPIF1<0，∴SKIPIF1<0將SKIPIF1<0中SKIPIF1<0分別替換為SKIPIF1<0和SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0和SKIPIF1<0，即SKIPIF1<0SKIPIF1<0SKIPIF1<0∴SKIPIF1<0將SKIPIF1<0中SKIPIF1<0替換為SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù).對(duì)于A，SKIPIF1<0，故A正確；對(duì)于B，當(dāng)SKIPIF1<0時(shí)，由圖象可知其值域?yàn)镾KIPIF1<0，故B正確；對(duì)于C，由圖象知，其圖象的對(duì)稱(chēng)中心為點(diǎn)SKIPIF1<0（SKIPIF1<0Z），當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0為SKIPIF1<0圖象的對(duì)稱(chēng)中心，因此將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得函數(shù)SKIPIF1<0為奇函數(shù)，故C正確；對(duì)于D，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將SKIPIF1<0軸下方的圖象翻折至SKIPIF1<0軸上方，得到函數(shù)SKIPIF1<0的圖象，易知SKIPIF1<0的圖象過(guò)點(diǎn)SKIPIF1<0如圖，SKIPIF1<0的圖象與SKIPIF1<0的圖象有6個(gè)交點(diǎn)，所以方程SKIPIF1<0有6個(gè)不同實(shí)數(shù)解，故D錯(cuò)誤．故選：D.二、多選題6．（2022·廣東·高三階段練習(xí)）若函數(shù)SKIPIF1<0和SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0有意義，SKIPIF1<0與SKIPIF1<0都為SKIPIF1<0上單調(diào)遞增的奇函數(shù)，則（

）A．SKIPIF1<0為偶函數(shù) B．SKIPIF1<0為SKIPIF1<0上的單調(diào)遞增函數(shù)C．SKIPIF1<0為奇函數(shù) D．SKIPIF1<0為SKIPIF1<0上的單調(diào)遞增函數(shù)【答案】ACD【分析】根據(jù)單調(diào)性、奇偶性的定義與結(jié)論逐項(xiàng)分析判斷.【詳解】選項(xiàng)A：由偶函數(shù)的定義直接得，SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)，故正確；選項(xiàng)B：SKIPIF1<0在SKIPIF1<0上不一定是增函數(shù)，比如，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上都是奇函數(shù)且單調(diào)遞增，但SKIPIF1<0在SKIPIF1<0上不是單調(diào)遞增函數(shù)，故不正確；選項(xiàng)C：SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，故正確；選項(xiàng)D：因?yàn)楹瘮?shù)SKIPIF1<0定義域?yàn)镾KIPIF1<0，且SKIPIF1<0有意義，由復(fù)合函數(shù)單調(diào)性的判斷法則得，SKIPIF1<0在SKIPIF1<0上一定是增函數(shù)，故正確.故選：ACD.7．（2022·江蘇省灌南高級(jí)中學(xué)高三階段練習(xí)）若SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【分析】指數(shù)式換成對(duì)數(shù)式，解出SKIPIF1<0，逐個(gè)驗(yàn)證選項(xiàng).【詳解】由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，A選項(xiàng)正確；SKIPIF1<0，B選項(xiàng)錯(cuò)誤；SKIPIF1<0，C選項(xiàng)正確；SKIPIF1<0，D選項(xiàng)正確；故選：ACD8．（2022·重慶南開(kāi)中學(xué)高三階段練習(xí)）已知SKIPIF1<0、SKIPIF1<0為函數(shù)SKIPIF1<0的兩個(gè)不相同的零點(diǎn)，則下列式子一定正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABC【分析】分析可知直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，數(shù)形結(jié)合可得出SKIPIF1<0，利用基本不等式可判斷ABC選項(xiàng)，利用特殊值法可判斷D選項(xiàng).【詳解】令SKIPIF1<0可得SKIPIF1<0，則直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，且這兩個(gè)交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0、SKIPIF1<0，如下圖所示：由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，設(shè)SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，所以，SKIPIF1<0，對(duì)于A選項(xiàng)，SKIPIF1<0，A對(duì)；對(duì)于B選項(xiàng)，SKIPIF1<0，B對(duì)；對(duì)于C選項(xiàng)，SKIPIF1<0，則SKIPIF1<0，C對(duì)；對(duì)于D選項(xiàng)，取SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，D錯(cuò).故選：ABC.9．（2022·遼寧·東北育才學(xué)校高三階段練習(xí)）已知定義R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，又SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，且SKIPIF1<0，則（

）A．函數(shù)SKIPIF1<0的周期為12 B．SKIPIF1<0C．SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng) D．SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)【答案】ABD【分析】結(jié)合函數(shù)的對(duì)稱(chēng)性、奇偶性、周期性確定正確答案.【詳解】由SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng).由于SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0是奇函數(shù).所以SKIPIF1<0，所以SKIPIF1<0的周期為SKIPIF1<0，A選項(xiàng)正確.SKIPIF1<0，B選項(xiàng)正確.結(jié)合上述分析可知，SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0（SKIPIF1<0）對(duì)稱(chēng)，所以SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0（SKIPIF1<0）對(duì)稱(chēng)，所以SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0（SKIPIF1<0）對(duì)稱(chēng)，所以SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0（SKIPIF1<0）對(duì)稱(chēng)，令SKIPIF1<0，得SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，D選項(xiàng)正確，C選項(xiàng)錯(cuò)誤.故選：ABD三、填空題10．（2022·湖北·仙桃市田家炳實(shí)驗(yàn)高級(jí)中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镽，且SKIPIF1<0為奇函數(shù)，其圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0