江蘇省xx中學(xué)2022-2023學(xué)年高一上學(xué)期期中數(shù)學(xué)試題（含解析）

高一上學(xué)期期中數(shù)學(xué)試題一、單項(xiàng)選擇題：本大題共8小題，每小題5分，共40分．每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的．1.已知集合SKIPIF1<0且SKIPIF1<0，集合SKIPIF1<0，集合SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)集合SKIPIF1<0包含的元素特征，結(jié)合SKIPIF1<0的結(jié)果可得結(jié)果.【詳解】SKIPIF1<0，SKIPIF1<0.故選：D.2.已知SKIPIF1<0為實(shí)數(shù)，使“SKIPIF1<0”為真命題的一個(gè)充分不必要條件是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)全稱(chēng)量詞命題的真假性求得SKIPIF1<0的取值范圍，然后確定其充分不必要條件.【詳解】依題意，全稱(chēng)量詞命題：SKIPIF1<0為真命題，SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，所以SKIPIF1<0，所以使“SKIPIF1<0”為真命題的一個(gè)充分不必要條件是“SKIPIF1<0”.故選：B3.十六世紀(jì)中葉，英國(guó)數(shù)學(xué)家雷科德在《礪智石》一書(shū)中首先把“＝”作為等號(hào)使用，后來(lái)英國(guó)數(shù)學(xué)家哈利奧特首次命題正確是使用“＜”和“＞”符號(hào)，并逐漸被數(shù)學(xué)屆接受，不等號(hào)的引入對(duì)不等式的發(fā)展影響深遠(yuǎn)．若a，b，c∈R，則下列命題正確的是()A.若a＜b，則SKIPIF1<0 B.若a＞b＞0，則SKIPIF1<0C.若a＞b，則SKIPIF1<0 D.若SKIPIF1<0，則a＞b【答案】D【解析】【分析】舉反例說(shuō)明選項(xiàng)AC錯(cuò)誤；作差法說(shuō)明選項(xiàng)B錯(cuò)誤；不等式性質(zhì)說(shuō)明選項(xiàng)D正確.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，選項(xiàng)A錯(cuò)誤；SKIPIF1<0，所以SKIPIF1<0，所以選項(xiàng)B錯(cuò)誤；SKIPIF1<0時(shí)，SKIPIF1<0，所以選項(xiàng)C錯(cuò)誤；SKIPIF1<0時(shí)，SKIPIF1<0，所以選項(xiàng)D正確.故選：D4.設(shè)SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】觀察所求結(jié)構(gòu)知把SKIPIF1<0放到對(duì)數(shù)的真數(shù)部分作指數(shù)即可求解．【詳解】解：SKIPIF1<0，故選：C．5.設(shè)SKIPIF1<0為實(shí)數(shù)，若二次函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn)，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】【分析】根據(jù)二次函數(shù)的性質(zhì)求得正確答案.【詳解】二次函數(shù)SKIPIF1<0的開(kāi)口向上，對(duì)稱(chēng)軸為SKIPIF1<0，要使二次函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn)，則需SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C6.SKIPIF1<0世紀(jì)，在研究天文學(xué)的過(guò)程中，為了簡(jiǎn)化大數(shù)運(yùn)算，蘇格蘭數(shù)學(xué)家納皮爾發(fā)明了對(duì)數(shù)，對(duì)數(shù)的思想方法即把乘方和乘法運(yùn)算分別轉(zhuǎn)化為乘法和加法運(yùn)算，數(shù)學(xué)家拉普拉斯稱(chēng)贊“對(duì)數(shù)的發(fā)明在實(shí)效上等于把天文學(xué)家的壽命延長(zhǎng)了許多倍”．已知SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0所在的區(qū)間為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】利用指數(shù)和對(duì)數(shù)互化，結(jié)合對(duì)數(shù)運(yùn)算法則可求得SKIPIF1<0，由此可得SKIPIF1<0.【詳解】SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0.故選：C.7.已知奇函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且對(duì)任意兩個(gè)不相等的正實(shí)數(shù)SKIPIF1<0，都有SKIPIF1<0，在下列不等式中，一定成立的是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】利用題意得到SKIPIF1<0在SKIPIF1<0單調(diào)遞增，可得到SKIPIF1<0，結(jié)合奇函數(shù)即可得到答案【詳解】對(duì)任意兩個(gè)不相等的正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，因?yàn)镾KIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，且SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，故選：A8.已知函數(shù)SKIPIF1<0是定義域?yàn)閰^(qū)間SKIPIF1<0，且圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則滿(mǎn)足SKIPIF1<0的x的取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)給定的條件，可得SKIPIF1<0，再與已知聯(lián)立結(jié)合函數(shù)單調(diào)性及定義域解不等式作答.【詳解】因函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，則有SKIPIF1<0，而SKIPIF1<0，于是得SKIPIF1<0，即SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，于是得SKIPIF1<0，解得SKIPIF1<0，所以滿(mǎn)足SKIPIF1<0的x的取值范圍是SKIPIF1<0.故選：C二、多項(xiàng)選擇題：本大題共4小題，每小題5分，共20分．在每小題給出的四個(gè)選項(xiàng)中，不止一項(xiàng)是符合題目要求的，每題全選對(duì)者得5分，部分選對(duì)得2分，其他情況不得分．9.若“SKIPIF1<0”為真命題，“SKIPIF1<0”為假命題，則集合SKIPIF1<0可以是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】AD【解析】【分析】根據(jù)兩個(gè)命題的真假，判斷出集合SKIPIF1<0中元素的范圍，確定合適的答案.【詳解】由題意SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，即集合SKIPIF1<0中一定要有小于0的元素，且任何元素都小于3，故SKIPIF1<0，SKIPIF1<0滿(mǎn)足.故選：AD.10.下列說(shuō)法正確的是()A.“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件B.命題“SKIPIF1<0”的否定是“SKIPIF1<0”C.“SKIPIF1<0”是“SKIPIF1<0”的既不充分也不必要條件D.設(shè)SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的必要不充分條件【答案】ACD【解析】【分析】對(duì)于ACD，化簡(jiǎn)不等式即可判斷；對(duì)于B，利用全稱(chēng)命題的否定即可判斷【詳解】對(duì)于A，由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，所以“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件，故正確；對(duì)于B，命題“SKIPIF1<0”的否定是“SKIPIF1<0”，故不正確；對(duì)于C，由SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0，所以“SKIPIF1<0”是“SKIPIF1<0”的既不充分也不必要條件，故正確；對(duì)于D，由SKIPIF1<0解得SKIPIF1<0且SKIPIF1<0，所以“SKIPIF1<0”是“SKIPIF1<0”的必要不充分條件，故正確，故選：ACD11.設(shè)SKIPIF1<0為正實(shí)數(shù)，SKIPIF1<0，則下列不等式中對(duì)一切滿(mǎn)足條件的SKIPIF1<0恒成立的是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】AC【解析】【分析】根據(jù)特殊值以及基本不等式對(duì)選項(xiàng)進(jìn)行分析，從而確定正確選項(xiàng).【詳解】A選項(xiàng)，由基本不等式得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，A選項(xiàng)正確.B選項(xiàng)，SKIPIF1<0時(shí)，SKIPIF1<0，但SKIPIF1<0，B選項(xiàng)錯(cuò)誤.C選項(xiàng)，由基本不等式得SKIPIF1<0，，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，C選項(xiàng)正確.D選項(xiàng)，SKIPIF1<0時(shí)，SKIPIF1<0，但SKIPIF1<0，D選項(xiàng)錯(cuò)誤.故選：AC12.已知函數(shù)SKIPIF1<0，則()A.SKIPIF1<0奇函數(shù) B.SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C.方程SKIPIF1<0有兩個(gè)實(shí)數(shù)根 D.函數(shù)SKIPIF1<0的值域是SKIPIF1<0【答案】BCD【解析】【分析】求出函數(shù)的定義域，不關(guān)于原點(diǎn)對(duì)稱(chēng)可判斷A，分離常數(shù)后可得函數(shù)SKIPIF1<0的單調(diào)性可判斷B，解方程可判斷C，分離常數(shù)求解函數(shù)值域可判斷D.【詳解】A．函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，不關(guān)于原點(diǎn)對(duì)稱(chēng)，既不是奇函數(shù)也不是偶函數(shù)，故A錯(cuò)誤；B．SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，B正確；C．由題可得SKIPIF1<0是方程SKIPIF1<0的一個(gè)根，SKIPIF1<0時(shí)，SKIPIF1<0(舍去)，SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，故C正確；D．SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)的值域?yàn)镾KIPIF1<0，故D正確.故選：BCD.三、填空題：本大題共4小題，每小題5分，共計(jì)20分．請(qǐng)把答案填寫(xiě)在答題卡相應(yīng)位置上．13.命題“SKIPIF1<0，SKIPIF1<0或SKIPIF1<0”的否定是____________．【答案】SKIPIF1<0，SKIPIF1<0【解析】【分析】由特稱(chēng)命題的否定形式可直接得到結(jié)果.【詳解】由特稱(chēng)命題的否定知：原命題的否定為SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0.14.已知三個(gè)不等式：①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0，用其中兩個(gè)作為條件，剩下的一個(gè)作為結(jié)論，則可組成______個(gè)真命題．【答案】3【解析】【分析】根據(jù)題意，結(jié)合不等式性質(zhì)分別判斷①、②、③作為結(jié)論的命題的真假性即可.【詳解】由不等式性質(zhì)，得SKIPIF1<0；SKIPIF1<0；SKIPIF1<0．故可組成3個(gè)真命題．故答案為：3.15.SKIPIF1<0的值為_(kāi)___________【答案】SKIPIF1<0##SKIPIF1<0【解析】【分析】將SKIPIF1<0和SKIPIF1<0配湊成完全平方的形式，代入所求式子中，結(jié)合對(duì)數(shù)運(yùn)算可求得結(jié)果.【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0.16.已知函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng)．若SKIPIF1<0，則SKIPIF1<0____________，若SKIPIF1<0，函數(shù)SKIPIF1<0的最小值記為SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)___________．【答案】①.SKIPIF1<0②.SKIPIF1<0【解析】【分析】根據(jù)函數(shù)的對(duì)稱(chēng)性，利用特殊值SKIPIF1<0得到方程組，求出SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的關(guān)系，從而求出SKIPIF1<0，得到SKIPIF1<0，根據(jù)對(duì)稱(chēng)性?xún)H研究SKIPIF1<0時(shí)函數(shù)最小值，，令SKIPIF1<0，SKIPIF1<0，根據(jù)二次函數(shù)的性質(zhì)求出SKIPIF1<0，再根據(jù)SKIPIF1<0的取值范圍計(jì)算可得.【詳解】解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)槠鋱D象關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，因?yàn)槠鋱D象關(guān)于SKIPIF1<0對(duì)稱(chēng)，所以SKIPIF1<0，此時(shí)SKIPIF1<0，由對(duì)稱(chēng)性?xún)H研究SKIPIF1<0時(shí)函數(shù)最小值，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；故答案為：SKIPIF1<0；SKIPIF1<0四、解答題：本大題共6小題，共計(jì)70分．請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟．17.化簡(jiǎn)求值(需要寫(xiě)出計(jì)算過(guò)程)(1)若SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值；(2)SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)先取對(duì)數(shù)將SKIPIF1<0表示出來(lái)，代入計(jì)算即可；(2)直接計(jì)算即可.【小問(wèn)1詳解】SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0【小問(wèn)2詳解】原式SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<018.已知集合A＝{x||x|－2≤0}，集合SKIPIF1<0．(1)設(shè)a為實(shí)數(shù)，若集合C＝{x|x≥3a且x≤2a＋1}，且C?(A∩B)，求a的取值范圍：(2)設(shè)m為實(shí)數(shù)，集合SKIPIF1<0，若x∈(A∪B)是x∈D的必要不充分條件，判斷滿(mǎn)足條件的m是否存在，若存在，求m的取值范圍：若不存在，請(qǐng)說(shuō)明理由．【答案】(1)SKIPIF1<0；(2)存在；SKIPIF1<0.【解析】【分析】(1)根據(jù)解絕對(duì)值不等式的公式，結(jié)合分式的性質(zhì)、交集的定義、子集的性質(zhì)進(jìn)行求解即可；(2)根據(jù)必要不充分條件的定義，結(jié)合一元二次不等式的解法進(jìn)行求解即可.【小問(wèn)1詳解】SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，(1)由已知得SKIPIF1<0，①SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)滿(mǎn)足題意；②SKIPIF1<0時(shí)，SKIPIF1<0，要滿(mǎn)足題意需SKIPIF1<0綜上所述，a的取值范圍是SKIPIF1<0；【小問(wèn)2詳解】由已知得SKIPIF1<0，由題意得D是SKIPIF1<0的真子集SKIPIF1<0，所以SKIPIF1<0，要滿(mǎn)足題意需SKIPIF1<0(等號(hào)不同時(shí)成立)SKIPIF1<0答：滿(mǎn)足條件的m存在，取值范圍是SKIPIF1<0．19.設(shè)a，b，c為實(shí)數(shù)，且SKIPIF1<0，已知二次函數(shù)SKIPIF1<0，滿(mǎn)足SKIPIF1<0，SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的解析式：(2)設(shè)SKIPIF1<0，當(dāng)x∈[t，t＋2]時(shí)，求函數(shù)f(x)的最大值g(t)(用t表示)．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)由題意，根據(jù)多項(xiàng)式相等的條件，建立方程組，可得答案；(2)根據(jù)二次函數(shù)的性質(zhì)，由給定區(qū)間與對(duì)稱(chēng)軸之間的位置關(guān)系，利用分類(lèi)討論，可得答案.【小問(wèn)1詳解】由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0；【小問(wèn)2詳解】由SKIPIF1<0可知其對(duì)稱(chēng)軸為SKIPIF1<0軸，開(kāi)口向下，①當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0；③當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0綜上所述，SKIPIF1<0．20.某高校為舉辦百年校慶，需要SKIPIF1<0氦氣用于制作氣球裝飾校園，化學(xué)實(shí)驗(yàn)社團(tuán)主動(dòng)承擔(dān)了這一任務(wù)．社團(tuán)已有的設(shè)備每天最多可制備氦氣SKIPIF1<0，按計(jì)劃社團(tuán)必須在SKIPIF1<0天內(nèi)制備完畢．社團(tuán)成員接到任務(wù)后，立即以每天SKIPIF1<0的速度制備氦氣．已知每制備SKIPIF1<0氦氣所需的原料成本為SKIPIF1<0百元．若氦氣日產(chǎn)量不足SKIPIF1<0，日均額外成本為SKIPIF1<0(百元)；若氦氣日產(chǎn)量大于等于SKIPIF1<0，日均額外成本為SKIPIF1<0(百元)．制備成本由原料成本和額外成本兩部分組成．(1)寫(xiě)出總成本SKIPIF1<0(百元)關(guān)于日產(chǎn)量SKIPIF1<0的關(guān)系式(2)當(dāng)社團(tuán)每天制備多少升氦氣時(shí)，總成本最少？并求出最低成本．【答案】(1)SKIPIF1<0(2)當(dāng)社團(tuán)每天制備SKIPIF1<0氦氣時(shí)，總成本最少，最低成本為SKIPIF1<0百元【解析】【分析】(1)根據(jù)生產(chǎn)天數(shù)要求，可確定SKIPIF1<0的取值范圍；計(jì)算可得日產(chǎn)量不足SKIPIF1<0和大于等于SKIPIF1<0時(shí)，SKIPIF1<0氦氣的平均成本，由此可得關(guān)系式；(2)分別在SKIPIF1<0、SKIPIF1<0的情況下，利用基本不等式和二次函數(shù)求最值的方法可求得最小值，綜合兩種情況可得結(jié)論.【小問(wèn)1詳解】若每天生產(chǎn)SKIPIF1<0氦氣，則需生產(chǎn)SKIPIF1<0天，SKIPIF1<0，則SKIPIF1<0；若氦氣日產(chǎn)量不足SKIPIF1<0，則SKIPIF1<0氦氣的平均成本為SKIPIF1<0百元；若氦氣日產(chǎn)量大于等于SKIPIF1<0，則SKIPIF1<0氦氣的平均成本為SKIPIF1<0百元；SKIPIF1<0.【小問(wèn)2詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào))，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0；綜上所述：當(dāng)社團(tuán)每天制備SKIPIF1<0氦氣時(shí)，總成本最少，最低成本SKIPIF1<0百元.21.設(shè)定義在SKIPIF1<0上的函數(shù)SKIPIF1<0，對(duì)任意SKIPIF1<0，恒有SKIPIF1<0．若SKIPIF1<0時(shí)，SKIPIF1<0．(1)判斷SKIPIF1<0的奇偶性，并加以證明；(2)判斷SKIPIF1<0的單調(diào)性，并加以證明；(3)設(shè)SKIPIF1<0為實(shí)數(shù)，若SKIPIF1<0，不等式SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0是奇函數(shù)；證明見(jiàn)解析(2)SKIPIF1<0為減函數(shù)；證明見(jiàn)解析(3)SKIPIF1<0【解析】【分析】(1)令SKIPIF1<0可求得SKIPIF1<0，令SKIPIF1<0可得SKIPIF1<0，由此可得奇偶性；(2)設(shè)SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，由此可得單調(diào)性；(3)利用單調(diào)性可將恒成立的不等式化為SKIPIF1<0，利用二次函數(shù)性質(zhì)可求得SKIPIF1<0，由此可得SKIPIF1<0的取值范圍.【小問(wèn)1詳解】令SKIPIF1<0，則SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0為定義在SKIPIF1<0上的奇函數(shù).【小問(wèn)2詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為定義在SKIPIF1<0上的減函數(shù).小問(wèn)3詳解】由SKIPIF1<0得：SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0；SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值，最大值為SKIPIF1<0