湖北省黃岡市2022-2023學(xué)年高一上學(xué)期元月期末數(shù)學(xué)試題（含解析）

高一上學(xué)期期末數(shù)學(xué)試題一?選擇題：共8小題，每小題5分，共40分.在每小題給出的四個選項中，只有一項是符合題目要求的.1.命題“SKIPIF1<0”的否定為()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)全稱量詞命題的否定是存在量詞命題直接寫出即可.【詳解】因為全稱量詞命題的否定是存在量詞命題，所以命題“SKIPIF1<0”的否定為“SKIPIF1<0”.故選:D.2.已知集合SKIPIF1<0則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】解不等式SKIPIF1<0得集合SKIPIF1<0，求函數(shù)SKIPIF1<0的定義域得集合SKIPIF1<0，再求SKIPIF1<0即可.【詳解】由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0有意義滿足SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0SKIPIF1<0，故選：D3.下列函數(shù)中最小正周期為SKIPIF1<0且是奇函數(shù)的為()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)正切函數(shù)的周期與奇偶性可判斷AB，根據(jù)誘導(dǎo)公式化簡CD的解析式，再根據(jù)正余弦函數(shù)的奇偶性可判斷.【詳解】SKIPIF1<0的最小正周期為SKIPIF1<0,故A錯誤；SKIPIF1<0為非奇非偶函數(shù)，故B錯誤；SKIPIF1<0，易知為奇函數(shù)，且最小正周期為SKIPIF1<0，故C正確；SKIPIF1<0為偶函數(shù)，故D錯誤.故選:C.4.衡量病毒傳播能力的一個指標(biāo)叫做傳播指數(shù)SKIPIF1<0，它指的是在自然情況下(沒有外力介人，同時所有人都沒有免疫)一個感染者傳染的平均人數(shù).它的計算公式是：SKIPIF1<0確診病例增長率SKIPIF1<0系列間隔，其中系列間隔是指在一個傳播鏈中兩例連續(xù)病例的間隔時間(單位：天).根據(jù)統(tǒng)計，某種傳染病例的平均增長率為SKIPIF1<0，兩例連續(xù)病例間隔時間平均為4天.根據(jù)以上數(shù)據(jù)計算，若甲感染這種傳染病，則經(jīng)過4輪傳播后由甲引起的得病總?cè)藬?shù)(不含甲)為()A.81人 B.120人 C.243人 D.36人【答案】B【解析】【分析】根據(jù)SKIPIF1<0確診病例增長率SKIPIF1<0系列間隔，先求得SKIPIF1<0，然后求經(jīng)過4輪傳播后由甲引起的得病總?cè)藬?shù).【詳解】由題意得：SKIPIF1<0，所以經(jīng)過4輪傳播后由甲引起的得病的總?cè)藬?shù)約為：SKIPIF1<0.故選：B.5.已知SKIPIF1<0，則有()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】將SKIPIF1<0化到同一個單調(diào)區(qū)間上的同名函數(shù)比大小，再將SKIPIF1<0與SKIPIF1<0比大小.詳解】SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0為增函數(shù)，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，故選：C6.已知角SKIPIF1<0的終邊過點SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)三角函數(shù)的定義和同角三角函數(shù)的基本關(guān)系即可求解.【詳解】由三角函數(shù)的定義可得：SKIPIF1<0，也即SKIPIF1<0，由SKIPIF1<0可得：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0(舍去)，因為角SKIPIF1<0的終邊過點SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，故選：SKIPIF1<0.7.已知SKIPIF1<0是定義在R上的奇函數(shù)，SKIPIF1<0，對SKIPIF1<0，且SKIPIF1<0有SKIPIF1<0，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)題干條件得到函數(shù)SKIPIF1<0在R上的單調(diào)遞增，且SKIPIF1<0，換元后得到SKIPIF1<0，分三種情況，由單調(diào)性解不等式得到SKIPIF1<0，從而得到SKIPIF1<0.【詳解】因為對SKIPIF1<0，且SKIPIF1<0有SKIPIF1<0，所以SKIPIF1<0上，SKIPIF1<0單調(diào)遞增，因為SKIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0在R上的單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，顯然滿足SKIPIF1<0，當(dāng)SKIPIF1<0時，因為SKIPIF1<0，SKIPIF1<0在R上的單調(diào)遞增，所以當(dāng)SKIPIF1<0時，滿足SKIPIF1<0，當(dāng)SKIPIF1<0時，因為SKIPIF1<0，SKIPIF1<0在R上的單調(diào)遞增，所以當(dāng)SKIPIF1<0時，滿足SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.故選：B8.已知函數(shù)SKIPIF1<0若關(guān)于SKIPIF1<0的方程SKIPIF1<0有SKIPIF1<0個不同的實數(shù)根，則實數(shù)SKIPIF1<0的取值范圍為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】令SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象，分析可知關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0內(nèi)有兩個不等的實根，令SKIPIF1<0，利用二次函數(shù)的零點分布可得出關(guān)于SKIPIF1<0的不等式組，解之即可.【詳解】令SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示：因為關(guān)于SKIPIF1<0的方程SKIPIF1<0有SKIPIF1<0個不同的實數(shù)根，則關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0內(nèi)有兩個不等的實根，設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)有兩個不等的零點，所以，SKIPIF1<0，解得SKIPIF1<0.故選：A.二?多項選擇題：本題共4小題，每小題5分，共20分.在每小題給出的四個選項中，有多項符合題目要求.全部選對的得5分，部分選對的得2分，有選錯的得0分.9.下列計算結(jié)果為有理數(shù)的是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)特殊角的三角函數(shù)判斷A，根據(jù)對數(shù)的運算性質(zhì)與換底公式判斷BCD.【詳解】SKIPIF1<0，不是有理數(shù)，故A錯誤；SKIPIF1<0，是有理數(shù)，故B正確；SKIPIF1<0，是有理數(shù)，故C正確；SKIPIF1<0，是有理數(shù)，故D正確.故選:BCD10.若SKIPIF1<0，則使“SKIPIF1<0”成立的一個必要不充分條件是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD【解析】【分析】若SKIPIF1<0SKIPIF1<0，SKIPIF1<0,則SKIPIF1<0是SKIPIF1<0的必要不充分條件，解指數(shù)不等式可判斷A；取SKIPIF1<0可判斷B；C選項中利用SKIPIF1<0可判斷；D選項中利用指數(shù)函數(shù)的值域進(jìn)行判斷.【詳解】對于A，由SKIPIF1<0可得SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的必要不充分條件，故A正確；對于B，當(dāng)SKIPIF1<0時，SKIPIF1<0，此時SKIPIF1<0，得不到SKIPIF1<0，故B錯誤；對于C，SKIPIF1<0時，SKIPIF1<0，此時SKIPIF1<0,故“SKIPIF1<0”不是使“SKIPIF1<0”成立的充分條件.因為SKIPIF1<0,所以SKIPIF1<0.當(dāng)SKIPIF1<0時，必有SKIPIF1<0.所以“SKIPIF1<0”是使“SKIPIF1<0”成立的必要條件.故“SKIPIF1<0”是使“SKIPIF1<0”成立必要不充分條件，故C正確；對于D，當(dāng)SKIPIF1<0時，SKIPIF1<0，此時SKIPIF1<0,故“SKIPIF1<0”不是使“SKIPIF1<0”成立的充分條件.當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0中至少有一個正數(shù)，不妨設(shè)SKIPIF1<0,則SKIPIF1<0,又因為SKIPIF1<0，則必有SKIPIF1<0,所以“SKIPIF1<0”是使“SKIPIF1<0”成立的必要條件.故“SKIPIF1<0”是使“SKIPIF1<0”成立必要不充分條件，故D正確.故選;ACD.11.函數(shù)SKIPIF1<0，以下正確的是()A.若SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0B.若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0C.當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0單調(diào)且在SKIPIF1<0不單調(diào)，則SKIPIF1<0.D.當(dāng)SKIPIF1<0時，若對任意的SKIPIF1<0有SKIPIF1<0成立，則SKIPIF1<0的最小值為SKIPIF1<0【答案】BCD【解析】【分析】由函數(shù)周期公式可判斷A；由題意得SKIPIF1<0，結(jié)合函數(shù)周期公式可判斷B；若SKIPIF1<0在SKIPIF1<0單調(diào)，則SKIPIF1<0且SKIPIF1<0，結(jié)合SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，驗證題設(shè)條件可判斷C；由題意得SKIPIF1<0，即SKIPIF1<0，求得SKIPIF1<0最小值可判斷D.【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故A錯誤；SKIPIF1<0，又SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故B正確；當(dāng)SKIPIF1<0時，若SKIPIF1<0在SKIPIF1<0單調(diào)，則SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，此時SKIPIF1<0在SKIPIF1<0單調(diào)且在SKIPIF1<0不單調(diào)，故C正確；當(dāng)SKIPIF1<0時，SKIPIF1<0，又因為對任意的SKIPIF1<0有SKIPIF1<0成立，則SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0取最小值SKIPIF1<0，故D正確．故選：BCD.12.空曠的田野上兩根電線桿之間的電線有相似的曲線形態(tài).這些曲線在數(shù)學(xué)上稱為懸鏈線.懸鏈線在工程上有廣泛的應(yīng)用.在恰當(dāng)?shù)淖鴺?biāo)系中，這類函數(shù)的表達(dá)式可以為SKIPIF1<0(其中SKIPIF1<0為非零常數(shù))，則對于函數(shù)SKIPIF1<0以下結(jié)論正確的是()A.若SKIPIF1<0，則SKIPIF1<0為偶函數(shù)B.若SKIPIF1<0，則函數(shù)SKIPIF1<0的零點為0和SKIPIF1<0C.若SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值為2D.若SKIPIF1<0為奇函數(shù)，且SKIPIF1<0使SKIPIF1<0成立，則SKIPIF1<0的最小值為SKIPIF1<0【答案】ABD【解析】【分析】根據(jù)函數(shù)的奇偶性定義判斷A即可；利用函數(shù)零點的定義及指對運算即可求得函數(shù)SKIPIF1<0的零點，從而判斷B即可；根據(jù)SKIPIF1<0得SKIPIF1<0，討論SKIPIF1<0的符號從而確定函數(shù)值域，從而判斷C即可；根據(jù)含參不等式能成立，利用指數(shù)函數(shù)的性質(zhì)進(jìn)行參變分離，結(jié)合基本不等式求得最值，即可得SKIPIF1<0的取值范圍，從而判斷D即可.【詳解】解：對于A，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)定義域為SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0為偶函數(shù)，故A正確；對于B，若SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0，整理得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點為0和SKIPIF1<0，故B正確；對于C，若SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時等號成立；當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時等號成立；所以SKIPIF1<0，故C錯誤；對于D，若SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0使SKIPIF1<0成立，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0能成立，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時，即SKIPIF1<0時，等號成立，所以SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0，故D正確.故選：ABD.三?填空題：本大題共4小題，每小題5分，共20分13.函數(shù)SKIPIF1<0的定義域為__________.【答案】SKIPIF1<0【解析】【分析】由解析式可得SKIPIF1<0，求解即可.【詳解】由題意可得SKIPIF1<0,故SKIPIF1<0，即SKIPIF1<0.故函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為:SKIPIF1<0.14.已知函數(shù)SKIPIF1<0的圖象過定點SKIPIF1<0，且點SKIPIF1<0在指數(shù)函數(shù)SKIPIF1<0圖象上，則SKIPIF1<0__________.【答案】SKIPIF1<0【解析】【分析】由對數(shù)函數(shù)的圖象可得SKIPIF1<0，故可求SKIPIF1<0的解析式，根據(jù)對數(shù)的運算即可求解.【詳解】在SKIPIF1<0中，令SKIPIF1<0,可得SKIPIF1<0,故SKIPIF1<0.設(shè)SKIPIF1<0,由題意可得SKIPIF1<0,解得SKIPIF1<0.所以SKIPIF1<0，SKIPIF1<0.故答案為:SKIPIF1<0.15.已知SKIPIF1<0，則SKIPIF1<0的最小值為__________.【答案】SKIPIF1<0##1.6【解析】【分析】由SKIPIF1<0可得SKIPIF1<0，又SKIPIF1<0，再用“乘1法”即可求最小值.【詳解】因為SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時等號成立.故SKIPIF1<0的最小值為SKIPIF1<0.故答案為:SKIPIF1<0.16.已知SKIPIF1<0，SKIPIF1<0，若對SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0成立，則實數(shù)SKIPIF1<0的取值范圍為__________.【答案】SKIPIF1<0【解析】【分析】分析可知，SKIPIF1<0，求出SKIPIF1<0在SKIPIF1<0上的最小值為SKIPIF1<0，可知SKIPIF1<0對任意的SKIPIF1<0恒成立，利用參變量分離法可求得實數(shù)SKIPIF1<0的取值范圍.【詳解】若對SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0成立，則SKIPIF1<0，當(dāng)SKIPIF1<0時，令SKIPIF1<0，則SKIPIF1<0，由對勾函數(shù)的單調(diào)性可知，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以，當(dāng)SKIPIF1<0時，SKIPIF1<0，故當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0對任意的SKIPIF1<0恒成立，所以，SKIPIF1<0對任意的SKIPIF1<0恒成立，由對勾函數(shù)的單調(diào)性可知，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<0.四?解答題：本大題共6小題，共70分.解答應(yīng)寫出文字說明?證明過程或演算步驟.17.(1)已知SKIPIF1<0，求SKIPIF1<0的值.(2)已知SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.【解析】【分析】(1)利用誘導(dǎo)公式及同角三角函數(shù)的基本關(guān)系可得原式SKIPIF1<0，代值求解即可；(2)將SKIPIF1<0兩邊平方可求SKIPIF1<0，從而可求SKIPIF1<0，利用平方差公式可得SKIPIF1<0，故可求解.【詳解】(1)原式=SKIPIF1<0(2)SKIPIF1<0兩邊平方得SKIPIF1<0SKIPIF1<0.SKIPIF1<0∴SKIPIF1<018.設(shè)函數(shù)SKIPIF1<0.(1)若不等式SKIPIF1<0的解集為SKIPIF1<0，求SKIPIF1<0的值；(2)若SKIPIF1<0，且SKIPIF1<0都有SKIPIF1<0，求SKIPIF1<0的最大值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)一元二次不等式的解集即可求解；(2)根據(jù)題意可得函數(shù)關(guān)于直線SKIPIF1<0對稱，利用二次函數(shù)的對稱軸得出SKIPIF1<0，再結(jié)合基本不等式即可求解.【小問1詳解】依題意可知：SKIPIF1<0和SKIPIF1<0是方程SKIPIF1<0的兩根，則有SKIPIF1<0且SKIPIF1<0∴SKIPIF1<0【小問2詳解】由SKIPIF1<0知SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，即SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時等號成立.∴SKIPIF1<0的最大值為SKIPIF1<019.已知函數(shù)SKIPIF1<0為奇函數(shù).(1)求函數(shù)SKIPIF1<0的最大值與最小值，并分別寫出取最大值與最小值時相應(yīng)SKIPIF1<0的取值集合.(2)求函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間.【答案】(1)SKIPIF1<0時SKIPIF1<0取最小值SKIPIF1<0；SKIPIF1<0時SKIPIF1<0取最大值2;(2)SKIPIF1<0與SKIPIF1<0.【解析】【分析】(1)根據(jù)奇函數(shù)的性質(zhì)可得SKIPIF1<0，結(jié)合SKIPIF1<0可求SKIPIF1<0從而可得SKIPIF1<0，根據(jù)正弦函數(shù)的性質(zhì)即可求解；(2)SKIPIF1<0，根據(jù)正弦函數(shù)的單調(diào)性即可求解.【小問1詳解】依題意有SKIPIF1<0即SKIPIF1<0，為奇函數(shù)，滿足題意.當(dāng)SKIPIF1<0時SKIPIF1<0取最小值SKIPIF1<0；當(dāng)SKIPIF1<0時SKIPIF1<0取最大值2.【小問2詳解】依題意SKIPIF1<0，若SKIPIF1<0單調(diào)遞減，則SKIPIF1<0SKIPIF1<0SKIPIF1<0又SKIPIF1<0，令SKIPIF1<0得其減區(qū)間為SKIPIF1<0與SKIPIF1<0.20.某兒童玩具廠生產(chǎn)的某一款益智玩具去年年銷量為2百萬件，每件銷售價格為20元，成本16元.今年計劃投入適當(dāng)廣告費進(jìn)行促銷.預(yù)計該款玩具的年銷售量SKIPIF1<0百萬件與年廣告費用SKIPIF1<0百萬元滿足SKIPIF1<0，現(xiàn)已知每件玩具的銷售價為年平均每件玩具所占廣告費的SKIPIF1<0與原銷售價之和.(1)當(dāng)投入廣告費為2百萬元時，要使該玩具的年利潤不少于12百萬元，求SKIPIF1<0的取值范圍；(2)若SKIPIF1<0時，則當(dāng)投入多少百萬元浩費該玩具生產(chǎn)廠獲得最大利潤.【答案】(1)SKIPIF1<0；(2)當(dāng)廣告費2百萬時最大利潤為SKIPIF1<0萬元.【解析】【分析】(1)年利潤SKIPIF1<0，解SKIPIF1<0即可；(2)當(dāng)SKIPIF1<0時，SKIPIF1<0,利用函數(shù)的單調(diào)性即可求解.小問1詳解】當(dāng)SKIPIF1<0時SKIPIF1<0，銷售價為SKIPIF1<0，年利潤SKIPIF1<0，解得SKIPIF1<0.【小問2詳解】當(dāng)SKIPIF1<0時，年利潤SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時SKIPIF1<0,所以SKIPIF1<0.綜上：當(dāng)廣告費2百萬時最大利潤為SKIPIF1<0萬元.21.已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0圖象與SKIPIF1<0的圖象關(guān)于SKIPIF1<0對稱.(1)若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，求實數(shù)SKIPIF1<0的取值范圍；(2)不等式SKIPIF1<0在SKIPIF1<0上恒成立，求實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)依題意可得SKIPIF1<0，再根據(jù)復(fù)合函數(shù)的單調(diào)性可列出不等式，結(jié)合二次不等式恒成立求解即可；(2)把問題轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上恒成立，分離參數(shù)，轉(zhuǎn)化為最值比較即可.【小問1詳解】因為函數(shù)SKIPIF1<0圖象與SKIPIF1<0的圖象關(guān)于SKIPIF1<0對稱.所以SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，令SKIPIF1<0SKIPIF1<0則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0對SKIPIF1<0恒成立.SKIPIF1<0，且SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，符合題意；當(dāng)SKIPIF1<0時，SKIPIF1<0的對稱軸為SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，符合題意.故t的取值范圍為SKIPIF1<0.【小問2詳解】依題意有SKIPIF1<0且SKIPIF1<0不等式SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，SKIPIF1<0在SKIPIF1<0上恒成立，當(dāng)SKIPIF1<0時不等式成立，所以必須SKIPIF1<0在SKIPIF1<0上恒成立.SKIPIF1<0令SKIPIF1<0而SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0綜上：a的取值范圍為SKIPIF1<0.22.已知SKIPIF1<0為SKIPIF1<0上的偶函數(shù)，當(dāng)SKIPIF1<0時函數(shù)SKIPIF1<0.(1)求SKIPIF1<0并求SKIPIF1<0的解析式；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0的最大值為SKIPIF1<0，求SKIPIF1<0值并求使不等式SKIPIF1<0成立實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0，SKIPIF1<0【解析】【分析】(1)由SKIPIF1<0為SKIPIF1<0上的偶函數(shù)，得SKIPIF1<0，可求SKIPIF1<0的值；當(dāng)SKIPIF1<0時SKIPIF1<0，SKIPI