山西省運(yùn)城市2022-2023學(xué)年高一上學(xué)期期末數(shù)學(xué)試題（含答案詳解）

運(yùn)城市2022-2023學(xué)年第一學(xué)期期末調(diào)研測試高一數(shù)學(xué)試題本試題滿分150分，考試時(shí)間120分鐘注意事項(xiàng)：1.答題前，考生務(wù)必先將自己的姓名、準(zhǔn)考證號(hào)填寫在答題卡上，認(rèn)真核對(duì)條形碼上的姓名、準(zhǔn)考證號(hào)，并將條形碼粘貼在答題卡的指定位置上.2.答題時(shí)使用0.5毫米的黑色中性（簽字）筆或碳素筆書寫，字體工整、筆跡清楚.3.請(qǐng)按照題號(hào)在各題的答題區(qū)域（黑色線框）內(nèi)作答，超出答題區(qū)域書寫的答案無效.4.保持卡面清潔，不折疊，不破損.一、單選題（本大題共8小題，每小題5分，共40分.在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的）1.已知SKIPIF1<0，集合SKIPIF1<0，SKIPIF1<0，則下列關(guān)系正確的是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】解不等式得SKIPIF1<0，由集合的運(yùn)算與關(guān)系對(duì)選項(xiàng)逐一判斷，【詳解】由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，對(duì)于A，SKIPIF1<0，故A錯(cuò)誤，對(duì)于B，C，SKIPIF1<0，故B錯(cuò)誤，C正確，對(duì)于D，SKIPIF1<0，故D錯(cuò)誤，故選：C2.函數(shù)SKIPIF1<0在的零點(diǎn)所在區(qū)間是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)零點(diǎn)存在性定理即可求解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0和SKIPIF1<0在SKIPIF1<0上都是單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間是SKIPIF1<0.故選：B3.設(shè)SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的SKIPIF1<0SKIPIF1<0A.充分不必要條件 B.必要不充分條件C.充分必要條件 D.既不充分也不必要條件【答案】B【解析】【分析】根據(jù)三角函數(shù)的性質(zhì)，利用充分條件和必要條件的定義進(jìn)行判斷．【詳解】由SKIPIF1<0，可知SKIPIF1<0．SKIPIF1<0“SKIPIF1<0”是“SKIPIF1<0”的必要不充分條件．故選B．【點(diǎn)睛】本題主要考查充分條件和必要條件的應(yīng)用，利用三角函數(shù)的性質(zhì)是解決本題的關(guān)鍵，比較基礎(chǔ)．4.我國在文昌航天發(fā)射場用長征五號(hào)運(yùn)載火箭成功發(fā)射探月工程嫦娥五號(hào)探測器，順利將探測器送入預(yù)定軌道，經(jīng)過兩次軌道修正，嫦娥五號(hào)順利進(jìn)入環(huán)月軌道飛行，嫦娥五號(hào)從橢圓形環(huán)月軌道變?yōu)榻鼒A形環(huán)月軌道，若這時(shí)把近圓形環(huán)月軌道看作圓形軌道，嫦娥五號(hào)距離月表400千米，已知月球半徑約為1738千米，則嫦娥五號(hào)繞月每旋轉(zhuǎn)SKIPIF1<0弧度，飛過的路程約為（SKIPIF1<0）（）A.1069千米 B.1119千米 C.2138千米 D.2238千米【答案】D【解析】【分析】利用弧長公式直接求解.【詳解】嫦娥五號(hào)繞月飛行半徑為400+1738=2138，所以嫦娥五號(hào)繞月每旋轉(zhuǎn)SKIPIF1<0弧度，飛過的路程約為SKIPIF1<0（千米）.故選：D5.已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的性質(zhì)比較大小可得答案.【詳解】SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0，綜上所述：SKIPIF1<0.故選：D6.我國著名數(shù)學(xué)家華羅庚曾說：“數(shù)缺形時(shí)少直觀，形少數(shù)時(shí)難入微，數(shù)形結(jié)合白般好，隔離分家萬事休.”在數(shù)學(xué)的學(xué)習(xí)和研究中，有時(shí)可憑借函數(shù)的圖象分析函數(shù)解析式的特征.已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則函數(shù)SKIPIF1<0的解析式可能為（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)圖象函數(shù)為奇函數(shù)，排除D；再根據(jù)函數(shù)定義域排除B；再根據(jù)SKIPIF1<0時(shí)函數(shù)值為正排除A；即可得出結(jié)果.【詳解】由題干中函數(shù)圖象可知其對(duì)應(yīng)的函數(shù)為奇函數(shù)，而D中的函數(shù)為偶函數(shù)，故排除D；由題干中函數(shù)圖象可知函數(shù)的定義域不是實(shí)數(shù)集，故排除B；對(duì)于A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不滿足圖象；對(duì)于C，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，滿足圖象.故排除A，選C.故選：C7.已知SKIPIF1<0為第二象限角，且SKIPIF1<0，則SKIPIF1<0的值是（）ASKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】利用誘導(dǎo)公式可得出SKIPIF1<0的值，利用同角三角函數(shù)的基本關(guān)系可求得SKIPIF1<0的值，再利用誘導(dǎo)公式化簡所求代數(shù)式，代值計(jì)算即可得出所求代數(shù)式的值.【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，又因?yàn)镾KIPIF1<0為第二象限角，則SKIPIF1<0，因此，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.故選：A.8.函數(shù)SKIPIF1<0的圖像與函數(shù)SKIPIF1<0的圖像所有交點(diǎn)的橫坐標(biāo)之和等于（）A.8 B.10 C.12 D.14【答案】C【解析】【分析】由題意可得，兩個(gè)函數(shù)有公共對(duì)稱軸SKIPIF1<0，分別做出兩個(gè)函數(shù)圖像，結(jié)合圖像即可得到結(jié)果.【詳解】函數(shù)SKIPIF1<0的圖像與函數(shù)SKIPIF1<0的圖像有公共對(duì)稱軸SKIPIF1<0，分別做出兩個(gè)函數(shù)的圖像如圖所示，由圖像可知，兩個(gè)函數(shù)共有12個(gè)交點(diǎn)，且關(guān)于直線SKIPIF1<0對(duì)稱，則所有交點(diǎn)橫坐標(biāo)之和為SKIPIF1<0.故選：C二、多選題（本大題共4小題，每小題5分，共20分.全對(duì)得5分，少選得3分，多選、錯(cuò)選不得分）9.下列說法正確的是（）A.函數(shù)SKIPIF1<0的定義域是SKIPIF1<0B.函數(shù)SKIPIF1<0在其定義域上單調(diào)遞減C.函數(shù)SKIPIF1<0的值域是SKIPIF1<0D.函數(shù)SKIPIF1<0的圖象過定點(diǎn)SKIPIF1<0【答案】CD【解析】【分析】選項(xiàng)A.求出函數(shù)的定義域可判斷；選項(xiàng)B.函數(shù)SKIPIF1<0在其定義域上不是單調(diào)函數(shù)可判斷；選項(xiàng)C.由指數(shù)函數(shù)的性質(zhì)可判斷；選項(xiàng)D.由SKIPIF1<0時(shí)，SKIPIF1<0可判斷.【詳解】選項(xiàng)A.函數(shù)SKIPIF1<0的定義域是SKIPIF1<0,故不正確.選項(xiàng)B.函數(shù)SKIPIF1<0在其定義域上不是單調(diào)函數(shù)，故不正確.選項(xiàng)C.函數(shù)SKIPIF1<0的值域是SKIPIF1<0，故正確.選項(xiàng)D.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0過SKIPIF1<0，故正確.故選：CD10.下列說法正確的是（）A.函數(shù)SKIPIF1<0的最小值為2B.若正實(shí)數(shù)a，b滿足SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0C.關(guān)于x的不等式SKIPIF1<0的解集是SKIPIF1<0，則SKIPIF1<0D.函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的定義域?yàn)镽，則實(shí)數(shù)m的取值范圍是SKIPIF1<0【答案】BC【解析】【分析】A由三角函數(shù)的性質(zhì)，結(jié)合特殊情況判斷；B應(yīng)用基本不等式“1”的代換求最值；C由一元二次不等式的解集求參數(shù)a、b，即可判斷；D由對(duì)數(shù)函數(shù)、二次函數(shù)的性質(zhì)有SKIPIF1<0即可判斷.【詳解】A：當(dāng)SKIPIF1<0時(shí)，顯然SKIPIF1<0，故錯(cuò)誤；B：由SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，正確；C：根據(jù)不等式的解集可知1,2是方程SKIPIF1<0的根，所以SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，正確；D：由題意，SKIPIF1<0在R上恒成立，則SKIPIF1<0，解得SKIPIF1<0，錯(cuò)誤.故選：BC11.已知函數(shù)SKIPIF1<0，則（）A.SKIPIF1<0為函數(shù)SKIPIF1<0的一個(gè)周期B.對(duì)于任意的SKIPIF1<0，函數(shù)SKIPIF1<0都滿足SKIPIF1<0C.函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減D.SKIPIF1<0的值域?yàn)镾KIPIF1<0【答案】ABD【解析】【分析】計(jì)算SKIPIF1<0，可判定選項(xiàng)A；根據(jù)三角函數(shù)的誘導(dǎo)公式進(jìn)行化簡運(yùn)算，可判定選項(xiàng)B；化簡函數(shù)SKIPIF1<0，結(jié)合三角函數(shù)的性質(zhì)，可判定選項(xiàng)C；根據(jù)SKIPIF1<0，且SKIPIF1<0，結(jié)合選項(xiàng)A、B及函數(shù)的單調(diào)性，可判定選項(xiàng)D.【詳解】對(duì)于A，由SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0的周期，故A正確；對(duì)于B，由SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B正確；對(duì)于C，由SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，根據(jù)正弦函數(shù)的性質(zhì)，可得函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故C錯(cuò)誤；對(duì)于D，函數(shù)SKIPIF1<0，由SKIPIF1<0，且SKIPIF1<0，所以函數(shù)SKIPIF1<0的最小值為SKIPIF1<0；由SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最大值為SKIPIF1<0；由B選項(xiàng)知函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，又SKIPIF1<0為SKIPIF1<0的周期，SKIPIF1<0的值域?yàn)镾KIPIF1<0，故D正確.故選：ABD.12.已知函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0為自然對(duì)數(shù)的底數(shù)），則（）A.函數(shù)SKIPIF1<0至多有SKIPIF1<0個(gè)零點(diǎn)B.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，總有SKIPIF1<0成立C.函數(shù)SKIPIF1<0至少有SKIPIF1<0個(gè)零點(diǎn)D.當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有SKIPIF1<0個(gè)不同實(shí)數(shù)根【答案】ABCD【解析】【分析】分別解方程SKIPIF1<0、SKIPIF1<0，取SKIPIF1<0，可判斷A選項(xiàng)；利用分段函數(shù)的單調(diào)性可判斷B選項(xiàng)；對(duì)實(shí)數(shù)SKIPIF1<0的取值進(jìn)行分類討論，確定函數(shù)SKIPIF1<0在SKIPIF1<0不同的取值下，SKIPIF1<0的零點(diǎn)個(gè)數(shù)，可判斷C選項(xiàng)；當(dāng)SKIPIF1<0時(shí)，解方程SKIPIF1<0，可判斷D選項(xiàng).【詳解】對(duì)于A選項(xiàng)，令SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，可得SKIPIF1<0.故當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，所以，函數(shù)SKIPIF1<0至多有SKIPIF1<0個(gè)零點(diǎn)，A對(duì)；對(duì)于B選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以，故當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，B對(duì)；對(duì)于C選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上無零點(diǎn)，在SKIPIF1<0上有唯一零點(diǎn)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)SKIPIF1<0，在SKIPIF1<0上無零點(diǎn)，綜上所述，函數(shù)SKIPIF1<0至少有一個(gè)零點(diǎn)，C對(duì)；對(duì)于D選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.令SKIPIF1<0，則方程SKIPIF1<0為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，該方程無解，由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0.綜上所述，方程SKIPIF1<0的解集為SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有SKIPIF1<0個(gè)不同實(shí)數(shù)根，D對(duì).故選：ABCD.三、填空題（本大題共4小題，每小題5分，共20分）13.命題“SKIPIF1<0，SKIPIF1<0”的否定是______.【答案】SKIPIF1<0，SKIPIF1<0【解析】【分析】由存在量詞命題的否定可得出結(jié)論.【詳解】命題“SKIPIF1<0，SKIPIF1<0”為存在量詞命題，該命題的否定為“SKIPIF1<0，SKIPIF1<0”.故答案為：SKIPIF1<0，SKIPIF1<0.14.已知點(diǎn)SKIPIF1<0在角SKIPIF1<0的終邊上，則SKIPIF1<0______.【答案】SKIPIF1<0【解析】【分析】首先求SKIPIF1<0，再由齊次式化簡求值.【詳解】由題意可知SKIPIF1<0，∴SKIPIF1<0.故答案為：SKIPIF1<0.15.已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0______.【答案】SKIPIF1<0【解析】【分析】推導(dǎo)出函數(shù)SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)，再結(jié)合函數(shù)SKIPIF1<0的周期性和奇偶性可求得SKIPIF1<0的值.【詳解】因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且滿足SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，所以，函數(shù)SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0.16.已知函數(shù)SKIPIF1<0的部分圖象如圖所示，且SKIPIF1<0在SKIPIF1<0上恰有一個(gè)最大值和一個(gè)最小值，則SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【解析】【分析】由SKIPIF1<0，推出SKIPIF1<0，從而知SKIPIF1<0，再由SKIPIF1<0，求得SKIPIF1<0的取值范圍，并結(jié)合正弦函數(shù)的圖象與性質(zhì)，即可得解．【詳解】由圖知SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0，知SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上恰有一個(gè)最大值和一個(gè)最小值，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0．四、解答題（本大題共6小題，第17題10分，第18-22題每小題12分，共70分.解答應(yīng)寫出必要的文字說明、證明過程或演算步驟.）17.設(shè)集合SKIPIF1<0，集合SKIPIF1<0.（1）若SKIPIF1<0，求SKIPIF1<0；（2）設(shè)條件SKIPIF1<0，條件SKIPIF1<0，若SKIPIF1<0是SKIPIF1<0成立的必要條件，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】（1）SKIPIF1<0（2）SKIPIF1<0【解析】【分析】（1）解不等式化簡SKIPIF1<0，根據(jù)并集的概念可求出結(jié)果；（2）將SKIPIF1<0是SKIPIF1<0成立的必要條件，轉(zhuǎn)化為SKIPIF1<0，根據(jù)子集關(guān)系列式可求出結(jié)果.【小問1詳解】由SKIPIF1<0，解得SKIPIF1<0，可得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0解得SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.【小問2詳解】由SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0.∵SKIPIF1<0是SKIPIF1<0成立的必要條件，∴SKIPIF1<0，由于SKIPIF1<0，所以有：SKIPIF1<0，解得：SKIPIF1<0.∴實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.18.已知函數(shù)SKIPIF1<0.（1）求SKIPIF1<0的單調(diào)遞增區(qū)間；（2）設(shè)SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值.【答案】（1）SKIPIF1<0（2）SKIPIF1<0【解析】【分析】（1）利用三角恒等變換化簡SKIPIF1<0，根據(jù)正弦函數(shù)的性質(zhì)求得答案；（2）根據(jù)SKIPIF1<0求得SKIPIF1<0，結(jié)合SKIPIF1<0，求得SKIPIF1<0，再利用誘導(dǎo)公式求得答案.【小問1詳解】SKIPIF1<0SKIPIF1<0∵SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0的遞增區(qū)間為SKIPIF1<0.【小問2詳解】由SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，于是SKIPIF1<0.19.已知函數(shù)SKIPIF1<0，現(xiàn)有下列3個(gè)條件：①相鄰兩個(gè)對(duì)稱中心的距離是SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0.（1）請(qǐng)選擇其中兩個(gè)條件，求出滿足這兩個(gè)條件函數(shù)SKIPIF1<0的解析式；（2）將（1）中函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長度，再把橫坐標(biāo)縮小為原來的SKIPIF1<0（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象，請(qǐng)寫出函數(shù)SKIPIF1<0的解析式，并求其在SKIPIF1<0上的值域.【答案】（1）選擇見解析，SKIPIF1<0（2）SKIPIF1<0，SKIPIF1<0【解析】【分析】（1）根據(jù)題意，結(jié)合周期公式，選擇相應(yīng)的條件，代入函數(shù)解析式即可求解；（2）根據(jù)圖象變換規(guī)則即可得到函數(shù)SKIPIF1<0的解析式，結(jié)合正弦函數(shù)的性質(zhì)即可求解.【小問1詳解】選①②，因?yàn)橄噜弮蓚€(gè)對(duì)稱中心的距離為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.選①③，因?yàn)橄噜弮蓚€(gè)對(duì)稱中心的距離為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.選②③，由題意SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.【小問2詳解】將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長度，可得SKIPIF1<0的圖象，再將橫坐標(biāo)縮小為原來的SKIPIF1<0（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象.∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，即函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0.20.某醫(yī)藥公司研發(fā)的一種新藥，如果成年人按規(guī)定的劑量服用，由監(jiān)測數(shù)據(jù)可知，服用后6小時(shí)內(nèi)每毫升血液中含藥量SKIPIF1<0（單位：微克）與時(shí)間SKIPIF1<0（單位：小時(shí)）之間的關(guān)系滿足如圖所示的曲線，當(dāng)SKIPIF1<0時(shí)，曲線是二次函數(shù)圖象的一部分，當(dāng)SKIPIF1<0時(shí)，曲線是函數(shù)SKIPIF1<0圖象的一部分，根據(jù)進(jìn)一步測定，每毫升血液中含藥量不少于2微克時(shí)，治療有效.（1）試求服藥后6小時(shí)內(nèi)每毫升血液中含藥量SKIPIF1<0與時(shí)間SKIPIF1<0之間的函數(shù)關(guān)系式；（2）問服藥多久后開始有治療效果？治療效果能持續(xù)多少小時(shí)？（精確到0.1）（參考數(shù)據(jù)SKIPIF1<0）【答案】（1）SKIPIF1<0（2）0.3小時(shí)后，5.2小時(shí)【解析】【分析】（1）當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，再將SKIPIF1<0代入即可求出SKIPIF1<0的值，當(dāng)SKIPIF1<0時(shí)，將點(diǎn)SKIPIF1<0的坐標(biāo)代入函數(shù)表達(dá)式SKIPIF1<0即可求出SKIPIF1<0的值，則可寫出答案；（2）分段求出SKIPIF1<0時(shí)，對(duì)應(yīng)的SKIPIF1<0的取值范圍，即可寫出答案.【小問1詳解】當(dāng)SKIPIF1<0時(shí)，由圖象可設(shè)SKIPIF1<0，將點(diǎn)SKIPIF1<0的坐標(biāo)代入函數(shù)表達(dá)式，解得SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，將點(diǎn)SKIPIF1<0的坐標(biāo)代入函數(shù)SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0.【小問2詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，故服藥0.3小時(shí)之后開始有治療效果，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，綜上，SKIPIF1<0，所以服藥后的治療效果能持續(xù)5.2小時(shí).21.已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0是奇函數(shù).（1）求實(shí)數(shù)SKIPIF1<0，SKIPIF1<0的值；（2）判斷SKIPIF1<0在SKIPIF1<0上的單調(diào)性并用定義證明；（3）若對(duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】（1）SKIPIF1<0，SKIPIF1<0（2）單調(diào)遞減，證明見解析（3）SKIPIF1<0【解析】【分析】（1）由SKIPIF1<0，SKIPIF1<0求出SKIPIF1<0，再驗(yàn)證SKIPIF1<0是奇函數(shù)；（2）函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，設(shè)SKIPIF1<0，按照作差SKIPIF1<0、變形、判斷符號(hào)、下結(jié)論這幾個(gè)步驟證明即可；（3）利用奇偶性轉(zhuǎn)化為SKIPIF1<0，根據(jù)單調(diào)性化為SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，再根據(jù)正弦函數(shù)值域以及二次函數(shù)知識(shí)可求出結(jié)果.【小問1詳解】由題意，定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0是奇函數(shù)，得SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，經(jīng)檢驗(yàn)知，SKIPIF1<0是奇函數(shù).故SKIPIF1<0.【小問2詳解】由（1）知，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù).證明如下：設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.∴函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù).【小問3詳解】由SKIPIF1<0，且SKIPIF1<0是奇函數(shù)，得SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0上是減函數(shù)，所以SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，由SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0，故得實(shí)數(shù)SKIPIF1<0的取值范圍SKIPIF1<0.22.已知函數(shù)SKIPIF1<0