廣西南寧市2022-2023學(xué)年高一上學(xué)期期末數(shù)學(xué)模擬試題（含答案詳解）

2022年廣西南寧高一期末數(shù)學(xué)模擬卷一.單選題（每小題5分，8小題共40分）.1.命題“SKIPIF1<0，SKIPIF1<0”的否定是（）A.SKIPIF1<0，SKIPIF1<0 B.SKIPIF1<0，SKIPIF1<0C.SKIPIF1<0，SKIPIF1<0 D.SKIPIF1<0，SKIPIF1<0【答案】D【解析】【分析】直接利用全稱命題的否定是特稱命題，將任意改成存在，并將結(jié)論否定即可.【詳解】根據(jù)全稱命題的否定的定義可知，命題“SKIPIF1<0，SKIPIF1<0”的否定是SKIPIF1<0，SKIPIF1<0.故選：D.2.已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】化簡(jiǎn)集合A,B，根據(jù)補(bǔ)集、交集運(yùn)算即可求解.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.故選：A3.已知銳角SKIPIF1<0的終邊上一點(diǎn)SKIPIF1<0，則銳角SKIPIF1<0＝A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【詳解】∵銳角SKIPIF1<0的終邊上一點(diǎn)SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0＝70°故選C4.已知正數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為（）A.SKIPIF1<0 B.2 C.SKIPIF1<0 D.6【答案】B【解析】【分析】化簡(jiǎn)SKIPIF1<0，再利用基本不等式求解.【詳解】由題得SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等.所以SKIPIF1<0的最小值為2.故選：B【點(diǎn)睛】方法點(diǎn)睛：利用基本不等式求最值時(shí)，常用到常量代換，即把所求代數(shù)式中的某一常量換成已知中的代數(shù)式，再利用基本不等式求解.5.我國(guó)古代數(shù)學(xué)家趙爽的弦圖是由四個(gè)全等的直角三角形與-一個(gè)小正方形拼成的一個(gè)大正方形(如圖).如果小正方形的邊長(zhǎng)為SKIPIF1<0，大正方形的邊長(zhǎng)為SKIPIF1<0，直角三角形中較小的銳角為SKIPIF1<0，則SKIPIF1<0（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】設(shè)出直角三角形中較短的直角邊，利用勾股定理求出x的值，從而求出sinθ，cosθ的值，再利用兩角和與差的三角函數(shù)公式即可算出結(jié)果．【詳解】直角三角形中較短的直角邊為x，則：x2+（x+2）2＝102，解得：x＝6，∴sinθSKIPIF1<0，cosθSKIPIF1<0，∴sin（SKIPIF1<0）﹣cos（SKIPIF1<0）＝﹣cosθ﹣（cosθcosSKIPIF1<0）SKIPIF1<0sinθ﹣（SKIPIF1<0）cosθSKIPIF1<0，故選：D．【點(diǎn)睛】本題考查的知識(shí)點(diǎn)是兩角和與差的余弦公式，誘導(dǎo)公式，難度不大，屬于基礎(chǔ)題．6.若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0大小關(guān)系正確的是（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】利用指，對(duì)，冪函數(shù)的性質(zhì)，以及和特殊值1比較大小，判斷選項(xiàng).【詳解】SKIPIF1<0；SKIPIF1<0，SKIPIF1<0；SKIPIF1<0．故選：SKIPIF1<0．7.定義域在R上的函數(shù)SKIPIF1<0是奇函數(shù)且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的值為（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)函數(shù)的奇偶性和周期性進(jìn)行求解即可.【詳解】因?yàn)镾KIPIF1<0，所以函數(shù)的周期為SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0是奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，故選：A二.多選題（每小題5分，部分對(duì)2分，選錯(cuò)0分，4小題，共20分）.8.若函數(shù)SKIPIF1<0且滿足對(duì)任意的實(shí)數(shù)SKIPIF1<0都有SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)解析式及滿足的不等式SKIPIF1<0，可知函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù)，由分段函數(shù)單調(diào)性的性質(zhì)，結(jié)合指數(shù)函數(shù)與一次函數(shù)單調(diào)性的性質(zhì)，即可得關(guān)于SKIPIF1<0的不等式組，解不等式組即可求得SKIPIF1<0的取值范圍.【詳解】函數(shù)SKIPIF1<0滿足對(duì)任意的實(shí)數(shù)SKIPIF1<0都有SKIPIF1<0，所以函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù)，則由指數(shù)函數(shù)與一次函數(shù)單調(diào)性可知應(yīng)滿足SKIPIF1<0，解得SKIPIF1<0，所以數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0，故選：A【點(diǎn)睛】本題考查根據(jù)分段函數(shù)單調(diào)性求參數(shù)的取值范圍，在滿足各段函數(shù)單調(diào)性的情況下，還需滿足整個(gè)定義域內(nèi)的單調(diào)性，屬于中檔題.9.（多選題）下列計(jì)算正確的是（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.已知SKIPIF1<0，則SKIPIF1<0【答案】BC【解析】【分析】根據(jù)根式運(yùn)算和指數(shù)冪的運(yùn)算法則求解判斷.【詳解】A.SKIPIF1<0，故錯(cuò)誤；B.SKIPIF1<0，故正確；C.SKIPIF1<0，故正確；D.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，故錯(cuò)誤；故選：BC10.（多選）在同一平面直角坐標(biāo)系中，函數(shù)SKIPIF1<0與SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）的圖象可能是（）A. B.C. D.【答案】AC【解析】【分析】SKIPIF1<0為指數(shù)函數(shù)，分SKIPIF1<0與SKIPIF1<0兩種情況討論，從而判斷出圖象的可能結(jié)果.【詳解】若SKIPIF1<0，則函數(shù)SKIPIF1<0是R上的增函數(shù)，函數(shù)SKIPIF1<0的圖象的對(duì)稱軸方程為SKIPIF1<0且SKIPIF1<0，故A符合，B不符合；若SKIPIF1<0，則函數(shù)SKIPIF1<0是R上減函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象與y軸的負(fù)半軸相交，故C符合，D不符合.故選：AC.11.已知函數(shù)SKIPIF1<0，則（）A.函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0B.直線SKIPIF1<0是SKIPIF1<0圖象的一條對(duì)稱軸C.SKIPIF1<0的值域?yàn)镾KIPIF1<0D.若SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，則SKIPIF1<0的取值范圍是SKIPIF1<0【答案】BC【解析】【分析】根據(jù)函數(shù)SKIPIF1<0的周期是函數(shù)SKIPIF1<0周期的一半，可判斷A選項(xiàng)；將SKIPIF1<0代入函數(shù)解析式求值，判斷是否為函數(shù)的對(duì)稱軸；對(duì)于C：將函數(shù)SKIPIF1<0化簡(jiǎn)得到SKIPIF1<0，接著利用換元法求得值域即可；對(duì)于D選項(xiàng)：SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，可得SKIPIF1<0或SKIPIF1<0，最后求得SKIPIF1<0的取值范圍.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的最小正周期為SKIPIF1<0，而函數(shù)SKIPIF1<0周期為SKIPIF1<0，故A錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以直線SKIPIF1<0是SKIPIF1<0圖象的一條對(duì)稱軸，故B正確；SKIPIF1<0化簡(jiǎn)整理得：SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，二次函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，故C正確；SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0，故D錯(cuò)誤.故選：BC.【點(diǎn)睛】(1)應(yīng)用公式時(shí)注意方程思想的應(yīng)用，對(duì)于sinα＋cosα，sinα－cosα，sinαcosα這三個(gè)式子，利用(sinα±cosα)2＝1±2sinαcosα可以知一求二．(2)關(guān)于sinα，cosα的齊次式，往往化為關(guān)于tanα的式子．12.設(shè)函數(shù)SKIPIF1<0，若實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0則下列結(jié)論恒成立的是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】ABC【解析】【分析】由函數(shù)零點(diǎn)與方程的根的關(guān)系，作出函數(shù)的圖象，然后利用作差法比較大小，即可求解.【詳解】解：由題意，實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，結(jié)合圖象，可得SKIPIF1<0，即SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0和SKIPIF1<0恒成立，即A、B恒成立；又由SKIPIF1<0，所以SKIPIF1<0，所以C恒成立；又由SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的符號(hào)不能確定，所以D不恒成立，故選：ABC.【點(diǎn)睛】本題主要考查了函數(shù)與方程的綜合應(yīng)用，以及對(duì)數(shù)函數(shù)圖象的應(yīng)用，其中解答中正確作出函數(shù)的圖象，得到SKIPIF1<0的關(guān)系式是解答的關(guān)鍵，著重考查了分析問題和解答問題的能力，屬于中檔題.三?填空題（每小題5分，4小題共20分）.13.函數(shù)SKIPIF1<0的定義域?yàn)開___.【答案】SKIPIF1<0【解析】【分析】根據(jù)解析式有意義可得出關(guān)于實(shí)數(shù)SKIPIF1<0的不等式組，進(jìn)而可求得函數(shù)SKIPIF1<0的定義域.【詳解】對(duì)于函數(shù)SKIPIF1<0，有SKIPIF1<0，解得SKIPIF1<0.因此，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.14.若函數(shù)SKIPIF1<0的一個(gè)周期是SKIPIF1<0，則SKIPIF1<0的取值可以是___________.（寫出一個(gè)即可）.【答案】4（答案不唯一）【解析】【分析】先利用三角函數(shù)恒等變換公式對(duì)函數(shù)化簡(jiǎn)，然后利用周期公式求解即可【詳解】SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0的最小正周期SKIPIF1<0，從而SKIPIF1<0，解得SKIPIF1<0，且SKIPIF1<0故答案為：4（答案不唯一）15.已知函數(shù)SKIPIF1<0，那么SKIPIF1<0________；若存在實(shí)數(shù)a，使得SKIPIF1<0，則a的個(gè)數(shù)是_______________．【答案】①.1②.4【解析】【分析】（1）直接代入求值即可；（2）運(yùn)用換元法，結(jié)合函數(shù)的圖象，分類討論求出a的個(gè)數(shù).【詳解】（1）SKIPIF1<0（2）令SKIPIF1<0，即滿足SKIPIF1<0，①t=1，即a=±1時(shí)，經(jīng)檢驗(yàn)，均滿足題意；②t<1，即?1<a<1或a>1時(shí)，SKIPIF1<0，由SKIPIF1<0，解得t=0或1（舍去）；再由SKIPIF1<0解得a=0或2；③t>1，即a<?1時(shí)，SKIPIF1<0，由t=2?t，解得t=1（舍去）；綜上所述：共有4個(gè)a．【點(diǎn)睛】本題考查了求函數(shù)值，考查了方程有解求實(shí)數(shù)個(gè)數(shù)問題，考查了分類討論法、換元法。16.如下圖，SKIPIF1<0是邊長(zhǎng)為SKIPIF1<0的正三角形，記SKIPIF1<0位于直線SKIPIF1<0左側(cè)的圖形的面積為SKIPIF1<0，現(xiàn)給出函數(shù)SKIPIF1<0的四個(gè)性質(zhì)，其中說法正確的是__________．①SKIPIF1<0②SKIPIF1<0在SKIPIF1<0上單調(diào)遞增③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值④對(duì)于任意SKIPIF1<0，都有SKIPIF1<0【答案】②④【解析】分析】先分析出SKIPIF1<0,再根據(jù)分段函數(shù)性質(zhì)依次判斷即可【詳解】由題可知,SKIPIF1<0所在直線為SKIPIF1<0,SKIPIF1<0所在直線為SKIPIF1<0則當(dāng)SKIPIF1<0時(shí),SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0；則SKIPIF1<0,①當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故①錯(cuò)誤；②易知,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞增,且SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,故②正確；③因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增,則無最大值,故③錯(cuò)誤；④由題,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,故④正確；故答案為②④【點(diǎn)睛】本題考查分段函數(shù)的應(yīng)用,考查二次函數(shù)單調(diào)性與最值問題,考查求函數(shù)值,考查運(yùn)算能力四.解答題（第17題10分，18-22每小題12分，共70分）.17.已知集合SKIPIF1<0，SKIPIF1<0.（1）當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0；（2）若SKIPIF1<0是SKIPIF1<0的充分不必要條件，求實(shí)數(shù)m的取值范圍.【答案】（1）SKIPIF1<0（2）SKIPIF1<0或SKIPIF1<0【解析】【分析】（1）解一元二次不等式，再根據(jù)并集運(yùn)算求解；（2）根據(jù)充分不必要關(guān)系確定SKIPIF1<0真包含于SKIPIF1<0即可求解.【小問1詳解】由SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，所以SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0.小問2詳解】因?yàn)镾KIPIF1<0是SKIPIF1<0的充分不必要條件，所以SKIPIF1<0真包含于SKIPIF1<0，由（1）知SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0.18.如圖，以O(shè)x為始邊作角SKIPIF1<0與SKIPIF1<0，它們的終邊分別與單位圓相交于P，Q兩點(diǎn)，已知點(diǎn)P的坐標(biāo)為SKIPIF1<0．（1）求SKIPIF1<0的值；（2）若SKIPIF1<0，求SKIPIF1<0的值．【答案】（1）SKIPIF1<0（2）SKIPIF1<0【解析】【分析】(1)由三角函數(shù)的定義首先求得SKIPIF1<0的值，然后結(jié)合二倍角公式和同角三角函數(shù)基本關(guān)系化簡(jiǎn)求解三角函數(shù)式的值即可；(2)由題意首先求得SKIPIF1<0的關(guān)系，然后結(jié)合誘導(dǎo)公式和兩角和差正余弦公式即可求得三角函數(shù)式的值.【詳解】（1）由三角函數(shù)定義得SKIPIF1<0，SKIPIF1<0，∴原式SKIPIF1<0．（2）∵SKIPIF1<0，且SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．∴SKIPIF1<0．【點(diǎn)睛】本題主要考查三角函數(shù)的定義，二倍角公式及其應(yīng)用，兩角和差正余弦公式的應(yīng)用等知識(shí)，意在考查學(xué)生的轉(zhuǎn)化能力和計(jì)算求解能力.19.在下列三個(gè)條件中任選一個(gè)，補(bǔ)充在下面的問題中，并作答．①SKIPIF1<0的最小正周期為SKIPIF1<0，且SKIPIF1<0是偶函數(shù)：②SKIPIF1<0圖象上相鄰兩個(gè)最高點(diǎn)之間的距離為SKIPIF1<0，且SKIPIF1<0；③直線SKIPIF1<0與直線SKIPIF1<0是SKIPIF1<0圖象上相鄰的兩條對(duì)稱軸，且SKIPIF1<0．問題：已知函數(shù)SKIPIF1<0，若．（1）求SKIPIF1<0，SKIPIF1<0的值；（請(qǐng)先在答題卡上寫出所選序號(hào)再做答）（2）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后，再將得到的函數(shù)圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來的4倍，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象，求SKIPIF1<0在SKIPIF1<0上的最小值和最大值．【答案】（1）SKIPIF1<0，SKIPIF1<0（2）最小值為1，最大值為2【解析】【分析】(1)根據(jù)①②③所給的條件，以及正余弦函數(shù)的對(duì)稱性和周期性之間的關(guān)系即可求解；(2)根據(jù)函數(shù)的伸縮平移變換后的特點(diǎn)寫出SKIPIF1<0的解析式即可.小問1詳解】選條件①：∵SKIPIF1<0的最小正周期為SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；又SKIPIF1<0是偶函數(shù)，∴SKIPIF1<0對(duì)SKIPIF1<0恒成立，得SKIPIF1<0對(duì)SKIPIF1<0恒成立，∴SKIPIF1<0，∴SKIPIF1<0（SKIPIF1<0），又SKIPIF1<0，∴SKIPIF1<0；選條件②：∵函數(shù)SKIPIF1<0圖象上相鄰兩個(gè)最高點(diǎn)之間的距離為SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0；又SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0（SKIPIF1<0），又SKIPIF1<0，∴SKIPIF1<0；選條件③：∵直線SKIPIF1<0與直線SKIPIF1<0是SKIPIF1<0圖象上相鄰的兩條對(duì)稱軸，∴SKIPIF1<0，即SKIPIF1<0．∴SKIPIF1<0；又SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0（SKIPIF1<0），又SKIPIF1<0，∴SKIPIF1<0；【小問2詳解】由（1）無論選擇①②③均有SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后，得到SKIPIF1<0的圖象，將SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來的4倍，縱坐標(biāo)不變，得到SKIPIF1<0的圖象，∵SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0上單調(diào)遞減．又∵SKIPIF1<0SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0的最小值為1，最大值為2；綜上：SKIPIF1<0，最小值=1，最大值=2.20.某汽車公司為測(cè)量某型號(hào)汽車定速巡航狀態(tài)下的油耗情況，選擇一段長(zhǎng)度為SKIPIF1<0的平坦高速路段進(jìn)行測(cè)試，經(jīng)多次測(cè)試得到一輛汽車每小時(shí)耗油量SKIPIF1<0單位：SKIPIF1<0與速度SKIPIF1<0單位：SKIPIF1<0的一些數(shù)據(jù)如下表所示.SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0為了描述汽車每小時(shí)耗油量SKIPIF1<0與速度SKIPIF1<0的關(guān)系，現(xiàn)有以下三種函數(shù)模型供選擇：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0.（1）請(qǐng)選出你認(rèn)為最符合實(shí)際的函數(shù)模型，并求出相應(yīng)的函數(shù)解析式；（2）這輛車在該測(cè)試路段上以什么速度行駛才能使總耗油量最少SKIPIF1<0【答案】（1）SKIPIF1<0，SKIPIF1<0（2）SKIPIF1<0【解析】【分析】SKIPIF1<0根據(jù)題意可知，代入數(shù)據(jù)列得關(guān)于SKIPIF1<0的方程組，解方程組即可，故可得解析式.SKIPIF1<0設(shè)這輛汽車在該測(cè)試路段的總耗油量為SKIPIF1<0單位：SKIPIF1<0，行駛時(shí)間為SKIPIF1<0單位：SKIPIF1<0，由題意得SKIPIF1<0，根據(jù)二次函數(shù)的性質(zhì)求出最值.【小問1詳解】由題意可知，符合本題的函數(shù)模型必須滿足定義域?yàn)镾KIPIF1<0，且在SKIPIF1<0上單調(diào)遞增.函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以不符合題意SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以不符合題意；函數(shù)SKIPIF1<0，且SKIPIF1<0中的SKIPIF1<0，即定義域不可能為SKIPIF1<0，也不符合題意所以選擇函數(shù)模型SKIPIF1<0.由已知數(shù)據(jù)得SKIPIF1<0解得SKIPIF1<0所以SKIPIF1<0.【小問2詳解】設(shè)這輛車在該測(cè)試路段的總耗油量為SKIPIF1<0，行駛時(shí)間為SKIPIF1<0.由題意得:SKIPIF1<0，因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值SKIPIF1<0.所以這輛車在該測(cè)試路段上以SKIPIF1<0的速度行駛才能使總耗油量最少，最少為SKIPIF1<0.21.定義在非零實(shí)數(shù)集上的函數(shù)SKIPIF1<0對(duì)任意非零實(shí)數(shù)SKIPIF1<0，SKIPIF1<0都滿足SKIPIF1<0.（1）求SKIPIF1<0的值；（2）求SKIPIF1<0的解析式；（3）設(shè)函數(shù)SKIPIF1<0，求SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值SKIPIF1<0.【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【解析】【分析】（1）分別令SKIPIF1<0，SKIPIF1<0和SKIPIF1<0，SKIPIF1<0，可得出關(guān)于SKIPIF1<0和SKIPIF1<0的方程組，即可解出SKIPIF1<0的值；（2）令SKIPIF1<0，則SKIPIF1<0，再用SKIPIF1<0替換SKIPIF1<0可得出SKIPIF1<0，利用加減消元法可解出SKIPIF1<0，即可得出函數(shù)SKIPIF1<0的解析式；（3）由題意得出SKIPIF1<0，然后分SKIPIF1<0和SKIPIF1<0，分析二次函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性，即可得出函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值SKIPIF1<0的表達(dá)式.詳解】（1）令SKIPIF