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高一上學(xué)期期中數(shù)學(xué)試題一、選擇題:本題共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.已知集合SKIPIF1<0,則下列說法正確的是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】求出集合A,然后根據(jù)元素與集合,集合與集合的關(guān)系即得..【詳解】SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以ABD錯(cuò)誤,C正確.故選:C.2.SKIPIF1<0的一個(gè)充分不必要條件是()A.SKIPIF1<0或SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)分式的性質(zhì),結(jié)合充分不必要條件的定義進(jìn)行求解即可.【詳解】由SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,顯然SKIPIF1<0能推出SKIPIF1<0,但SKIPIF1<0不一定能推出SKIPIF1<0,選項(xiàng)CD都推不出SKIPIF1<0,選項(xiàng)A能推出SKIPIF1<0,SKIPIF1<0也能推出SKIPIF1<0或SKIPIF1<0,故選:B3.已知函數(shù)SKIPIF1<0的對(duì)應(yīng)關(guān)系如表所示,函數(shù)SKIPIF1<0的圖象是如圖所示,則SKIPIF1<0的值為()SKIPIF1<0123SKIPIF1<043-1A.-1 B.0 C.3 D.4【答案】A【解析】【分析】根據(jù)函數(shù)圖象和表格直接求解即可.【詳解】由圖可知SKIPIF1<0,所以SKIPIF1<0,又由表可知SKIPIF1<0,所以SKIPIF1<0.故選:A4.已知函數(shù)SKIPIF1<0是冪函數(shù),且在SKIPIF1<0上單調(diào)遞增,則SKIPIF1<0()A.3 B.-1 C.1或-3 D.-1或3【答案】A【解析】【分析】根據(jù)冪函數(shù)的概念及性質(zhì)即得.【詳解】因?yàn)镾KIPIF1<0是冪函數(shù),所以SKIPIF1<0,解得SKIPIF1<0或3;又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,不符合題意,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,符合題意,故SKIPIF1<0.故選:A.5.已知函數(shù)SKIPIF1<0是SKIPIF1<0上的減函數(shù),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】C【解析】【分析】由題可得函數(shù)在SKIPIF1<0及SKIPIF1<0時(shí),單調(diào)遞減,且SKIPIF1<0,進(jìn)而即得.【詳解】由題意可知:SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,即SKIPIF1<0;SKIPIF1<0在SKIPIF1<0上也單調(diào)遞減,即SKIPIF1<0;又SKIPIF1<0是SKIPIF1<0上的減函數(shù),則SKIPIF1<0,∴SKIPIF1<0,解得SKIPIF1<0.故選:C.6.已知偶函數(shù)f(x)在區(qū)間SKIPIF1<0單調(diào)遞增,則滿足SKIPIF1<0的x取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由偶函數(shù)性質(zhì)得函數(shù)在SKIPIF1<0上的單調(diào)性,然后由單調(diào)性解不等式.【詳解】因?yàn)榕己瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0區(qū)間SKIPIF1<0上單調(diào)遞減,故SKIPIF1<0越靠近SKIPIF1<0軸,函數(shù)值越小,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,解得:SKIPIF1<0.故選:A.7.因工作需求,張先生的汽車一周需兩次加同一種汽油.現(xiàn)張先生本周按照以下兩種方案加油(兩次加油時(shí)油價(jià)不一樣),甲方案:每次購(gòu)買汽油的量一定;乙方案:每次加油的錢數(shù)一定.問哪種加油的方案更經(jīng)濟(jì)?()A.甲方案 B.乙方案 C.一樣 D.無法確定【答案】B【解析】【分析】設(shè)兩次加油的油價(jià)分別為SKIPIF1<0,SKIPIF1<0(SKIPIF1<0,且SKIPIF1<0),分別計(jì)算兩種方案的平均油價(jià),然后比較即得.【詳解】設(shè)兩次加油的油價(jià)分別為SKIPIF1<0,SKIPIF1<0(SKIPIF1<0,且SKIPIF1<0),甲方案每次加油的量為SKIPIF1<0;乙方案每次加油的錢數(shù)為SKIPIF1<0,則甲方案的平均油價(jià)為:SKIPIF1<0,乙方案的平均油價(jià)為:SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即乙方案更經(jīng)濟(jì).故選:B.8.已知函數(shù)SKIPIF1<0在其定義域內(nèi)為偶函數(shù),且SKIPIF1<0,則SKIPIF1<0()A.0 B.2021 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】先由偶函數(shù)的性質(zhì)求得SKIPIF1<0,再由SKIPIF1<0求得SKIPIF1<0,由此得到SKIPIF1<0的解析式,觀察所求式子容易考慮SKIPIF1<0的值,求之可解得結(jié)果.【詳解】因?yàn)镾KIPIF1<0是偶函數(shù),所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,又因?yàn)镾KIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0.故選:D.二、選擇題:本題共4小題,每小題5分,共20分.在每小題給出的選項(xiàng)中,有多項(xiàng)符合題目要求,全部選對(duì)的得5分,部分選對(duì)的得2分,有選錯(cuò)的得0分.9.已知SKIPIF1<0,則下列說法正確的是()A.若SKIPIF1<0,則SKIPIF1<0 B.若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0C.若SKIPIF1<0,則SKIPIF1<0 D.若SKIPIF1<0,則SKIPIF1<0【答案】BC【解析】【分析】利用不等式的性質(zhì)可判斷AC,根據(jù)作差法可判斷BD.【詳解】對(duì)于A選項(xiàng),若SKIPIF1<0,則SKIPIF1<0,故A錯(cuò)誤;對(duì)于B選項(xiàng),因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故B正確;對(duì)于C選項(xiàng),因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,故C正確;對(duì)于D選項(xiàng),因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故D錯(cuò)誤.故選:BC.10.下列結(jié)論正確的是()A.若SKIPIF1<0,則SKIPIF1<0 B.若SKIPIF1<0,則SKIPIF1<0C.若SKIPIF1<0,則SKIPIF1<0 D.若SKIPIF1<0,則SKIPIF1<0【答案】ABC【解析】【分析】根據(jù)基本不等式和對(duì)勾函數(shù)逐項(xiàng)分析判斷.【詳解】對(duì)于A選項(xiàng),若SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立),故A正確;對(duì)于B選項(xiàng),因?yàn)镾KIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立),所以B正確;對(duì)于C選項(xiàng),因?yàn)镾KIPIF1<0,令SKIPIF1<0,SKIPIF1<0,對(duì)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,則SKIPIF1<0,故C正確;對(duì)于D選項(xiàng),若SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立),故D錯(cuò)誤.故選:ABC.11.德國(guó)著名數(shù)學(xué)家狄利克雷是解析數(shù)學(xué)的創(chuàng)始人,以其名字命名的函數(shù)稱為狄利克雷函數(shù),其解析式為SKIPIF1<0,則下列關(guān)于狄利克雷函數(shù)SKIPIF1<0的說法錯(cuò)誤的是()A.對(duì)任意實(shí)數(shù)SKIPIF1<0,SKIPIF1<0B.SKIPIF1<0既不是奇函數(shù)又不是偶函數(shù)C.對(duì)于任意的實(shí)數(shù)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0D.若SKIPIF1<0,則不等式SKIPIF1<0的解集為SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)題意結(jié)合奇偶性、一元二次不等式的解法逐項(xiàng)分析判斷.【詳解】若SKIPIF1<0是有理數(shù),則SKIPIF1<0;若SKIPIF1<0是無理數(shù),則SKIPIF1<0,故A正確;若SKIPIF1<0是有理數(shù),則SKIPIF1<0也是有理數(shù),此時(shí)SKIPIF1<0;若SKIPIF1<0是無理數(shù),則SKIPIF1<0也是無理數(shù),此時(shí)SKIPIF1<0;即SKIPIF1<0為偶函數(shù),故B錯(cuò)誤;若SKIPIF1<0是無理數(shù),取SKIPIF1<0,則SKIPIF1<0是無理數(shù),此時(shí)SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,故C錯(cuò)誤;若SKIPIF1<0是有理數(shù),則SKIPIF1<0的解集為SKIPIF1<0;若SKIPIF1<0是有理數(shù),SKIPIF1<0,顯然不成立,故D錯(cuò)誤.故選:BCD.12.設(shè)矩形SKIPIF1<0(SKIPIF1<0)的周長(zhǎng)為定值SKIPIF1<0,把SKIPIF1<0沿SKIPIF1<0向SKIPIF1<0折疊,SKIPIF1<0折過去后交SKIPIF1<0于點(diǎn)SKIPIF1<0,如圖,則下列說法正確的是()A.矩形SKIPIF1<0的面積有最大值 B.SKIPIF1<0的周長(zhǎng)為定值C.SKIPIF1<0的面積有最大值 D.線段SKIPIF1<0有最大值【答案】BC【解析】【分析】根據(jù)基本不等式的性質(zhì),結(jié)合圖形折疊的性質(zhì),結(jié)合對(duì)鉤函數(shù)的性質(zhì)逐一判斷即可.【詳解】設(shè)SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0.矩形SKIPIF1<0的面積SKIPIF1<0,因?yàn)镾KIPIF1<0,所以無最大值.故A錯(cuò).根據(jù)圖形折疊可知SKIPIF1<0與SKIPIF1<0全等,所以SKIPIF1<0周長(zhǎng)SKIPIF1<0.故B正確.設(shè)SKIPIF1<0,則SKIPIF1<0,有SKIPIF1<0,即SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),取最大值.故C正確.SKIPIF1<0,因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,所以當(dāng)SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)函數(shù)有最小值,無最大值.故D錯(cuò)誤.故選:BC.【點(diǎn)睛】關(guān)鍵點(diǎn)睛:利用基本不等式的性質(zhì)、對(duì)鉤函數(shù)的性質(zhì)是解題的關(guān)鍵.三、填空題:本題共4小題,每小題5分,共20分.13.計(jì)算:SKIPIF1<0_________.【答案】7【解析】【分析】根據(jù)指數(shù)的運(yùn)算法則計(jì)算即可.【詳解】原式SKIPIF1<0.故答案為:7.14.已知SKIPIF1<0是一次函數(shù),且SKIPIF1<0,則SKIPIF1<0_________.【答案】SKIPIF1<0##SKIPIF1<0【解析】【分析】根據(jù)待定系數(shù)法設(shè)SKIPIF1<0,代入整理得SKIPIF1<0,對(duì)比系數(shù)列式求解.【詳解】設(shè)SKIPIF1<0,因?yàn)镾KIPIF1<0,則SKIPIF1<0,可知SKIPIF1<0,解得SKIPIF1<0,故SKIPIF1<0.故答案為:SKIPIF1<0.15.已知SKIPIF1<0,若正數(shù)SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的最小值為_________.【答案】SKIPIF1<0【解析】【分析】首先判斷函數(shù)的奇偶性,即可得到SKIPIF1<0,再利用乘“1”法及基本不等式計(jì)算可得.【詳解】解:因?yàn)镾KIPIF1<0定義域?yàn)镾KIPIF1<0,且SKIPIF1<0,所以SKIPIF1<0為奇函數(shù),因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0時(shí)取等號(hào).故答案為:SKIPIF1<016.對(duì)于區(qū)間SKIPIF1<0,若函數(shù)SKIPIF1<0同時(shí)滿足:①SKIPIF1<0在SKIPIF1<0上是單調(diào)函數(shù);②函數(shù)SKIPIF1<0,SKIPIF1<0的值域是SKIPIF1<0,則稱區(qū)間SKIPIF1<0為函數(shù)SKIPIF1<0的“保值”區(qū)間.(1)寫出函數(shù)SKIPIF1<0的一個(gè)“保值”區(qū)間為_________;(2)若函數(shù)SKIPIF1<0存在“保值”區(qū)間,則實(shí)數(shù)SKIPIF1<0的取值范圍為_________.【答案】①.SKIPIF1<0##[0,0.5]②.SKIPIF1<0【解析】【分析】(1)由條件可知SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)函數(shù),根據(jù)SKIPIF1<0的值域判斷出SKIPIF1<0,由此得到SKIPIF1<0從而求解出SKIPIF1<0的值;(2)設(shè)存在的“保值”區(qū)間為SKIPIF1<0,考慮兩種情況:SKIPIF1<0,SKIPIF1<0,根據(jù)單調(diào)性得到關(guān)于SKIPIF1<0等式,然后根據(jù)二次函數(shù)的性質(zhì)即得.【詳解】(1)因?yàn)镾KIPIF1<0,所以SKIPIF1<0的值域?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,解得SKIPIF1<0,所以一個(gè)“保值”區(qū)間為SKIPIF1<0;(2)設(shè)保值區(qū)間為SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上為增函數(shù),所以SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0為方程SKIPIF1<0的2個(gè)不等實(shí)根,設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0;若SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上為減函數(shù),所以有SKIPIF1<0,兩式相減:SKIPIF1<0,代入得:SKIPIF1<0,所以方程SKIPIF1<0有2個(gè)不等實(shí)根SKIPIF1<0,SKIPIF1<0,從而有SKIPIF1<0,解得得SKIPIF1<0;綜上所述:SKIPIF1<0.故答案:SKIPIF1<0;SKIPIF1<0.四、解答題:本題共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.已知SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0,SKIPIF1<0;(2)求圖中陰影部分表示集合.【答案】(1)SKIPIF1<0,SKIPIF1<0(2)陰影部分表示的集合是SKIPIF1<0或SKIPIF1<0.【解析】【分析】(1)先根據(jù)一元二次不等式求解集合B,再根據(jù)集合的并集、交集運(yùn)算求解;(2)根據(jù)題意理解可得陰影部分表示的集合是SKIPIF1<0,根據(jù)補(bǔ)集運(yùn)算求解.【小問1詳解】由SKIPIF1<0,解得:SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0;【小問2詳解】由題意可知:陰影部分表示的集合是SKIPIF1<0SKIPIF1<0或SKIPIF1<0.18.已知函數(shù)SKIPIF1<0,SKIPIF1<0.(1)若函數(shù)值SKIPIF1<0時(shí),其解集為SKIPIF1<0,求SKIPIF1<0與SKIPIF1<0的值;(2)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集中恰有兩個(gè)整數(shù),求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0或SKIPIF1<0.【解析】【分析】(1)根據(jù)二次不等式的解法及韋達(dá)定理即得;(2)分SKIPIF1<0,SKIPIF1<0,SKIPIF1<0討論,然后結(jié)合條件即得.【小問1詳解】由題意可知SKIPIF1<0的解集為SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0;【小問2詳解】由SKIPIF1<0,可得SKIPIF1<0,①當(dāng)SKIPIF1<0時(shí),不等式的解集為SKIPIF1<0,若SKIPIF1<0的解集中恰有兩個(gè)整數(shù)解,則SKIPIF1<0;②當(dāng)SKIPIF1<0時(shí),不等式的解集為SKIPIF1<0,若SKIPIF1<0的解集中恰有兩個(gè)整數(shù)解,SKIPIF1<0;③當(dāng)SKIPIF1<0時(shí),不等式的解集為SKIPIF1<0,不合題意;綜上所述,實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0或SKIPIF1<0.19.已知函數(shù)SKIPIF1<0.(1)根據(jù)函數(shù)單調(diào)性的定義證明SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減;(2)若SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)镾KIPIF1<0,求SKIPIF1<0的取值范圍.【答案】(1)證明見解析;(2)SKIPIF1<0.【解析】【分析】(1)利用作差法證得SKIPIF1<0,由此可證得SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減;(2)先求得雙勾函數(shù)SKIPIF1<0與SKIPIF1<0時(shí)SKIPIF1<0的取值,結(jié)合圖像,可知區(qū)間SKIPIF1<0的子集與全集情況,由此求得SKIPIF1<0的取值范圍.【小問1詳解】任取SKIPIF1<0,不妨設(shè)SKIPIF1<0,因?yàn)镾KIPIF1<0SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減.【小問2詳解】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立),所以SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,結(jié)合雙勾函數(shù)SKIPIF1<0的圖象可知,SKIPIF1<0或SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值為SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的最大值為SKIPIF1<0;故SKIPIF1<0的取值范圍為SKIPIF1<0..20.已知函數(shù)SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞增,且對(duì)任意的SKIPIF1<0都滿足SKIPIF1<0.(1)判斷并證明函數(shù)的奇偶性;(2)若SKIPIF1<0對(duì)所有的SKIPIF1<0均成立,求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)函數(shù)是奇函數(shù),證明見解析;(2)SKIPIF1<0.【解析】【分析】(1)利用賦值法得到SKIPIF1<0,由此證得函數(shù)的奇偶性;(2)利用函數(shù)奇偶性與單調(diào)性推得SKIPIF1<0,進(jìn)而得到SKIPIF1<0,利用復(fù)合函數(shù)的單調(diào)性證得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,由此求得SKIPIF1<0的取值范圍.【小問1詳解】函數(shù)是奇函數(shù).證明如下:因?yàn)閷?duì)任意的SKIPIF1<0都有SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0是奇函數(shù).【小問2詳解】因?yàn)镾KIPIF1<0,SKIPIF1<0恒成立,又因?yàn)镾KIPIF1<0在定義域SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0恒成立,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0恒成立,因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減,SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,所以復(fù)合函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0,故SKIPIF1<0,即SKIPIF1<0.21某城市對(duì)居民生活用水實(shí)行“階梯水價(jià)”,計(jì)費(fèi)方法如下表所示.每戶每月用水量水價(jià)不超過SKIPIF1<0的部分2.5元/SKIPIF1<0超過SKIPIF1<0但不超過SKIPIF1<0的部分6元/SKIPIF1<0超過SKIPIF1<0的部分9元/SKIPIF1<0(1)求用戶每月繳納水費(fèi)SKIPIF1<0(單位:元)與每月用水量SKIPIF1<0(單位:SKIPIF1<0)的函數(shù)關(guān)系式;(2)隨著生活水平的提高,人們對(duì)生活的品質(zhì)有了更高的要求,經(jīng)驗(yàn)表明,當(dāng)居民用水量在一定范圍內(nèi)時(shí),若隨性用水,用水量增加,生活越方便;若時(shí)刻想著節(jié)約用水,生活也會(huì)麻煩.?dāng)?shù)據(jù)表明,人們的“幸福感指數(shù)”SKIPIF1<0與繳納水費(fèi)SKIPIF1<0及“生活麻煩系數(shù)”SKIPIF1<0存在以下關(guān)系:SKIPIF1<0(其中SKIPIF1<0),當(dāng)某居民用水量在SKIPIF1<0時(shí),求該居民“幸福感指數(shù)”SKIPIF1<0的最大值及此時(shí)的用水量.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0的最大值為SKIPIF1<0,此時(shí)用水量為SKIPIF1<0.【解析】【分析】(1)根據(jù)已知條件,分段求解函數(shù)關(guān)系式即可;(2)根據(jù)題意寫出SKIPIF1<0與SKIPIF1<0的關(guān)系式,再求其最大值即可.【小問1詳解】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;可知SKIPIF1<0與SKIPIF1<0的函數(shù)關(guān)系式為SKIPIF1<0.【小問2詳解】由題意可知:當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,于是SKIPIF1<0,所以當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0取得最大值SKIPI
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