高一上學(xué)期期末數(shù)學(xué)試題一．單選題(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SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.SKIPIF1<0的值為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.已知對(duì)數(shù)函數(shù)SKIPIF1<0且SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.函數(shù)SKIPIF1<0的圖象大致為()A. B.C. D.5.已知SKIPIF1<0，則()ASKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<06.已知SKIPIF1<0，則SKIPIF1<0的值為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<07.圖(1)是某條公共汽車線路收支差額SKIPIF1<0關(guān)于乘客量SKIPIF1<0的圖象，圖(2)(3)是由于目前本條路線虧損，公司有關(guān)人員提出的兩種扭虧為盈的建議，則下列說法錯(cuò)誤的是()A.圖(1)中的點(diǎn)A表示當(dāng)乘客量為0時(shí)，虧損1.5個(gè)單位B.圖(1)中的點(diǎn)B表示當(dāng)乘客量為3時(shí)，既不虧損也不盈利C.圖(2)的建議為降低成本同時(shí)提高票價(jià)D.圖(3)的建議為保持成本同時(shí)提高票價(jià)8.函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、多選題：(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.下列說法錯(cuò)誤的是()A.命題“SKIPIF1<0”的否定為“SKIPIF1<0”B.命題“SKIPIF1<0都有SKIPIF1<0”的否定為“SKIPIF1<0使得SKIPIF1<0”C.“SKIPIF1<0”是“SKIPIF1<0”充要條件D.“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件10.下列函數(shù)既是偶函數(shù)，又在SKIPIF1<0上單調(diào)遞增的是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<011.若SKIPIF1<0，則下列說法正確的是()A.SKIPIF1<0的最大值為SKIPIF1<0 B.SKIPIF1<0的最小值是SKIPIF1<0C.SKIPIF1<0的最大值為SKIPIF1<0 D.SKIPIF1<0的最小值為SKIPIF1<012.函數(shù)SKIPIF1<0，則下列結(jié)論正確的是()A.當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0B.不論SKIPIF1<0為何值，函數(shù)SKIPIF1<0既沒有最小值，也沒有最大值C.不論SKIPIF1<0為何值，函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸都有交點(diǎn)D.存在實(shí)數(shù)SKIPIF1<0，使得函數(shù)SKIPIF1<0為R上的減函數(shù)三．填空題：(本題共4小題，每小題5分，共20分．)13.已知扇形的面積為2，扇形圓心角的弧度數(shù)是2，則扇形的弧長為________14.不等式SKIPIF1<0的解集是________．15.生物入侵是指生物由原生存地侵入到另一個(gè)新環(huán)境，從而對(duì)入侵地的生態(tài)系統(tǒng)造成危害的現(xiàn)象，若某入侵物種的個(gè)體平均繁殖數(shù)量為SKIPIF1<0，一年四季均可繁殖，繁殖間隔SKIPIF1<0為相鄰兩代間繁殖所需的平均時(shí)間．在物種入侵初期，可用對(duì)數(shù)模型SKIPIF1<0來描述該物種累計(jì)繁殖數(shù)量SKIPIF1<0與入侵時(shí)間SKIPIF1<0(單位：天)之間的對(duì)應(yīng)關(guān)系，且SKIPIF1<0，在物種入侵初期，基于現(xiàn)有數(shù)據(jù)得出SKIPIF1<0．據(jù)此估計(jì)該物種累計(jì)繁殖數(shù)量是初始累計(jì)繁殖數(shù)量的SKIPIF1<0倍所需要的時(shí)間為________天．(結(jié)果保留一位小數(shù)．參考數(shù)據(jù)：SKIPIF1<0)16.已知函數(shù)SKIPIF1<0對(duì)于任意SKIPIF1<0，都有SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是________．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.設(shè)全集SKIPIF1<0，集合SKIPIF1<0，其中SKIPIF1<0．(1)若“SKIPIF1<0”是“SKIPIF1<0”成立的必要不充分條件，求SKIPIF1<0的取值范圍；(2)若命題“SKIPIF1<0，使得SKIPIF1<0”是真命題，求SKIPIF1<0的取值范圍．18.在平面直角坐標(biāo)系SKIPIF1<0中，角SKIPIF1<0與角SKIPIF1<0的頂點(diǎn)均為坐標(biāo)原點(diǎn)SKIPIF1<0，始邊均與SKIPIF1<0軸的非負(fù)半軸重合，角SKIPIF1<0的終邊過點(diǎn)SKIPIF1<0，將SKIPIF1<0繞原點(diǎn)SKIPIF1<0按順時(shí)針方向旋轉(zhuǎn)SKIPIF1<0后與角SKIPIF1<0的終邊SKIPIF1<0重合．(1)寫出角SKIPIF1<0與角SKIPIF1<0的關(guān)系，并求出SKIPIF1<0的值：(2)求SKIPIF1<0值．19.已知函數(shù)SKIPIF1<0的部分圖象如圖所示．(1)求函數(shù)SKIPIF1<0的解析式：(2)將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位后，得到函數(shù)SKIPIF1<0的圖象，求函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)減區(qū)間．20.已知二次函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0取何值時(shí)，不等式SKIPIF1<0對(duì)一切實(shí)數(shù)SKIPIF1<0都成立：(2)若SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)恰有一個(gè)零點(diǎn)，求實(shí)數(shù)SKIPIF1<0取值范圍．21.某蔬菜種植基地共有蔬菜種植大棚SKIPIF1<0個(gè)，用于種植普通蔬菜，平均每個(gè)大棚年收入為SKIPIF1<0萬元．為適應(yīng)市場(chǎng)需求，提高收益，決定調(diào)整原種植方案，將SKIPIF1<0個(gè)大棚改種速生蔬菜，其余大棚繼續(xù)種植普通蔬菜．經(jīng)測(cè)算，調(diào)整種植方案后，種植普通蔬菜的每個(gè)大棚年收入比原來提高SKIPIF1<0，種植速生蔬菜的每個(gè)大棚年收入為SKIPIF1<0萬元．(1)當(dāng)SKIPIF1<0時(shí)，要使蔬菜種植大棚全年總收入不少于原來的SKIPIF1<0，求SKIPIF1<0的取值范圍(2)當(dāng)SKIPIF1<0時(shí)，求蔬菜種植大棚全年總收入的最大值．