湖北省重點高中2020屆高三數(shù)學上學期期中試題文

/12/12/湖北省重點高中2020屆高三數(shù)學上學期期中試題文考試時間：2019年11月12日下午15：00－17：00試卷滿分：150分★祝考試順利★注意事項：1.答卷前，考生務(wù)必將自己的學校、考號、班級、姓名等填寫在答題卡上。2.選擇題的作答：每小題選出答案后，用2B鉛筆把答題卡上對應(yīng)題目選項的答案信息點涂黑，如需改動，用橡皮擦干凈后，再選涂其他答案標號，答在試題卷、草稿紙上無效。3.填空題和解答題的作答：用0.5毫米黑色簽字筆直接答在答題卡上對應(yīng)的答題區(qū)域內(nèi)，答在試題卷、草稿紙上無效。4.考生必須保持答題卡的整潔。考試結(jié)束后，將試題卷和答題卡一并交回。第I卷選擇題(共60分)一、選擇題(本大題共12小題，每小題5分，共60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.已知復數(shù)z＝i3(3－i)，則z＝A.1＋3iB.1－3iC.－1＋3iD.－1－3i2.已知集合A＝{－2，－1，0，1，}，B＝{x|x2－4≤0}，則A∩B＝A.{－1，0，1，2}B.{0，1，2}C.{－1，0，1}D.{－2，－1，0，1，2}3.產(chǎn)品質(zhì)檢實驗室有5件樣品，其中只有2件檢測過某成分含量。若從這5件樣品中隨機取出3件，則恰有2件檢測過該成分含量的概率為A.B.C.D.4.已知向量a，b滿足a·b＝1，|b|＝2則(3a－2b)·b＝A.5B.－5C.6D.65.函數(shù)y＝|x|＋1的圖象與圓x2＋(y－1)2＝4所圍成圖形較小部分的面積是A.B.C.D.π6.已知方程表示焦點在x軸的雙曲線，則m的取值范圍是A.－2<m<－1B.－3<m<－2C.1<m<2D.2<m<37.已知l，m，n是三條不重合的直線。其中命題“若l//m且l⊥n則m⊥n”是真命題。若把l，m，n中的任意兩條直線換成平面，另一條保持不變，則所得到的所有新命題中，真命題的個數(shù)是A.0B.1C.2D.38.如圖所示的程序框圖，若輸入x的數(shù)值是19，則輸出的y值為A.－124B.124C.26D.09.已知f(x)＝lnx＋1，0<a<b，若，則關(guān)于l，m，n的關(guān)系式中，正確的是A.m＝n<lB.mB.m＝n＞lC.lC.l＝n<mD.l＝n＞m10.已知非零實數(shù)a，b，c不全相等，則下列說法正確的個數(shù)是(1)如a，b，c成等差數(shù)列，則能構(gòu)成等差數(shù)列(2)如a，b，c成等差數(shù)列，則不可能構(gòu)成等比數(shù)列(3)如果a，b，c成等比數(shù)列，則能構(gòu)成等比數(shù)列(4)如a，b，c成等比數(shù)列，則不可能構(gòu)成等差數(shù)列A.1個B.2個C.3個D.4個11.在△ABC中，“△ABC是鈍角三角形”是“cosC＝2sinAsinB”的()條件A.必要不充分B.充要C.充分不必要D.既不充分也不必要12.已知函數(shù)，且函數(shù)g(x)滿足g(x)＋f(4－x)＝5，則函數(shù)y＝f(x)－g(x)的零點個數(shù)為A.0B.4C.3D.2第Ⅱ卷非選擇題(共90分)二、填空題(本大題共4小題，每小題5分，共20分。)13.某金屬零件的三視圖，如圖所示(單位：m)，則該零件的體積為cm3。14.已知實數(shù)m是區(qū)間[0，4]上的隨機數(shù)，則方程x2＋3x＋3m－3＝0有異號兩根的概率為。15.已知函數(shù)，則f(x)的最小正周期是，最小值是。16.某制藥廠生產(chǎn)A，B兩種藥品均需用甲，乙兩種原料。已知生產(chǎn)1噸每種藥品所需原料及每天原料的可用限額，如下表所示。如果生產(chǎn)1噸A，B產(chǎn)品可獲利潤分別為4萬元，5萬元，則該制藥廠每天可獲最大利潤為萬元。三、解答題(本大題共6小題，共70分，解答應(yīng)寫出文字說明、證明過程或者演算步驟。第17－21題為必做題，每個試題考生必須作答，第22，23題為選做題，考生根據(jù)要求作答。)(一)必做題：共60分17.(本小題滿分12分)某電信運營公司為響應(yīng)國家5G網(wǎng)絡(luò)建設(shè)政策。擬實行5G網(wǎng)絡(luò)流量階梯定價。每人月用流量中不超過kGB(一種流量計算單位)的部分按2元/GB收費；超出kGB的部分按4元/GB收費。從用戶群中隨機調(diào)查了10000位用戶，獲得了他們某月的流量使用數(shù)據(jù)。整理得到如下的頻率分布直方圖：(I)若k為整數(shù)，依據(jù)本次調(diào)查，為使80%以上用戶在該月的流量價格為2元/GB。k至少定為多少?(II)假設(shè)同組中的每個數(shù)據(jù)用該組區(qū)間的右端點值代替，當k＝3時，試估計用戶該月的人均流量費。18.(本小題滿分12分)已知等差數(shù)列{an}和正項等比數(shù)列{bn}滿足a1＝b1＝2。a2＋a3＝10，b2b4＝a18。(I)求數(shù)列{an}，{bn}的通項公式；(II)設(shè)數(shù)列{cn}中，cn＝an＋bn，求和：c1＋c3＋c5＋…＋c2n－1。19.(本小題滿分12分)如圖，直三棱柱ABC－A1B1C1的底面是邊長為4的正三角形，M，N分別是BC，CC1的中點。(I)證明：平面AMN⊥平面B1BCC1；(II)若直線A1C與平面A1ABB1所成的角為30°，試求三棱錐M－ANC的體積。20.(本小題滿分12分)設(shè)橢圓與兩坐標軸的交點分別為A(a，0)，B(0，b)(a>b>0)，點O為坐標原點，點M滿足，OM所在直線的斜率為。(I)試求橢圓的離心率e；(Il)設(shè)點C的坐標為(0，－b)，N為線段AC的中點，證明MN⊥AB。21.(本小題滿分12分)已知函數(shù)f(x)＝(x＋a)lnx，曲線y＝f(x)在點(1，f(1))處的切線與直線x＋2y＝0垂直。(I)求a的值；(Il)令，是否存在自然數(shù)n，使得方程f(x)＝g(x)在(n，n＋1)內(nèi)存在唯一的根?如果存在，求出n，如果不存在，請說明理由。(二)選做題：共10分。請考生在第22，23題中任選一題作答，如果多做，則按所做的第一題計分。22.(本小題滿分10分)選修4－4：坐標系與參數(shù)方程在直角坐標系xOy中，直線l的參數(shù)方程為(t為參數(shù))，以原點為極點，x軸正半軸為極軸，建立極坐標系，⊙C的極坐標方程為ρ＝4sinθ。(I)寫出⊙C的直角坐標方程；(II)P為直線l上的一動點，當P到圓心C的距離最小時，求P的直角坐標。23.(本小題滿分10分)選修4－5：不等式選講已知關(guān)于x的不等式|x＋a|<b的解集為{x|4<x<6}。(1)求實數(shù)a，b的值；(2)求的最大值。2019秋高三文數(shù)參考答案及評分細則一、選擇題：本大題共12小題，每小題5分，共60分.1.答案：C，注意是求z的共軛復數(shù).2.答案：D，集合，故3.答案：B，列舉后容易知道，基本事件總數(shù)有10種，恰有2件檢測過該成分含量的事件共有3種，所以所求概率為4.答案：B，5.答案：D，如下圖，所圍成的圖形的面積，6.答案：B，易知即7.答案：C，①l不變，有l(wèi)∥α且l⊥βα⊥β；②m不變，有m∥α且α⊥βm⊥β；③n不變，有α∥β且n⊥αn⊥β；分析知①，③正確.8.答案：A，9.答案：C，由對數(shù)運算的性質(zhì)知，，所以l=n，又為增函數(shù)，時，，所以m>l,所以有10答案：C，（1）錯，（2）（3）（4）對11.答案：A，假設(shè)C為鈍角，則，，顯然充分性不成立，又由可知，即，此時有，即A為鈍角或B為鈍角，從而△ABC為鈍角三角形，必要性成立12.答案：D，由知，令，則所以有，即的圖像關(guān)于直線對稱.當時，；當時，。作出的圖像可知，當時，有兩個零點.二、填空題（本大題共4小題，每小題5分，共20分）13.答案：14答案：答案:,解析：由題意有:故最小正周期為,最小值為.答案：解析：設(shè)每天生產(chǎn)A藥品x噸，B藥品y噸，利潤，則有作出可行域知，z在點處取得最大值.三、解答題：本大題共6小題，共70分。解答應(yīng)寫出必要的文字說明。證明過程或者演算步驟。第17-21題為必考題，每個試題考生必須作答，第22,23題為選考題，考生根據(jù)要求作答。（一）必考題：共60分17.解：（I）由直方圖可知，用戶所用流量在區(qū)間內(nèi)的頻率依次是0.1,0.15,0.2,0.25,0.15，········································································································3分所以該月所用流量不超過3GB的用戶占85%，所用流量不超過2GB的用戶占45%，故k至少定為3；·····································································································································6分（II）由所用流量的頻率分布圖及題意，用戶該月的人均流量費用估計為：2×1×0.1+2×1.5×0.15+2×2×0.2+2×2.5×0.25+3×2×0.15+(3×2+0.5×4)×0.05+(3×2+1×4)×0.05+(3×2+1.5×4)×0.05=5.1元······················································································12分18.解：（I）設(shè)等差數(shù)列的公差為d，因為，所以又，所以d=2，即，··································································3分設(shè)正項等比數(shù)列的公比為q，因為即，由，知，所以·················································································································6分（II）······················································································8分設(shè)，則····································································································································12分解：（I）證明:如圖,由直三棱柱知，··············································2分又M為BC的中點知AM⊥BC,又,所以·······································4分又AM?平面AMN,所以平面AMN⊥平面B1BCC1·······································································6分(II)如圖：設(shè)AB的中點為D,連接A1D,CD.因為△ABC是正三角形,所以CD⊥AB.由直三棱柱知CD⊥AA1.所以CD⊥平面A1ABB1,所以∠CA1D為直線A1C與平面A1ABB1所成的角.即∠CA1D=30°,···8分所以A1C=2CD=2×=,所以A1D=6,在Rt△AA1D中，AA1=，NC=·······························································································10分三棱錐的體積即為三棱錐的體積,所以V=···································································12分20.解：（I）由，，知,··································2分由kOM=知······································································································4分所以,,所以e=.·································································6分(II)證明:由N是AC的中點知,點N,所以，···············································8分又，所以·····················································10分由（I）知，即，所以=0，即MN⊥AB.····················································································································12分21.解:(I)易知切線的斜率為2,即,又,所以a=1;······························4分(II)設(shè),當時,.又所以存在,使得.······················································································································6分又··························································································8分所以當時,,當[2,)時,即時,為增函數(shù),所以時,方程在內(nèi)存在唯一的根.·························································