瀘州市高2018級第一次教學質量診斷性考試數(shù)學（理科）本試卷分第I卷（選擇題）和第II卷（非選擇題）兩部分.第I卷1至2頁，第II卷3至4頁.共150分.考試時間120分鐘.注意事項：1.答題前，先將自己的姓名、準考證號填寫在試卷和答題卡上，并將準考證號條形碼粘貼在答題卡上的指定位置.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題的答案標號涂黑.3.填空題和解答題的作答：用簽字筆直接答在答題卡上對應的答題區(qū)域內，作圖題可先用鉛筆繪出，確認后再用0.5毫米黑色簽字筆描清楚，寫在試題卷、草稿紙和答題卡上的非答題區(qū)域均無效.4.考試結束后，請將本試題卷和答題卡一并上交.第I卷（選擇題共60分）一、選擇題：本大題共有12個小題，每小題5分，共60分．每小題給出的四個選項中，只有一項是符合要求的．4x≤0}，Bx|x2n1,nN，則ABAxx1．已知集合{|2A．3B．1,3C．1,3,4D．1,2,3,42．“sincos”是“cos20”的A．充分但不必要條件C．充要條件B．必要但不充分條件D．既不充分也不必要條件3．已知log5，bln1，c1.51.1，則a，b，c的大小關系正確的是a23A．bcabacC．acbD．abcB．4．我國的5G通信技術領先世界，5G技術的數(shù)學原理之一是著名的香農(Shannon)公式，香農提出并嚴格證明了“在被高斯白噪聲干擾的信道中，計算最大信息傳送速率C的公式SCWlog(1)”，其中W是信道帶寬（赫茲），S是信道內所傳信號的平均功率（瓦），2NSN是信道內部的高斯噪聲功率（瓦），其中叫做信噪比．根據(jù)此公N式，在不改變W的前提下，將信噪比從99提升至，使得C大約增加33了60%，則的值大約為(參考數(shù)據(jù):100.21.58)A．1559C．1579B．3943D．251222正視圖側視圖5．右圖為某旋轉體的三視圖，則該幾何體的側面積為A．10C．9B．8D．102俯視圖高三·理數(shù)第1頁共4頁6．函數(shù)3xy（其中e是自然對數(shù)的底數(shù)）的圖象大致為xeexyyyyxxxOxOOOA．B．C．D．x0)與x軸的兩個交點，且Ax7．已知,0，,0兩點是函數(shù)Bxfx()2sin()(A、612兩點間距離的最小值為，則的值為B3A．2B．3C．4D．5()滿足f(2xfx)()，f(2x)f(x)x[0,1]，當8.定義在上的函數(shù)R時，，fxx()2fxf(x)的圖象與g(x)|x|的圖象的交點個數(shù)為B．4C．5則函數(shù)A．3D．69．在長方體ABCDABCD中，,分別為,,分別為,EFOMCDBC的中點，BDEF的中11111111點，則下列說法錯誤的是D1EC1CA.四點B、D、E、F在同一平面內MA1FB1B.三條直線BF，DE，CC1有公共點DC.直線與直線OF不是異面直線ACOA1BD.直線上存在點使,,三點共線ACNMNO12logx0的兩根分別為，，則下列關系正確的是xx10．已知方程2x12A．12xxxx2B．0xx1C．xx1D．12121212角形，且平面平和△BDC是邊長為11．已知三棱錐中，2的等邊三ABCDBACABD△面，BCD該三棱錐外接球的表面積為2016A．4C．8B．D．331311fx12．已知函數(shù)()ax3x2(a0)，若存在實數(shù)x(1,0)0且，使()()，fxfx0220則實數(shù)的a取值范圍為22A．(,5)B．(,3)(3,5)331818C．(,6)D．(,4)(4,6)77高三·理數(shù)第2頁共4頁第II卷（非選擇題共90分）注意事項：(1)非選擇題的答案必須用0.5毫米黑色簽字筆直接答在答題卡上，作圖題可先用鉛筆繪出，確認后再用0.5毫米黑色簽字筆描清楚，答在試題卷和草稿紙上無效.(2)本部分共10個小題，共90分.二、填空題（本大題共4小題，每小題5分，共20分．把答案填在答題紙上）2x3,x≤013．已知函數(shù)f(x),則f(f(1))的值___________．21,x0xsinx(x[0,])與x軸所圍圖形的面積為．14．曲線y15．在平面直角坐標系xOy中，角與角均以Ox為始邊，它們的終邊關于y軸對稱．若tan1，則tan()．316．如圖，棱長為1的正方體ABCD-A1B1C1D1中，P為線段A1B上的動點（不含端點），有下列結論：D1C1①平面⊥平面ADPA1AP;11A1B1②多面體CDPD的體積為定值；1③直線與BC所成的角可能為3；DPD1CP④APD△能是鈍角三角形.1BA其中結論正確的序號是（填上所有序號）．三、解答題：共70分.解答應寫出文字說明、證明過程或演算步驟.第17～21題為必考題，每個試題考生都必須作答.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．（本題滿分12分）已知函數(shù)f(x)3sinx2cos2x1．2（Ⅰ）若f()23f(tan)，求的值；6（Ⅱ）將函數(shù)f(x)圖象上所有點的縱坐標保持不變，橫坐標變?yōu)樵瓉淼?倍得函數(shù)g(x)2的圖象，若關于x的方程g(x)m0在[0,]上有解，求m的取值范圍．218．（本題滿分12分）已知曲線f(x)kxsinxb在點(,f())處的切線方程為2xy30．22（Ⅰ）求k，b的值；（Ⅱ）判斷函數(shù)f(x)在區(qū)間(0,)上零點的個數(shù)，并證明．2高三·理數(shù)第3頁共4頁19.（本題滿分12分）a，b，c，已知asin(AB)csinBC．在△ABC中，角，，的對邊分別為ABC2（Ⅰ）求A；（Ⅱ）已知，求AD．c3，b1，邊BC上有一點D滿足S3S△ADC△ABD20.（本題滿分12分）如圖，在四棱錐S—ABCD中，底面ABCD是菱形，G是線段AB上一點(不含A,B），在平面SGD內過點G作GP//平面SBC交SD于點P．S（Ⅰ）寫出作點P、GP的步驟(不要求證明)；BAD，ABSASBSD2，P是SD（Ⅱ）若的中3點，求平面SBC與平面SGD所成銳二面角的大小．AD21．（本題滿分12分）GCm1,e，e是自然，其中已知1B函數(shù)fxxmlnxmx對數(shù)的底數(shù).（Ⅰ）求函數(shù)fx的單調遞增區(qū)間；11,e恒成立時k的最大值為kxn對xcx（Ⅱ）設關于的不等式fxxxlnx1,e求nc的取值范圍.（，），kRn（二）選考題：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22．（本題滿分計分.10分）選修4-4：坐標系與參數(shù)方程在平面直角坐標系xOy中，曲線C是圓心在（0，2），半徑為2的圓，曲線C的參數(shù)12x22cost方程為y22sin(t)（t為參數(shù)且0≤t≤），以坐標原點O為極點，x軸正半軸為極24軸建立極坐標系．(Ⅰ)求曲線C的極坐標方程；1（Ⅱ）若曲線C與兩坐標軸分別交于A,B兩點，點P為線段AB上任意一點，直線OP與求OM的最大值．OP2曲線C交于點M（異于原點），123．（本題滿分10分）選修4-5：不等式選講若a0,b0且2ab23ab，已知ab有最小值為k.(Ⅰ)求k的值；（Ⅱ）若xR使不等式xmx2km成立，求實數(shù)的取值范圍．0高三·理數(shù)第4頁共4頁瀘州市高2018級第一次教學質量診斷性考試數(shù)學（理科）參考答案及評分意見評分說明：1．本解答給出了一種或幾種解法供參考，如果考生的解法與本解答不同，可根據(jù)試題的主要考查內容比照評分參考制訂相應的評分細則．2．對計算題，當考生的解答在某一步出現(xiàn)錯誤時，如果后繼部分的解答未改變該題的內容和難度．可視影響的程度決定后繼部分的給分，但不得超過該部分正確解答應得分數(shù)的一半；如果后繼部分的解答有較嚴重的錯誤，就不再給分．3．解答右側所注分數(shù)，表示考生正確做到這一步應得的累加分數(shù)．4．只給整數(shù)分數(shù)，選擇題和填空題不給中間分．一、題號1答案B2345678910C11D12DAACDABAC二、填空題13．33415．16．①②④14．2三、解答題17．解：（Ⅰ）因為f(x)3sinx2cos2x123sinxcosx··················································································1分2sin(x)，···················································································2分6因為f()23f()，所以sin()23sin，·······························3分663sincos123sin，·························································4分所以22即33sincos，··········································································5分所以tan3；··············································································6分9（Ⅱ）f(x)圖象上所有點橫坐標變?yōu)樵瓉淼?倍得到函數(shù)g(x)的圖象，2所以函數(shù)g(x)的解析式為g(x)f(2x)2sin(2x)，······························8分65，···············································10分因0x，所以≤2x≤2666高三·理數(shù)第5頁共4頁故m的取值范圍為[1,2].········································12分f(x)ksinxkxcosx，···································································2分所以1g(x)2，18．解：（Ⅰ）因為f()ksin2cos2所以kk，··························································3分b，························································4分22k又因為f()ksinb2222點(,f())處的切線方程為2xy30．2所以2k2，···························································································5分b3.··································································································6分（Ⅱ）f(x)在(0,)上有且只有一個零點，························································7分2因為f(x)2sinx2xcosx，···································································8分當x(0,)時，f(x)0，······································································9分2所以f(x)在x(0,)上為單調遞增函數(shù)且圖象連續(xù)不斷，·····························10分2因為f(0)30，f()30，·······················································11分2f(x)在(0,)上有且只有一個零點．····················································12分2所以（Ⅱ）因為f(x)2sinx2xcosx,設g(x)sinxxcosx,g(x)2cosxxsinx0恒成立．·······································7分當x(,)時，2所以g(x)在(,)上單調遞減，································································8分2又g()0,g()0，所以t(,)使得g(t)0，·····································9分22所以f(x)在(,t]為單調遞增函數(shù)，在[t,)為單調遞減函數(shù)，························10分2因為2f()0,f()0，········································································11分所以f(x)在(,)上有且只有一個零點．···················································12分219．解：（Ⅰ）由AB+CBCAA可得sin(AB)sin(C)sinC,sinsincos,222)csinBCA，得asinCccos,············································2分又asin(AB22A得sinsinsinCcos,··························································3分AC由正弦定理2A因sinC0，所以sinAcos,2AAA則2sincoscos,············································································4分222cos0因0A,所以·······································································5分A222高三·理數(shù)第6頁共4頁A1sin=,即=,則A．·····························································6分2226A所以3（Ⅱ）解法一：設ABD的AB邊上的高為，h1BADC的AC邊上的高為h,2,c3,b1,因S3Sh1ABDADC所以31bh,1Dchh2C22A12所以hh，AD是ABC角A的內角平分線，·········································8分1230，所以BAD34因S3S，可知S，··················································10分ABCSABDADCABD1sin3031ABACsin60，所以ABAD24233.·················································································12分所以AD4解法二：設BAD=(0)，則=,················································7分DAC33因S3S,c3,b1ADCABD所以cADsin311sin()，bAD223sinsin()，········································································8分3所以所以sin3cos1sin，所以tan3，223因0所以30，BAD33因S3S可知S························································10分ABCS4ABDADCABD1sin3031ABACsin60所以ABAD24233，················································································12分所以AD4解法三：設ADx,BDA=，則=,ADC在ABC中由c3,b1及余弦定理可得：a22bccosAbc22因a7，························································································7分因S3SABDADC=37，·········································································8分可知BD3DC4在ABD中AB2BD2AD22BDADcos，即963AD237ADcos，··························································10分162在ADC中，17AD27ADcos()，162即17AD2+7ADcos，····························································11分162高三·理數(shù)第7頁共4頁33．················································································12分所以AD420.解：（Ⅰ）第一步：在平面ABCD內作GH‖BC交CD于點H；·······································2分第二步：在平面內作HP‖SC交SD于P；·············································4分SCD第三步：連接GP，點P、GP即為所求．······················································5分（Ⅱ）解法一:因是SD的中點，HP//SC，所以H是CD的中點，···························6分P而GH//BC，所以G是AB的中點.·····················································7分連AC,GD交于O，連SO，設S在底面ABCD的射影為M，因為SASBSDMAMBMD，即M為ABD的外心，，所以所以M與O重合，·········································································8分233262，所以SO，SD，OC2AC43，3因OD33OGOE,OS分別為x，y，z軸建立空過O作OE//GB交BC于E，以,間直角坐標系，則(0,0,26),B(3,1,0),C(3233,2,0)，·······················9分S33,1,26所以(SB),BC(3,1,0)，設平面的法向量為3SBC3n(x,y,z)，zSnSB33xy26z03，則PnBC3xy0AD取z2，則x1,y3，所以(1,3,2)．····················10分M(O)HGnC又GBx平面，SGDyB故GB(0,1,0)為平面的法向量，···············································11分SGD設平面與平面所成銳二面角的大小為，SBCSGD|nGB|32，則cos|n||GB|62(0,)，所以.····························································12分因為24成銳二面角的大小為．4故平面與平面所SBCSGD解法二：延長DG,CB交與I，連接SI，取SI的中點K，連接GK,BK，，GP因為GP//平面，平面平面SGDSI平面，SGDSBCSBC所以GP//SI，··············································································7分又P是SD的中點，則G是DI的中點，故GIGDGS，················································8分GKSI所以，又GB平面，SIDBKGCSID所以為二面角的平面角.···········································10分SO2IO222,SG2SK21，SGI在中，SGGI3,SI則SK2,從而GK又GE1，BGGK，故BKG，4高三·理數(shù)第8頁共4頁故平面與平面所成銳二面角的大小為．SGD·····························12分SBC4mlnxmx0,m1,e，121.解：（Ⅰ）因為fxxxmx2mx1,因x0,m1,e······································1分11所以fxx2xx21m0,2時,fx的增區(qū)間為，······················2分所以①當m40即2240即2me時,方程2②當m10的兩根為xmxmm24，xmm4，2x1222fx的增區(qū)間為0,x,x,，·····························································4分12綜上①當1m0,2時，fx的增區(qū)間為，m242m2m②當2me時，fx的增區(qū)間為0,,m4,，2········································································································5分（Ⅱ）原不等式km1lnxxxlnxn.·····················································6分x因m1,ex1,e,,所以m1lnxxxlnxn1lnxxxlnxn,xx1lnxxxlnxn,···································································7分令gxxlnxxn即gxlnxxn，即px11，，令pxx2x所以px在x1,e上遞增；····································································8分①當p10即n1時,n1,e,所以n1，因為1,e上遞增，，所以在gx當x1,epx0,即gx0,cgxg1n，所以minnc2n2，···············································································9分故時，②當pe0ne1,e即x1,epx0,即gx0，因為,所以gx在1,en2ge，e上遞減，所以cgxmin故ncn2ne1,e21························································10分eee③當p1pe0即n1,e1時,又pxlnxxn在1,e上遞增，,使得x1,epx0,即0nxlnx所以存在唯一實數(shù)，000則當x1,x時px0，即gx0，當xx,e時px0即gx0，故gx在00x1,x上減，xx,e上增，00高三·理數(shù)第9頁共4頁1lnxxxxnlnx1ln00cgxgx所以.························11分x0000x0min0所以nclnx1xlnxx1，0x0000x0x210，0x設uxx11x1,e），則ux1（'0x0x20001e1,e,nc2,eux在上遞增所以.所以2.···································································12分nc2,e1綜上所述e22.解：(Ⅰ)解法一：設曲線C與過極點且垂直于極軸的直線相交于異于極點的點E，且曲1線C上任意點F(,)，邊接OF，EF，則OF⊥EF，·····································2分1在△OEF中，4cos()4sin，······················································4分2解法