高考文科數(shù)學試卷及答案

數(shù)本試卷分第Ⅰ卷（選擇題）和第Ⅱ卷（非選擇題）兩部分，第Ⅰ卷第1至第2頁，第Ⅱ卷第3至第4頁．全卷滿分150分，考試時間120分鐘．．．．．，BAS4R2P(AB)P()P(B)R4，BAVπR33P(B)P()P(B)R第I卷（選擇題共60分）52,AB）IB2,1(CA)UB(,0)．A．R．AUB(0,)．C)IB2,1RAB(2,4)AC(1,3)，,則（）．）））D）,,,nm）．若,,‖,,‖若mnmnB．‖,‖,‖．若mmn‖,‖,‖m．若m則n0ax22x10a）3,7BACABC則）．．．．3643(x)(x1)1(x0)f．f21(x)1x1(x1)(x)1f(x)1x1(x1)1B．x1(x2)D．f1(x)1x1(x2)．f1(1x)aaxLax,aa,L,a8則）80180,18234D5sin(2x)y）3xxxD．．x．C．6126121(x)2xx0),f(x)f則（）x（10CD(x2)y1ll22（）33,]33(,)33．[3,3]．(3,3)．[D．33x00（）若Ay從－21ayx2xyaA()3474．1．D54882()．CA．CA．CAD．CA2866282328262825數(shù)學（文科）第Ⅱ卷（非選擇題共90分）44x21(x)（f．(x2x2y213＝（nnn5a}a4nnaaLaanbnnN*a,b（）則ab，，22n12n（,B,C,DAC213,AD,平面BCD,BCCD,AB6,若8B,C則6（(x)cos(2x)2sin(x)sin(x)f344(x)(x)ff[,]122（3.1232（分OOABCDABCD1的ABC,OA2M為OA,,4M與；BDABC（a3(x)xx(ax1,af3232(x)x1在af(x)xxa1a(0,)xf'2（a,aca1c,cN,a,c0caa*0n1nnan11,c，bn(1a),nNba*S；n22nnnn0a1nN*0c1n（xy22:abF4.Cxa2b2C(2,0),F傾斜角為C于AB14222;(2,0),,F于AB和DE,求的C1一.3[3,)二.4－1三.Qf(x)cos(2x)2sin(x)sin(x)（）344132x2xxxxx)22132x2xxx2222132x2x2x22sin(2x)6∴周期T2Qx[,2x[,]（）122636[,]123[,]32f(x)sin(2x)6f(x)1當x331f()3Qf()()fx又∴當x22212232f(x)在區(qū)間[,]上的值域為[122（1310333101010100027(i1,2,3)（）設A張卡片帶有后鼻音“iiP(A),則i71C17CC23C3P(A)P(A)3,4012023103C3107111P(AA)P(A)P(A)40120602323O９（Q‖AB,∴MDC與MCDP,作∵BCDQCDA2∵∴=P42BDP1∵MDMAAD2∴cosMDP,MDCMDP22，MD23與3‖∵∴點ABA作，∵CD,CD∴平面OAP,∵平面OAP∴∵OP∴平面又A1322∵41，2222222222gg2322∴B3232)CD,y,z作AP222,BPD(OM,222z設與所成的角為,O22,∵(,22gMABMD1∴cos,23ABMD∴與3DAP∵,(222,,xyCB222n(x,y,z)則nOP0,nODgg02y2z02即22xy2z02n(0,4,2)22取zuBd則d為n(0,4,2),n∵OB(1,0,2)∴23,d.n2B3解:f(x)ax3x(a1)f(x)1f(1)0在x'2'3a10,∴a1即aax(x2)x2x0a(0,)3x(a1)xxa1a(0,)22即a22(a)a(x2)x2x(aR)g(a)(aR)設g,xR，22(0,)g(a)0，g(0)0ax22x0∴2x0，即|2xxx3x(a1)xxa1a(0,)ax22(x2)x2x0a(0,)即a22x2xx2x22a(0,)0ax22x22∴2x0|2xxx解c28aa2c24∴4b2a2b2c2x2y2∴1C84F2(2,0)Ce21:x4設l為橢圓的左準線。則ll于A,BBl于B作AA，l與x)1111點A2∴2112()2112212∴2cos122cos12242∴ABAFBF。2cos2cos2cos112當tanyk(x:k22y8(12k)x8kx8(k1)0x得222222(x,y),B(x,y)設Ax,x.1122128k8(k22∴xx,xx.12k12k122122AB(xx)(yy)k)(xx)k)[(xx)4xx]22222212121212128k32(k1)42(1k)222k)[()2]212k12k12k222422∵ktan,222AB22當242∴22（的傾斜角為ABAB,4242，2222424222122222222242當443解∵a1c(a1)n1n∴a1當1a1acn∴a1(a1)cn1a(ac11a1an1nnn∴(can1nN。1()aa*nn2ca1(1)(1)L(1)(1)nacaca1acn12n1nn1n2∴a(ac1n1nn1aa1∴(cn1nN*。1()aaann1bna)cn()n1n2n111SbbLb2()Ln()2n2212n12n1211S()2()Ln()23n1222n11111∴S()L()n()2nn122211212n111∴S1()L()n()2[1()]n()2n1nnn222222n1∴S2(2n)()n2n(ac1an1n0(ac110