2019全国II卷理科数学高考真题【2020新】.pdf

20195122B0.5345125601A={x|x25x+60}B={x|x10}AB=A(1)B(21)C(31)D(3+)2z=3+2izABCD3ABuuur=(2,3)ACuuur=(3t)||BCuuur=1ABBCuuuruuur=A3B2C2D342019132L2LMMR2Lrr121223()()MMMRrRrrR.rR34532333(1)rA21MRMB212MRMC2313MRMD2313MRM59911779ABCD6abAln(a-b)0B3a3bCa3-b30Dab7ABCD8y2=2px(p0)2231xyppp=A2B3C4D892(42)Af(x)=cos2xBf(x)=sin2xCf(x)=cosxDf(x)=sinx10(02)2sin2=cos2+1sin=A15B55C33D25511FC22221(0,0)xyababOOF222xyaPQPQOFCA2B3C2D512()fxR(1)2()fxfx(0,1]x()(1)fxxx(,]xm8()9fxmA9,4B7,3C5,2D8,3452013100.97200.98100.99__________14()fx0x()eaxfx.(ln2)8fa__________15ABC,,ABC,,abc.6,2,3bacBABC_________1612481_________________237017212223601712ABCDA1B1C1D1ABCDEAA1BEEC11BEEB1C12AE=A1EBECC1181211110:102.0.50.4.10:10X.1PX=22X=4.1912{an}{bn}a1=1b1=01434nnnaab1434nnnbba.1{an+bn}{anbn}2{an}{bn}.201211lnxfxxx.1f(x)f(x)2x0f(x)y=lnxA(x0lnx0)exy.2112A(-2,0)B(2,0)M(x,y)AMBM-12.MC.1CC2CPQPPExEQECG.iPQGiiPQG.10222322[44]10O000(,)(0)M:4sinCl(4,0)AOMP.10=30l2MCPOMP.23[45]10()|||2|().fxxaxxxa11a()0fx2(,1)x()0fxa.20191A2C3C4D5A6C7B8D9A10B11A12B130.98143156316262117111BC11ABBABE11ABBA11BCBE1BEECBE11EBC21190BEBRtABE11RtABE45AEBAEAB12AAABDDAuuurx||DAuuurDxyzC010B1101C012E101(1,0,0)CBuuur(1,1,1)CEuuur1(0,0,2)CCuuuurEBCn=xyx0,0,CBCEuuuruuurnn0,0,xxyzn=(0,1,1).1ECCm=xyz10,0,CCCEuuuuruuurmm20,0.zxyzm=1101cos,||||2nmnmnm1BECC32181X=2101022PX=2=0.50.4+10.510.4=0.52X=41010441[0.510.4+10.50.4]0.50.4=0.1191114()2()nnnnabab111()2nnnnababa1+b1=lnnab112114()4()8nnnnabab112nnnnababa1b1=lnnab1221112nnnab21nnabn111[()()]222nnnnnnaababn111[()()]222nnnnnnbababn201fx01U1+212()0(1)f'xxx()fx011+fe=e110e122222e1e3(e)20e1e1ffx1+x1fx1=01101x1111111()ln()01xfxfxxxfx0111xfx20ln01exxBlnx001xy=ex0()0fx0001ln1xxxAB00000000000111ln111ln1xxxxxkxxxxxxy=ex001(ln,)Bxx01xlnyx00(,ln)Axx01xlnyx00(,ln)Axxy=ex2111222yyxx221(||2)42xyxCx2iPQk(0)ykxk22142ykxxy2212xk2212uk(,),(,),(,0)PuukQuukEuQG2k()2kyxu22(),2142kyxuxy22222(2)280kxukxku(,)GGGxyuGx22(32)2Gukxk322GukykPG322212(32)2ukukkukkukPQPGPQGiii2||21PQuk2221||2ukkPGkPQG222218()18(1)||12(12)(2)12()kkkkSPQPGkkkkt=k+1kk0t2k=12812tSt[2+t=2k=1S169PQG16922100,MC0304sin233||||cos23OPOA(,)QlPRtOPQcos||23OP(2,)3Pcos23lcos232(,)PRtOAP||||cos4cos,OPOA4cosPOMAPOM,42P4cos,,42231a=1()=|1|+|2|(1)fxxxxx1x2()2(1)0fxx1x()0fx()0fx(,1)2()=0fa1a1a(,1)x()=()+(2)()=2()(1)0fxaxxxxaaxxa[1,)