2011年高考真题（河北省）-理科数学（河北省） Word版含解析

2011年高考全国卷I理科数学试题详细解析1izz1，z为zzziiiiz1i,z1(1i)(1i)(1i)1i.z.2x(yx2x2(xR)y(yy444x(xR)y4x(0)22yyy22x(中，y0且反解x得x【精讲精析】选.在函数y，所以4x2y2x(y(x.4（3）下面四个条件中，使bb1b1ab＞＞ab3322P使Pab,abS1SS24kanad2，nn1k2knkSaaSk2kk2k1.SSaa2a(2kd2(2k1)224k5.k2kk2k11＞cosx(f(x)0)yf(x)313963个f(x)y3326kk1k(kZ)6.minC.3lA,AClB,BDl,DD2361333AC平面D作DEBC于C.BCDE,BCACCDE为DBDDC126在RtBCDDE.BC3323441种种种种4B1C1462C24e和2x131223与.y2e2x,|2yx22yr01211S。2335f(x)2f(x)2x(1x)=f()=21-214141252A.5111.2()f()f()f222y24xy24cosAFB=x45353545D.y24x2x425401,4.x，消y得xxxyA在x22224532.M600M4N(C)11(D)13作示意M的面积为4，易得MA23，22RtOMNOMN30。故MNOMcos303..MN相交于CD，MNMC9413圆Nr22r13圆NSD21|b1,ab,ac,bca,b,c|a||c|232|c|Aa,b,c,,,，|c|x)x.9CC.rnrnnC(x)由T得xC,xCCC.r21820182029r12020，525由，sincos43525512,tan得cos,25242.123x2y2FF:-M.29FAFAF|=122.|AF||MF|1,|AF||AF2a6|AF||MF|2|AF|6.2221211ABCDABCDBBCCBEEB,CF=FC12341111面AEFABC.2交3CAP90FCA2FCCA23tanFCA232sinAcosCD由AC90AsinAcosC，acbsinAsinC2sinB，ACCC2sin(C)2B，又是ABCC45B或(C45)B180ABC(90C)(C45)C180。15C.1lX(10.5)0.3pp0.6。1(10.5)(10.6)0.8.11(10.5)(10.6)0.2PP11。XB(100,0.2).EX1000.220X（）如图，四棱锥S中，AB//CD,BCCDSABABBC2,CDSD1.;SBCAB2SASDAD222SASDSB知SDS又SASBSDD做Dz，平面SAB.平面ABCD1A(2,-1,0),B(2,1,0),C(0,1,0),S(,0,3)22n|ABn|2321|cosAB,n.|AB||n|27721arcsin.7110且1.aa1a1an1n1nan1an,记Sb,证明：S1.bn1nnnknk11111.{}1a1a1an1nnab}nn1{}11an11(n1)1.(nN)1a1an1n1abnnn11an111n1nnnn1n1111111nSb()()()11.n1n1nk1223nk1y2C:x1O在yF22OAOBOP0.PPPP在C明APB,AQBNA(x,y),B(x,y)2112y22x1x214x222x10l:y26262x1,x44221xx,xx122412由OAOBOP0.得P((xx),(yy))12122(xx),212(yy)(2x12x1)2(xx)211212122(2()2122所以点P在C上。y(1y(222x()x()kk2212y(1y(21kk122x()x()22123(xx)4(xx)2321219233xx12(xx)212y12y1122x2x()1kk1kk22y1y112122x2x()221(xx)4(xx)122121233xx12(xx)212APB,AQB222P(,1)Q(,1)lyx22212121,)lyx设M(4224221l、lN(,)88122213118|()(1),22288322|AB1(2)|xx2213242211)()338|AM,|(,2248283118|NA|AM||MN|22|NA||NP|NQ|,|NA||NB|故|NP|.N2x0x)＞f(x)0；f(x)x21到的91()＜取ppe212(x2x(x2)2x2f(x)0x1(x(1x)2f(x)在()当x0f(x)f(0)0。p2xf(0)0x)f(x)x21()911019ln19ln(1)192故110102210919().e2e2xxxxxx,x012nnn12ni