2018靜安區(qū)高考數(shù)學(xué)一模試卷

一.45nlim(1)n11in2ii11ix2y2(1916從53(x)a23aaRyf(x)yf(x)1f（x1在(xa)x7a(2,3)ax(a1)y30aO12,yRe、exyx2P的1(x,y)OM(1,2)M83(5a)x1x1(x)a0a1，R的af（axx131(x)|xxx)|yf(x)f222a0aay(x)ax4x1xRf(f(x))0a的f2二.45312a}a1）2n1323112或333RA{x|ylog(1x)}B{x||x1|(C)B（設(shè)全集U，，）3U(0,1](0,1)(1,2)[1,2)l、m都在平面l和m）x2y2|yx2C:與1）44(,1](1,)(1,)(,1][1,0)(1,)三.5ABCABC中，AA4與AA.BC1111113ABCABC111BCAACC.111ABCm(aB)n(b)，中ABCabc在n且m∥n，m.BA；2ABABx.xABCDA、PQBCCDPAB、，tant.C、P、QABCDSS.|zi|3i|mizxxRyRCCz（，(x,y)mx3y60(lNCP、QM.lCl與m||23lAMANtttt.n已知數(shù)列a}a（n,aN.*nnana、a、aa124kN且ak（k、a、aka*13ka}ana}.nn一.x2y21i(3,0)0391641(,3]U[,)7427{,,,}123126a3[3,5)二.BDCA三.162811ACBB1）1B=是2分與111113BC因?yàn)?，所?43，4分11113于是，三棱柱體積VSHSAA16344836分4ABC1（2）過B作BDAC，D為垂足，則BD平面，C11D1是C8分111827DC1134所以Carcsin分111377arctanarccos(，)741628(aB),n(bA且mn，aAbcoscos0）mB分absincossincosB,即Asin2sin2又又AABB2RsinAsinB02,2BABC中2A2B22AB或AB或AB即分2BabcosAcosBmn且，，若Amn分AB2sinAsinBsinAsinBxsinAsinBsinAcosA分xsinAsinBsinAcosA設(shè)sinAcosAt，t2A)4分A12siA()20A44442t21t212sinAcosAsinAcosA分21t222x在tt，tx221t2111tt21tt2t22tx)1628QCDP45AB）PAB,tanBPt1tCPt，；1t1t11t1tQ01)t，,;1t2=分221t1t1t2CPQ=t))分1t1tt11t=1S分)22分221t12t12t112當(dāng)t分S22分142636ylCQMPAOxNm13:C,k,分m3031則k故kkm與m.l分01lll3xy30）l與x1;分xk(xly.分x231.分k3431k由CM.分分21k14340y.lx或x5(()A(當(dāng)l與M,N又x35(0,)故t5.分3k(xy得(3)4當(dāng)l的斜率存在時(shí),設(shè)直線l的方程為y,代入圓的方程x22xxkk2k)x(2k6k)xk6k50x2,2222則121kM2kkkkkk222yk(x即,M(,)分1k1k1kMM222yk(xk1kkk12(,1,k)AM1k1k1kx3y6222k65k55k5得N(,)(則,k).1k1k1k1k1kk5kkk)kk)22故t（）5.kk)kk)kk)2225,tl且t.分CACM知mB,CN,ACm,又CMl,,C,N,BMCNt5.分142737124a2a，a，，1a2a44a1a2a∵a，，aaa，分1241422aa2a2a．分13ka3a，k，aa，，1a3aka1k(a)aa∵a，a，a，分23131kk9a9ka9aka6ka(39)(9)ak2,ka22(9)39,(3)99∵akakaka，a21分當(dāng)k10,a39k11當(dāng),39,k設(shè)tk9N*la3N*,則a.9分tl,k∵alt,36kk9,l1,2,3,4,6,9,12,18,36l或者列舉45)分分naaak，tnnktktaaa111ankt1)nakataknka