2019年江蘇卷數(shù)學(xué)高考真題

2019年普通高等學(xué)校招生全國統(tǒng)一考試（江蘇卷）注意事項4第1第11nnx,x,,xsxx2xx．i212nnini1i1柱體的體積VShhS1Sh錐體的體積VSh35B{x|xx}Aa，AB▲.a▲i.S▲.76xx2y▲32221▲▲...y21(b3，4)7x2b2▲.{anN)aaa0,S27*n若SS▲8.nn2589ABCDE為CCABCD▲.111114x(x0)Py▲AxAA▲P+x..O6中，D是E與若，AB則▲.AC23π2▲.π44(xg(x)f(x)g(x)f(x).f當xR，(xx1k(0,2]f(x)1(x2，gx1()9],1x22(x)g(x)于xf有8k▲.C．2=2=c3sinAcosBsin(B)ab2ABCE．111AB；111C．1x2y2ab:F1a2b2(x1)y4aFF作xxlF:222C222FC．12215=．12CEOOlOOBl和DP和QD和dQ(x)(xa)(xb)(xc),a,b,cRf==a=f'(x){、f'(x)為4ab1,c1≤．271.(nN)aaa,a4a4a0aa}*n245324n122bSbn(nN)b}*n1Sbbnnnnn1bn(nN)mc}*mcbcnk1kkm2019年普通高等學(xué)校招生全國統(tǒng)一考試（江蘇卷）分.537[y2x{1,6}10212,3(e,1)34.分.2c,b2,cosBacba，32c)c(13222222Bc2.232cc3c.3sinAcosB，ababcosBsinBcosB2sinB.sinAsinBbb45cosB(2sinB)cos41cosBcos2B222B2.255B0BB0B.π255sinBcosB.2ABCAB，1111111AB，1111AB.111E為.ABC.1111.1CAAC=，111111A.11CAC.1111.C.FFFF12125532FF()2=x=，21212221222212由bacb2222x2y21C.43x2y21：，43xA2將F+y=22211y5x6x110，222115x1或x.11122x2y將xBy，5511123(,)y(x1)又F：.552243y(x4137由x1或x7x6x0.2x2y21431.Ex23323將x1y(1)xy(1,)E.42x2y21：.2121122112xx.211x32Fy.x2y2114332Ey.23(1,)E.2.作.8，84cosPBDsinABE.105BD12415PB.cosPBD5.處.AEED10，AD22222270.O....O.PPBAB1P設(shè)，113PDPBsinPBDPBcosEBA159；51111△PPBPBPB15.11.由（2）知，要使得，點Q只有位于點C的右側(cè)，才能符合規(guī)劃要求.當時，CQQAAC156321.22223綜上，當⊥，點Q位于點C右側(cè)，且=3時，d最小，此時P，Q兩點間的距離.321..xy223??.44，34yx3253.PB(134)(93)15.?22.?.?3x6(4x4)：y.41523345，，22244O....O≥F的.P為PBABPP設(shè)l?1111△PPBPBPB15.11.2(a4)(93)15(a4)4343（由AQ=22.43?（PQ4321(13)17321.3.abcf(x)(xa)(xb)(xc)(xa)3．8(4a)8a2．f3cb，(x)(xa)(xb)x(ab)xb(2ab)xab2，f2322ab2ab(x)3(xb)x()0f'xxbx或f'．332aba,b,，都在集合ab，32ab1,a3,b3．3f(x)(x3)(x3)2，f'(x)3(x3)(x1)．令f'(x)0x3x1或x10f'(x)f(x)(x)+–+f(1)(13)(13)32．f2a0,c1f(x)x(xb)(x1)x(b1)xbx，32f'(x)3x2(b1)xb．20b14(b1)12b(2b1)30，22x,xxx．則f'(x)1212b1bb1b1bb122由f'(x)0x,x．3312x,xx(x,)x11222f'(x)f(x)+0–0+Mfx．(x)f1Mfxxbx3121112bb1xb12(1)bb[3xbxb]x1213999112bb1b1)2(1)2bb3bb1227927bbbb22(bb1)3bb1)244M．272727270b1x(0,1)1．當x(0,1)f(x)x(xb)(x1)x(x2．1(x)x(x1),x(0,1)(x)3x(x．令g231令g'(x)0x3111x3g'(x)g(x)+0–114g(x)g(x)gx．3max344x(0,1)f(x)g(x)M．2727{，aqaq1naaaaqaqa1244由2451．1aq1a4a4a04aq4a0q22321111{a}.n12Sbb2bn0．nnn11221,Sbb.2b1由11b1122122bbnnSbbSn由1，bb)nnn1n1nbbnnbb2bSSb當n11nn，2bb2bbn1nnnn1nnn1bbb．n1n1n{.bnnN.*{=bnbnnN.=，kbk*k{，cqcq1nkq≤≤qkcbc.1kkmkkk當；kqlnklnklnq當．．kmkk1lnx1lnx設(shè)（=fx(x1)f'(x)xx2f'(x)0令e)x(1,e)+0–（）fxln2ln8ln9ln3ln33f(k)f(3)．2663lnklnqk3kq，取q3kkkqk1．m若，mk35qm．m數(shù)學(xué)Ⅱ(附加題)C31A22A；2.,B43.A42.設(shè)xR，解不等式|x.(1x)aaxaxax,n4,nNa2aa.n2n*23012n24(13)ab3a,bN*ab.n22{(0,0),(1,0),(2,0),,(n,0)}（本小題滿分分）在平面直角坐標系中，設(shè)點集A，nB,(N.nCnnnnABC令MM.nnnnn.n數(shù)學(xué)Ⅱ(附加題)參考答案．]31A，223131A2222233123112115==．23222122106A31f()4.222f)01,4.令A(yù)12．]，（2，ABO423(232)5.24=22sin()3，l4(32,)l．24(322)sin()2.B(2,)又lB242．]1x12x2-；xx31≤2<xxxx21當>xxxx21x|xx.3解：（1）因為(1x)CCxCxCxn4，n0122nnnnnnn(n1)n(nn2)aC,aC23，262n3nn(nn2)(n3)aC4．244n2aa，a2324n(nn2)n(n1)n(nn2)(n3)[]22，6224n5．n5．(13)(13)n5CC3C(C(C(C(0515252353454555ab3．a(chǎn),bN*aC3C9C76,C3C9C44，052545b153555b3a2．222CC(C(C(C(C(50515252353454555CC3C(C(C(C(0515252353454555．a(chǎn),bN*(13)3．a(chǎn)b5b(ab3)(ab3)(13)(13)(2)32a2．2555n1X1，22，5．7744XP(X,P(X，CC22662222P(X,P(X．CC2266(ab)B(cd)M（和nP(Xn)1P(Xn)XndnXnbb0d1②若，則AB()1acn21，所以Xn當且僅當n21，此時2a0cnanc0或2b0d2AB34nnac()4n2(1)4n以Xn220cnanc0或2n4a2b1d2④若，則AB()1acn21，所以Xn當且僅當n21，此時2a0cnanc0或2nXXn21和4n242P(Xn,P(Xn22．C2C22n42n46P(Xn)1P(XnP(Xn122．C22n41x12x2-；xx31≤2<xxxx21當>xxxx21x|xx.3解：（1）因為(1x)CCxCxCxn4，n0122nnnnnnn(n1)n(nn2)aC,aC23，262n3nn(nn2)(n3)aC4．244n2aa，a2324n(nn2)n(n1)n(nn2)(n3)[]22，6224n5．n5．(13)(13)n5CC3C(C(C(C(0515252353454555ab3．a(chǎn),bN*aC3C9C76,C3C9C44，052545b153555b3a2．222CC(C(C(C(C(50515252353454555CC3C(C(C(C(0515252353454555．a(chǎn),bN*(13)3．a(chǎn)b5b(ab3)(ab3)(13)(13)(2)32a2．2555n1X1，22，5．7744XP(X,P(X，CC22662222P(X,P(X．CC2266(ab)B(cd)M（和nP(Xn)1P(Xn)XndnXnbb