河北省衡水中学2021届高三上学期二调数学试题

一8540分.在每小题给出的四个选项中，只有一项是符合题目要求的.x|x22x，|13，BCx|x2,nNxA(AB)Cx（）20,20,2,42,4y2sinxy2cosx）44444(x)x(xa)bf10byf(xf（）12a1anS，a，a与aS22n244n1）6或62或63cos（）33313136333f(x)yfx()）f(x)xtanxfx()xsin2x.11(x)x2xf(x)xxf22m,}mnf(x)min3log,logxx，1420ab）fafbab当6891SS设Sn(,nNSaSan（）n*2n12100nnn111198100323211114950C3232二多选题：本题共4小题，每小题5分共20分.在每小题给出的选项中，有多项符合题目要求全部选对的得5分，有选错的得0分，部分选对的得3分.aa是递增数列，满足anS75nn）00d1085n时当nS时Snnfxyfx,ny和上,nyfx1xya，2a）32561032()sinfx(03)xx48g(x)的g(x)f(x)2cos2x）4(x)g(x)g(x)是g的最小正周期为x8,0g(x)点是58(x)x),f223fxfxfx1,的,xx31261212fx线x）,fx62fx,06fx12,44fx将倍2(6ygxg(x)cosx到三填空题：本题共4小题，每小题5分，共20分.,9,aaban2aSabn25nn.(x)asinxcosx(a0,0)ffx0,7,x0,kx(x)x[x]g(x)f[x]x1,x0.xfxgx5,57kBABC在AC；baac22AbAaab四､670分解答应写出文字说明､证明过程或演算步骤.ABCCacosCccosAbcosB.；B，，D.,a1,a0n2a2n1SSaSn1nnn1nS2S；n1nan343SACAMAMNcABCA，，ab，B，CB3c8M，N是BC边上的两个三等分点，AM和ABC.aa22f(x)x3x,g(x)3x,aR32222fxa3）若函数(x)f(x)g(x)x2x0,2在x0.a22甲a，SnnnS,S,S.12343nbaa3ba，TnT.1213nnnnnaq1S,S,S132.xf(x)f(x)()y在x()fx为fxf(x)(x)f(x)fDD(x)fD(x)ex1rxr(x)(x)(xexfex0f(x)；tfD(x)1tiD一8540分.在每小题给出的四个选项中，只有一项是符合题目要求的.x|x22x，|13，BCx|x2,nNxA(AB)Cx（）20,20,2,42,4Bx|x2x0A2x0x2∴Ax|13Bxx0x4∴ABx0x4∴ACx|x2,nN(AB)C0,2∴2sinxyy2cosx）44444A22)yxx222)，yxx22y2sinxy2x44(x)x(xa)bf10byf(xf（）12C10由yf(xyf(x)x1fa,b.yf(x1)1.xf(x)yf10(1,0).yfxa2a2由()，fxx2bxb24a12得，f(1)1ab0a2，b1a1anS，a，a与aS22n244n1）6或62或6Cadn1S411aa122aad2a得，11d34a1daa4adn241a0或a811，d1d5a0,d1S012364，1a8,d51S8(3)2724．．n3cos（）33313133633Dxx.333xxsinx，33313coscos2xcos2xcos2x2sinx12.333f(x)yfx()）f(x)xtanxfx()xsin2x11(x)x2xf(x)xxf22C.f(x)xtanx项ADx，1(x)xcosxf(x)fD2()12cos2[1,3]，fxxC，f(x)1cos2x0C.C.m,}mnf(x)min3log,logxx，1420abab）fafb当689Bfxfafbab3214.fxfafblogalogb3，214loglog3logalogblog(ab)3.ab8ab24222Bm,}fx.1SSS设San(,N（）Sann*nn2n12100nn1111981003232111149503232A2由递推式求出数列的首项,当n时分n为偶数和奇数求出a,代入n1S(1)a,nN*然后利用等比数列的前n.n2nnn1(,nN，Saa*n2nn111当n当na；Sa21411112aSS((an1an22n1nnn1nnn11a(a(a.nn2nnnn112n111(n2)n（当aann2nn11111an（当a2a(2)n2n2n12n2n12nn1nn111111,aaa2,a,a，a22222222424121234111111aa2,,a,a2.aa1002423100993499SSSSaaaaaaaa1231234561111111111222235992221001112111112450110011.321001142ASa与nnnaf(n1)f(n)；ncab；nnn二多选题：本题共4小题，每小题5分共20分.在每小题给出的选项中，有多项符合题目要求全部选对的得5分，有选错的得0分，部分选对的得3分.的aSa是递增数列，满足an75nn）00da10n时85SS当n时nna，dSn.1n的Sad，naaa6d3a4dad，7511100,ABada1nddd7dSnnannn，22222217dn73CS由2n或4时nnd2d7d0时n8nn0n07S令S或n.2D22nnabnnn2是nn的.fxyfx,ny和,nyfx1xya，2a）35610232fxa.1x(x)af(x)2ax.f与211,22,xa0,a1.22a22a0ax112020xa0,a,222a0a2,2020x，.AB.()sinfx(03)xx48g(x)g(x)f(x)2cos2x）4(x)g(x)的最小正周期为是gx8,0g(x)g(x)5点是8由x为f8gx(x)(x)2k,kZ.又kx为f88420,32()sin2gxx2cos2因为，所以，所以x441sin2x，gx3225cos(2x)x.322,kkZ，，.,AkkZCBD44(x)x),f223fxfxfx1,的,xx31261212fx线x）,fx62fx,0612,fx44将fx26ygxg(x)cosx到f(x)2x.374312x,kkZ，k62,,(x)又f23312T23375T261时满足，此时2,f(x)2xk，33,x()2fxA错x6233362fx由2,0为B636x2x,f(x)f446364min1121sin,f(x)fsin1C6222maxsinx(x)将f2y3ysinxsinxcosx移6632即g(x)x.D三填空题：本题共4小题，每小题5分，共20分.,9,aaban2aSabn25nn.SS,n2a2an1用abnanSnS,n1nnnn1a,9,a25aab2，nSnnn21aSSab2nab2b2；当nn1n1nnn1当n22，aSabb011b0a又2,9,a5aa224②bb251,b1，ab2.a(x)asinxcosx(a0,0)ffx0,72fx，(x)asinxcosxa1sin(x)据f20,7由a212fxa.(x)asinxcosxa1sin(x)，f2fxa212，3a3(或，af(x)3sinxcosx2sinx当，6fxx2k,kZ620当x0和，k2当k17，xx62，3＝(＝yfxyax＋＝φ＋＝)bxyAbyA,x0,kx(x)x[x]g(x)f[]x1x,x0.xfxgx5,5个不同的实数根，则实数k7.11,54(x)(x)g，f(x)(x)g，fygx[5,0)yfx3与fxgx在区间[5,0)3fxgx在区间[0,5]ygxyfx4与0,5414,15,114k.yk511若在区间[0,5]4k,.5411,.54.BABC在AC；baac22AbAaab35,222B2acosBacAbAa12A.abAbac2Bbac2accosB，2222222acosBa22cB，csin(AB)2sinAcosBsinAsin2sincossin，CABAsin(BA)sinAsincoscossin2sincossin即，ABABABA,B0,BA()()2AB2；BAA或BAAB3A(0,)A0,A，31bAaBAA2AAA2cosA2，2cosAabABA2A111181x3xcosA,1()22fxx,,1x()40，令，fxx22x22x22x21252fx,11335,f(1)f(x),.又f222235,；.22四､670分解答应写出文字说明､证明过程或演算步骤.ABCCacosCccosAbcosB.；B，，D23）B）.3ABCAC知ADC，DCBABC和ADCABCDsinAcosCsinCcosA2sinBcosB，得sinB2sinBcosB.0B,sinB0，12cosBB.3，32492222ABC中，，B，37.在ADC7,1，，，，ABCDACAD2，3B3AD2DC2AC21DC271cos，32ADDC2DC23或DCDC211SS23.S2323ABCADC.,a1,a0S2aS2n1anSn1nnn1nS2S；n1nanSaSS，SS2a）∴SS2S122n1n1nn1n1nnn1nna2an2n1naSS，SSa），22n1n1nn1n1n2SSSS，2nn1nn1SS2S0，n1n1naS0，nn1S2S0；n1nS2S，n1n2S）S，n1nS2Sn2，nn1a2an2，n1nanS2Saa2a，即21211a101，得2,n1,a12n2,nn,2ana则aa12，232112，11n1aSa2n2求求Sann111aSSaa1由32nnn1na4a.nn343SACAMAMNABCA，，abcB，CB3c8M，N是BC边上的两个三等分点，和ABC.tANABBNABBNcosBt2BM222AC，BCBM，然后用余弦定理求得AM，ACt,AMACtBM得AM,ACAN323，BMt2t8，，c设BMBABNANABBN2ABBNcosB，222t)84t282t60即，222t2t80，t2.在ABMAMABBM2ABBMcosB84282cos6052，22222，ACABBC2ABBCcosB86286cos6052，222222，bAC2134392Rsinsin6033，B22393.RM，N是SBC43，AMN123ABCS，113S123ABBCsin608BC，B222△ABCBC6BM2.在ABMAMABBM2ABBMcosB84282cos6052，22222，ACABBC2ABBCcosB86286cos6052，222222，bAC2134392Rsinsin6033，B22393R.t3BCt，设BM在ABMAMABBM2ABBMcosB8t2tcos608tt，2222222ABCACABBC2ABBCcosB89t28tcos60649t24t，222222AMAC8tt649t24t，222t2，在ABMAMABBM2ABBMcosB84282cos6052，22222，ACABBC2ABBCcosB86286cos6052，222222，bAC2134392Rsinsin6033，B22393.Raa22f(x)x3x,g(x)3x,aR.32222fxa3）若函数(x)f(x)g(x)xx0,2在x0.2a223(,)()fxa939aa当a3f(x),1和1,3369aa93,1,1）.为533fx3和a3a.xx在03133(0)()(1)3,20,2axaxaxxax2322222）()36313，fxx2xax2a0fx，3当a(,)(x)ffx9a9a当a30x或x，1133a939af(x),1和1,33fx9a9a0，得1令1，x33933a93af(x)1,1.33()fx(,)aa939a3当af(x),1和1,339a93a1,1.为33133）由题意得(x)ax(ax3xa,x0,2.3222220在xx321232320,2，(0)()(1)3a2x,a2xax3ax2x132(ax3x0,x0,2，即ax322当x0.131x0,2ax2ax30当即，223x23x23a.23xx2x22222xx1x+22令tx2()1,(2,4]，htttt22510()1,(2,4]ht)hthttt，2t2t36625，a(x2)15x26,.a5.甲anS，nnS,S,S.12343naa3bab，TnT.1213nnnnnaq1S,S,S132..q21nbn32n1q，2S,S,SS,S,S.1231321Sa若q则，21111Saaaaa，22212111113Saaaaaaa，24431231111SS2SS,S,S.1231321aa3aa3，413111n14aan211n12n32nnnbnan4，1212221111T123nn则，32222n1231221111T123nn，322242n1231111221111112221nTnnn.1322222n2n132n12341223n243n2T21，2n2n1n4n2N,11T.n*n32n1S,S,Sa132n12nqb232nn.xf(x)(x)()y在xffx为()fxf(x)(x)f(x)fDD(x)fD(x)ex1rxr(x)(x)(xexfex0f(x)；tfD(x)1tiDxx()rx）(1)0)(,1]，xe.xr(x)ex1x(x)ex1r(x)e1rxxr(x)(x)(x1)ex)ff(x)f(x)是xxxe12xf(x)1f(x)(x1)elnxx1x1(x)1)在)在x1ftexxx1x1h(x)(1)et.xxx(x)ex1r(x)e1，rxxxx()rxe(1)则x，r(x)ex1xxx()rx(e1)0x0，令xr(x)ex1xr(x)0.x0f(x)(x1)exf(x))，tx1xe12x()((xexexefx，xxxxxxxe12xf(x)1，(x)ff(x)(x1)elnxxex1ln①1ln③xxexxx，1(x1)eln②xe1(x1)eln④xe2x2xxx(x)f，fx0()f