高一下數(shù)學(xué)期末考試(必修三必修五)含答案

5分.a}aa8qn14（A）1248ABCa2,b3,B60A（A）x002zxyyxy22024b0a1111ababb2a2616nn6n2到11613122312,bab2aba32332242m~m10014組6mm~mmn,n0.27,782nm5p4（B）125bABC,B,Ca,b,cB2Aa((3,22)1/8（10）已知數(shù)列a}3aa4(na91Snnn1nn1Sn6125n567855.a}aaa9aa.13524n40030003200（n100n.ABC.A2a2a2a2a41a23則ann123nnnABC,B,Ca,b,cC60o25c2ABC.667設(shè)a}0a2aa，.232n1（Ⅰ）求a}n（Ⅱ）設(shè)b}12ab}S.nnnnn2/8ABCABCABCl40,5050,60…，頻率組距0.0250.0150.010.005分?jǐn)?shù)405060708090100ax或2；axacbc23/8ABC,B,Ca,b,c(abc)(abc),.B31AC.C41qf()(1,0).aSan1nn1(1)aS；nn1bb，bf(b)(nN,n2)b*12nn1nn1T.241ca(1)c}的前項(xiàng)和為T(mén)n2bnnnnnn4/8BA2AA22AB22A264bsinBsin2A2sinAcosA2cosA2cos2cosA2cosasinAsinAsinA46b22cosA3即a11aa4a1(a1)a8()1n133n1nn1nn11Sn66()Sn637507．nn3125nn345436220．．．．．a(chǎn)bab，化簡(jiǎn)可得．由余弦定理可得c2，所以22222525ab2ab2ab2abababABC22即411333SC．22422d(ad)．a(chǎn)211222d(2d)10得3分由a212d80d2d4或d22(n1)22na6分n2n18分bn(ab)(ab)L(ab)Sn1122nn(aaLa)(bbLb)12n12n(24L2n)(12L2)n15/8n2n)12n2nn1分n22122Saa2aaaa或0．221得111nnn2Saa且2n1n1n12aaaaa(aaaa)022nnn1nn1nn1nn1因?yàn)閿?shù)列a}aa10aa1即nn1nn1n所以a}annna1分13n1111TL1232n2n221221(L2232)22n211111111L)212233n2n22222111L223111)(n1)nn2122222222212221L)(n1)nn22112232222221)(n1)nn2L21122321312n(3)2nn分22n22122而==；323sinθAC=＜，3323sin(θ)AB3ABC+33333=l2360.036/80.0250.0150.010.005f1(0.0250.01520.010.005)100.34(0.0150.030.0250.005)100.757545f55f65f75f85f95f123456＝0.165758595[70,80)[80,90)[90,100]，，181715143229p363570ax或x與xbax22123a＝，2＝.aaxacbc2即x2.axacbc2當(dāng)ax2當(dāng)axbc.2a()]naq)1n1()])(n)n1S11q11n11a()()n1而an111n1S(1)a分nn11b()bb1,fb分n111bnn1nn17/811{}21bbn1112(nn1b.分bn1nn111(()n1a()1,ca分nn122bnnnn111T12()3()Ln()2n1222n11111T2()3()Ln()23n2212221n11111T