數(shù)學(xué)參考答案一、單選題:log2a3+log2a11=log2a3a11=4,選C.取第五位數(shù)據(jù)2.21，選B.4.解：∵l的方向向量為=(2,1)，則l斜率為，因為直線與l垂直，所以斜率為－2又∵過A點，所以直線為2x＋y－1＝0.選D.5.解：以AB,AD,AA1為基底，則DEQ\*jc3\*hps23\o\al(\s\up2(–),C)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),C)EQ\*jc3\*hps23\o\al(\s\up2(–),1)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),1)2-4x-3>0開口向上有解，:f(x)=f(2-x)→f(x)關(guān)于x=1對稱.:f(x)關(guān)于x=1對稱.→f(x)在x<1為減函數(shù)二、多選題通項公式為Tk+1=CEQ\*jc3\*hps16\o\al(\s\up4(k),6)26k(1)kx62k,所以6-2k=0,k=3則第四項為常數(shù)項，B錯二項式系數(shù)最大項為中間項第四項，所以為CEQ\*jc3\*hps16\o\al(\s\up5(3),6)=20，C對選ACD.10.解：遞推公式取倒得構(gòu)造新數(shù)列得得到為公比為3，首項為3的等比數(shù)列→→則A對，C錯.選ABD.22三、填空題)13.解：正四面體內(nèi)半徑最大的球為內(nèi)切球兀R2=2兀N*kk-1-k(1-μ)k-1=k[μk-1-(1-μ)k-1]所以即則恒成立，只需2λ2-tλ+2大于的最大值。可得bk+1<bk,則bk的最大值為b1=1λλ|(λ2,λλ|(λ2,四、解答題:b2=(a+c)2-ac:b2=又據(jù)余弦定理b2=a2+c2-2accosB得，········································································2分:B=2EQ\*jc3\*hps22\o\al(\s\up6(兀),3)·······························································3分2-ac·····························································4分分:CΔABC≤3+2解得a=c=2······················································································11分:BD丄AC,又AD=:據(jù)勾股定理得BD=1······································································13分分(2)由題可知X=0,20,40,60·········································································5分P······································································6分分甲總得分的分布列：X0204060P1 4············································································10分分所以甲獲勝概率更大···············································································15分:E、F、N三點分別為PB、AO、PA的中點:在平面PAO中，NF//PO,又:NF丈平面POC，PO平面POC:NF//平面POC 2分同理，EN//平面POC·····································································4分又:NF∩EN=N，NF平面ENF，EN平面ENF，所以平面ENF//平面POC:EF平面EFN:EF//平面POC··························································6分建立以BG,BC,BQ分別為x,y,z的空間直角坐標(biāo)系.A,C,P假設(shè)在PC上存在點M使得OM丄AB,設(shè)M(x,y,z)EQ\*jc3\*hps15\o\al(\s\up4(→),CM)EQ\*jc3\*hps15\o\al(\s\up4(→),CP)→Mλ,2-3λ,2λEQ\*jc3\*hps17\o\al(\s\up0(→),BA)EQ\*jc3\*hps17\o\al(\s\up4(→),OM)λ-1,-3λ,2λ)5EQ\*jc3\*hps15\o\al(\s\up4(→),BA)EQ\*jc3\*hps15\o\al(\s\up4(→),OM)5EQ\*jc3\*hps18\o\al(\s\up12(→),BM)EQ\*jc3\*hps18\o\al(\s\up12(→),n)設(shè)直線BM與平面PAO所成角為θ,則分可得橢圓方程：1···································································3分（2）解:(ⅰ)設(shè)直線PF的方程為：x=my+,點Q(x1,y1),R(x2,y2)，則x2(ⅱ)①當(dāng)直線PF斜率為0時，不妨設(shè)R(-2,0),Q(2,0),P(t,0)②當(dāng)直線PF斜率不為0時，設(shè)P(x0,y0)，由(ⅰ)得-2··························14分所以點P在定直線l:x=上，MF平行直線l，點P到直線MF的距離 綜上可知，△PFM的面積為·······························································17分·················································································2分f,,(x)<0；所以f(x)為(0，+∞)的凸函數(shù)··············································3分所以f在為“凹函數(shù)”···················6分2n時，等號成立:h(x)最小值為····································