广东省各地2015届高三上学期期中考试文科数学试题汇总(二)9份 Word版含答案

D．2【答案解析】解析：因为AEADAB,AFABAD,所以xy02,2)(Ⅱ)12,2)32(sincos)322sin()421Sh1222424212x+8y6x+6y6x+10y3x+2y3x+5y（Ⅱ）求证：2a1a2an4（Ⅱ）21(11)an4n24n14(n2n)4nn1a12a2an423(11)]1(11)1nn1n14x解析：因为集合Mxx21，即Mxx1或1，又因为A.2B.222则1117244428a2b2A．22C．31D．2+1a2bec121.故选：D．AB?DFBE?ADBEDF=2创2cos120?AB?mABl鬃AD+lADmAB(1-l)BC?(1m)DC=(1-l)AD?(112在B角BC1acsinB=1创a42?2由余弦定理得b2=a2+c2-2accosB=12+42()-2创1x1cos,PBAB,AB2PBBC9436ABBCxxR24，p,4kp+5p,(k2kp?xp?2kpp(kÎZ)245p(kÎZ)=2cos2q-sin2q-2sinqcosq（3）过点B作BH^NC于H，则BH=BC2-CH2=22-12=312+4?333四边形BCDEa1a2a2a311anan12.a1a2a2a31111...anan11335571111111...11......................................11分2335571111...........................................................13分22n12a1a2a2a311anan12..................................14由=2m()m+4m+4则y2-y1=y1=2m+3=43=3．aaRe()()f?(x)1-ax，()()()(()()∴fx在-()(()()()（3）当a=1时，fx=x-1+x，令gx=fx-kx-1=1-kx+x，()()()()(()()()()()(求其极值；3）gx=fx-kx-1=1-kx+x，()()()(()A.x=B.x=C.x=-D.x=-C．3B．3D．3a2b2D．y37xD.2①函数fx=2x211.6512.4314.2215.3502423252510csinAsinC，即2102，400201241432362161015，234...n13031321S234...n13n3132332S2111...1n13n303132331113n223n，4431n31n132n32n5∴b2431243114324343443y1x0k2，∴MNkkk4k12k4k143，„„„9分1x21(x0)ex(xa2b212812422DDDB2Sna1n2n2nN*a1a2an41xlnxa231B2B3A4C5B6C7C8B9A10.D面ABCDEF1BD13，B1FBF2BB12(2)2226B1EB1D12D1E212(22)231EFBF32323，2a4（2）n2时，2Snnan1n3n2nan1an1a2a11n1n2111121223a1a2223111111717n1n42n4n4a1a2111212231212121213111(11111111)323n2nn1n111(1111)71(11)7nn142nn14x1(xx)yx1xx12x1xyx2xyx1xyax10x[1,)n2时，令xn，即lnn1n1nlnnlnnlnn1ln3ln2111nn111即n2时，lnn2332x=-px=-pC．3B．3D．3a2b2D．2①函数fx=2x211.6512.4314.2215.3502423252510c，即2sinAsinC72，400201241S234...n13n3132332S2111...1n13n303132331113n223n，4431n31n13∴b244343431243114324343443y1x0k2，∴MNkkk4k12k4k143，„„„9分1(x0)ex(x25B.yy12122，得T，10ABBCsinCsinAAB22，ABC22r5，且OCr3，所以OEOC2CE232522a(3sinx)2(sinx)22sin2xb(cosx)2(sinx)21262，26661501005501501006.93【答案解析】(1)见解析；(2)21BABMsin1201633933ABMa11a2SS2，n112112∴sn是等差数列，其中首项为s1a1sns1(n1nN),(n2nN)2n(n1)41a2b2PF(x12)2y12(x12)22x122x1QF22x2MF(12)2(6)2222(22)42(x1x2)x22y24x2x22(y2y2)0kPQy1y21P(1,6)Q(1,6)P(1,6)Q(1,6)xx4kb1212k2xx2b24设点P(x1,y1),Q(x2,y2)，则：1212k2|PF|22x1|QF|22x2e2,a2,xM1|MF|2221,21x2aa1a1aa1(a,)aalnaa2aa1a21aa1a1fxf11a1a1f1a1a21,21A．5435)b(1,cos)OP若Dsin211cos2OP（Ⅱ）f(x)2sin2xay2sinxa2所以VPBCDESBCDEh1，VQABCDSABCDh2,CPh18CQh23.2，得2x33x210„„10分①当k2时，f(x)xk2kk24xkk24.kk24kk24kk24kk24k2时，0x1kk4kk24f(2k)ln(2k)4k2k2ln(2k)0，又1x2kk4k，k2(k2)1x1,cos2ysinx1，x(0,)44(x>2)232an11;b23loga,(nN*)a6=1;a6或a04an