黑龙江省哈尔滨市2013届高考数学第二次模拟考试试题 理（扫描版）VIP免费

黑龙江省哈尔滨市2013届高考数学第二次模拟考试试题理（扫描版）1234562013年哈尔滨市第三中学第二次高考模拟考试数学试卷（理工类）答案及评分标准一、选择题：题号123456789101112答案BABBBCCDCBBD二、填空题：13.114.215.1,016.3,2三、解答题：17.(Ⅰ)整理得21nnaa………………………………4分7又11a得12nan………………………………6分(Ⅱ)由（1）知)121121(21nnbn……………………………8分所以12nnTn……………………………………12分18.解:(Ⅰ)第六组08.0p···························2分第七组06.0p···························4分估计人数为180··························6分(Ⅱ)X可能的取值为0,1,2,3.························7分425)0(3935CCxP4220)1(392514CCCxP4215)2(391524CCCxP422)3(3934CCxP所以X的分布列·············10分)(XE=34.·····················12分19.(Ⅰ),//CDAB,ADCD22ABCDAD，F分别为CD的中点，ABFD为矩形，BFABX0123P42521101452118·················2分EFDCECDE,,又EFABCDAB,//AEEEFBF,面BEF,AE面ABE,平面ABE⊥平面BEF·····················4分(Ⅱ)EFDCECDE,,又EFPD//，PDABCDAB,//又PDAB,所以AB面PAD,PAAB··················6分法一：建系AB为x轴，AD为y轴，AP为z轴,)0,2,0(),0,0,1(DB),0,0(aP，)0,2,2(C,)2,1,1(aE平面BCD法向量1(0,0,1)n�，平面EBD法向量)2,,2(2aan··········9分]22,21[452cos2a，可得]5152,552[a.·············12分法二：连AC交BF于点K,四边形ABCF为平行四边形，所以K为AC的中点，连EK,则PAEK//,EK面ABCD,EKBD,作BDKH于H点，所以BD面EKH,连EH,则EHBD,EHK即为所求·············9分在EHKRt中，515221HK,]3,1[25512tanaa9解得]5152,552[a·············12分20.(Ⅰ)由已知21143322222accbaba解得42a，32b，方程为13422yx·······3分(Ⅱ)设),(),,(2211yxByxA，则)3,2(),3,2(2211yxQyxP（1）当直线l的斜率存在时，设方程为mkxy13422yxmkxy联立得：0)3(48)43(222mkmxxk有22212212243)3(44380)43(48kmxxkkmxxmk①由以PQ为直径的圆经过坐标原点O可得：0432121yyxx·整理得：04)(4)43(221212mxxkmxxk②将①式代入②式得：22243mk,···········6分048,0,043222mmk又点O到直线mkxy的距离21kmd2222222221223414334143433411mmkkmkkmkkxxkAB··········8分10所以32322122mmdABSOAB··········10分(2)当直线l的斜率不存在时，设方程为mx（22m）联立椭圆方程得：4)4(322my代入0432121yyxx得到04)4(3322mm即552m，5152y3212121yymdABSOAB综上：OAB的面积是定值3又ODE的面积33221，所以二者相等.·······12分21.(Ⅰ)由原式bxxxln11，················1分令xxxxgln11)(，可得)(xg在1,0上递减，在,1上递增，所以0)1()(mingxg即0b···············3分（Ⅱ）)0(,ln2)(xxaxxfxxaxfln2,0)(得令，xxxhln)(设，时当exexh1)(maxea21当时，函数)(xf在),0(单调递增···············5分ea210若，xaxgxxaxxg12)(),0(,ln2)('axxg21,0)('，0)(),,21(,0)(),21,0(//xgaxxgaxax21时取得极小值即最小值时而当ea210021ln1)21(aag，必有根0)(/xf，)(xf必有极值，在定义...