欢迎来到天天文库
浏览记录
ID:58313476
大小:195.52 KB
页数:6页
时间:2020-09-05
《四川省2016届高中毕业班高考适应性考试(“卷中卷”大联考(三))数学(理)答案.pdf》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、四川省高2013级高中毕业班高考适应性考试理科数学参考答案及评分标准一、选择题(每小题5分,共50分)1.D2.D3.A4.B5.C6.C7.A8.C9.C10.D二、填空题(每小题5分,共25分)1211.12.13.k0<k<1或k>914.4441115.[,]7e三、解答题n16.解:(I)∵Sk3mn3Sa3km3ass18k27解得km1133223n3n1n则当n2时,ass333nnn122n又a13nNan3an则有3为常数,故由等比数列的定义可知,数列a是等比数列.······6分nan1n1(II
2、)abloga∴bnn3n1nn3234nn1则T...n23n1n333331234nn1T...n234nn133333322111n152n5则T(...)n23nn1n133333362312n5即T(5)(nN)···································12分nn4317.解:(I)由已知在RtBCD中,BD2,BC1,CD3即有CBD60,又易知ABD60,则ABC120在ABC中,由余弦定理可得:222ACABBC2ABBCcosABC
3、21理科数学参考答案及评分标准第1页共6页∴AC21··········································5分(II)设BDC,ADC90,且CD2cos则在ACD中,由正余弦定理可得222ACADCD2ADCDcos(90)2124cos83coscos(90)2124cos83sincosACAD且43sin(90)sin302124cos83sincos得482cos243即3tan23tan80;又tan>0,解得tan···········12
4、分318.解:(Ⅰ)全班有12个男生,8个女生,所以男、女各选1人的方法数m12896而这两名学生阅读名著本数之和为4的方法数n13417,所以所求概率为n7···························································3分m96(Ⅱ)由已知随机变量X的可能的取值有0,1,2,3,40413CC1CC84444P(X0)P(X1)44C70C3588223140CC18CC8CC1444444P(X2)P(X3)P(X4)444C35C35C70888X的分布列为X01234181881P
5、703535357081881则X的数学期望为Dx12342···············9分3535357022(Ⅲ)S<S·······················································3分1219.解:(I)证明:设AC与AC交于F点,连接EF11E,F分别是线段AB,AC的中点1EF∥BC,又EF平面AEC,BC平面AEC1111故BC∥平面AEC············································4分11理科数学参考答案及评分标准第2页共6页(II)①在正三角形ABC中,过E
6、作EHAC于H,连接AH1显然AC平面AEH,AC平面ACCA111∴平面AEH平面ACCA,且两个平面的交线为AH1111过E作EGAH于G,则EG平面ACCA1113在RtAAB中,由已知易得AE1,在正三角形ABC中,EH112AEEH211则在RtAEH中,EG1AE2EH27121即点E到平面ACCA的距离为117∵E是线段AB中点221∴点B到平面ACCA的距离d2EG·······················8分117②延长EB至R点,使EBBR1,连接RC∴BR∥AE,则BR平面ARC,即有BRRC1111在BRC中易得RC7,
7、BC221设直线BC与平面ACCA所成角为111d42则sin············································12分BC14120.解:(Ⅰ)由椭圆定义可知,曲线E是以(23,0)和(23,0)为焦点,长轴长为46的椭圆,设椭圆的半长轴长、半短轴长、半焦距分别为a、b、c.22a26,c23,则bac2322c2xy椭圆的离心率e
此文档下载收益归作者所有