高三双基数学(文科)参考答案_高三数学文科卷及答案

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2014年大连市高三双基考试

数学（文科）参考答案及评分标准

说明：

一、本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则．

二、对解答题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．

三、解答右端所注分数，表示考生正确做到这一步应得的累加分数．

四、只给整数分数，选择题和填空题不给中间分．

一．选择题

1．B；2．A ；3．C ；4．D；5．A；6．B ；7．D； 8．D；9．A；10．A ；11．C；12．D．

二．填空题

4,n1513． yx； 14． ；15．；

16.． 2n1,n23

三．解答题

17．解:(Ⅰ)f(x)cosx(sinxx)sinxcosx

2sin2x 2sin(2x

当2x3).············································································································· 4分 2

32k

2（kZ)，即x{x|xk35.············································· 6分 ,kZ}时，f(x)取最大值1212

(Ⅱ)f()A

2,可得sin(A)0，因为A为△ABC内角，所以A.········· 8分 23

3由余弦定理abc2bccosAbcbc，由a3,bc2，解得bc1.··················································································· 10分

22222

所以SABC

1.··························································································· 12分 bcsinA

418.解：（Ⅰ）22列联表如下：

1000(400200100300)2

47.619 ····································································· 4分

500500700300

∵47.61910.828，∴有99.9%的把握认为“生产的零件是否为优质品与分厂有关”.··································································································· 6分（Ⅱ）现用分层抽样方法（按优质品和非优质品分二层）从乙厂抽取五件零件，从乙厂抽取优质品3件，记为A,B,C，非优质品2件，记为a,b.从这五件零件中任意取出两件，基本事件空间{AB,AC,Aa,Ab,BC,Ba,Bb,Ca,Cb,ab} ··········································· 8分用A表示“至少有一件非优质品”这一事件，则A{Aa,Ab,Ba,Bb,Ca,Cb,ab}.···· 10分

7.·························································································································· 12分 10

19.解：(Ⅰ)取BC中点O，因为三角形ABC是等边三角形，所以AOBC，又因为面BCC'B'底面ABC，AO面ABC，面BCC'B'面ABC=BC，所以

AO面BCC'B'，又BB'面BCC'B'，所以AOBB'.又BB'AC，AOACA，AO面ABC，AC面ABC，P(A)

所以BB'底面ABC.·········································································································· 6分(Ⅱ)显然M不是A',B'，棱A'B'上若存在一点M，使得C'M//面BEF，过M作

MN//AA'交BE于N，连接FN，所以MN//CF，即C'M和FN共面.所以C'M//FN，所以四边形C'MNF为平行四边形，所以MN2，所以MN 是梯形 A'B'BE的中位线，M为A'B'的中点.······························································ 12分

uuurr

20.解：（I）设点P(x0,y0)，M(x,y)，则Q(x

0,0)，则由QP，得0

xx0),y0y

0)，即xx0,y0，··············································· 3分

因为x0y020，所以x2y20.············································································ 5分

(II)将曲线C与直线l联立:，消y得：

直线l与曲线C交于A、B两点,设A(x1,y1),B(x2,y2)

又,························································································································ 7分



4m2m220x1x2,x1x2 ·························································································· 8分

点O到直线AB:的距离,AB

1x2

····················································································· 10分

.当且仅当m30m，即m15时取等号.所以，三角形OAB面积的最大值为21．解：(Ⅰ)f'(x)

.············································································ 12分

1lnx，f'(x)0解得xe.··················································· 2分 2

f'(x)0解得0xe，f(x)在(0,e)上为增函数，f'(x)0解得xe，f(x)在e,上为减函数.所以f(x)在xe取极大值

1.····························································································· 5分 e

（Ⅱ）f(x)k(x)2等价于lnxkx2x3k0，设函数g(x)lnxkx2x3k(x1)

12kx22x1

g'(x)2kx2 ·········································································· 7分

由题意知 g(1)0,即k当k

.······················································································ 8分 2

时，设h(x)2kx2x1，2

其开口向上，对称轴x1，2k

·········································· 10分 h(1)2k10，所以h(x)0在x[1,)上恒成立.·所以g'(x)0在x[1,)上恒成立，即g(x)在x[1,)上为增函数，所以g(x)g(1)0.所以实数k的取值范围为(,].··················································································· 12分

22.证明：（Ⅰ）连接OG，∵EF为eO的切线，∴OGEF,∴OGAKGE90,∵CDAB，∴OAGHKA90，∵OAOG，∴OGAOAG, ∴KGEHKAGKE，∴KEGE.····································································· 5分

KGGEKE，

KDKGKG

∵DKGGKE，∴△KDG∽△KGE

∴AGDE，又∵AGDACD，∴ACDE.（Ⅱ）连接DG，∵KGKDgGE，∴

∴ACPEF.·························································································································· 10分 23.解：（I）圆C1的普通方程为：(x4)y16，则C1的极坐标方程为：8cos 圆C2的普通方程为：x(y2)4，则C2的极坐标方程为：4sin ·················· 5分(II)设P(,)，则有8co，解得tan

2，s4sin

sin,arcsin，所以P

点的极坐标为（555

··············································································································································· 10分

5131

x2x2

24.解：（I）原不等式等价于2或或 222

x11x353

x2

22

x3

解得原不等式解集为(,)U(3,)················································································ 5分