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2011年高考模拟试题(7)(文理合卷) 一.选择题(本大题共12题,每小题5分,共60分)1.(理)复数的值是A. B. C. D.(文)的值是A. w.w.w.k.s.5 u.c.o.m B. C. D.2.已知集合,则集合=A. w.w.w.k.s.5 u.c.o.mB.C.D.3.若,下列命题中正确的是A.若, 则 B.若, 则C.若, 则 w.w.w.k.s.5 u.c.o.m D.若 , 则4.已知函数的反函数是,则是A.奇函数且在上单调递减 B.偶函数且在上单调递增C.奇函数且在上单调递减 D.偶函数且在上单调递增5.已知实数、满足约束条件,则的最大值 A.24 w.w.w.k.s.5 u.c.o.m B.20 C.16 D.126.设向量,向量,则与的夹角是A. B. C. D. 7.若半径是的球与正三棱柱的各个面都相切,则球与正三棱柱的体积比是A. B. C. D. O8.若函数()的部分图象如图所示,则有A. B. C. D. 9.定义在上的偶函数对于任意的都有,且,则的值为A. B. C. D.10. 在边长为1的等边中,设( ) A B0 C D311. 在等差数列an中,其前n项和为Sn ,若a2,a10是方程x2+12x-8=0的两个根,那么S11的值为( ) A.44 B.-44 C.66 D.-6612. 设,二次函数的图像为下列之一,则的值为( )A1B1CD二.填空题(本大题共4小题,每小题4分,共16分)13.在的展开式中,的系数是15,则实数_.14.(理)已知,则_.(文)函数的单调递减区间是_.15.编号为1、2、3、4、5的五个人分别去坐编号为1、2、3、4、5的五个座位,其中有且只有两个人的编号与座位号一致的坐法有_种.16. 已知抛物线的准线与双曲线的左准线重合,则抛物线的焦点坐标为 .三.解答题(本大题共6小题,共76分)17.(本小题满分12分) 已知(为常数).(1)求的单调递增区间;(2)若在上的最大值与最小值之和为3,求的值.18.(本小题满分12分) 在举办的奥运知识有奖问答比赛中,甲、乙、丙同时回答一道有关奥运知识的问题,已知甲回答对这道题目的概率是,甲、丙两人都回答错的概率是,乙、丙两人都回答对的概率是.(1)求乙、丙两人各自回答对这道题目的概率.(2)(理)求回答对这道题目的人数的随机变量的分布列和期望.(文)求甲、乙、丙三人中至少有两人回答对这道题目的概率.19.(本小题满分12分)在三棱柱中, ,是的中点,F是上一点,且.(1) 求证:;(2) 求平面与平面所成角的正弦值.ABCDFA1B1C120(本小题满分12分)=, =,(1)求证:为等差数列; (2) 若,问是否存在, 对于任意(),不等式成立.21. (本小题满分12分)如图,已知点,直线:, 为平面上的动点,过作直线的垂线,垂足为点,且。()求动点的轨迹的方程;()过点的直线交轨迹于,22(本题满分14分)(理)(本小题满分12分)已知. (1) 如果函数的单调递减区间为,求函数的解析式; (2)若的导函数为,对任意,不等式恒成恒成立,求实数的取值范围.(文(本小题满分12分)已知. (1) 如果函数的单调递减区间为,求函数的解析式; (2)若的导函数为,对任意,不等式恒成立,求实数的取值范围.2011年高考模拟试题(6)(文理合卷) 参考解答一.选择题1.(理) B(文)A2. C3. B4. D 5. B 6. C7. B8.C9. A10.A11.D12. B二.填空题14.(理)1(文)15.2016. 三.解答题(本大题共6小题,共76分)17.解:(1),即 , 的单调递增区间是 6分(2),则 , . 12分18.(本小题满分12分) 解:(1)设乙、丙各自回答对的概率分别是,根据题意,得 解得 ,; 6分(2)(理)可能取值0,1,2,3,; ; .分布列如下:0123期望为 . 12分(文) 12分19.(本小题满分12分)ABCDFA1B1C1解:(1)因为,是的中点,所以.又,所以 又,所以 在中 ,在中 所以 , 即,所以。 6分(2)延长交于,则为所求二面角的棱.由得:。过作,且与交于,又 ,为所求二面角的平面角. 由和相似得:。又,所以 。即所求二面角的正弦值是. 12分20(本小题满分12分)解:(1) 为等差数列 6分(2) 12分21. (本小题满分12分)解:()设点,则,由得:,化简得 ()设直线的方程为:.设,又,联立方程组,消去得:,故由,得:,整理得:,。22(本题满分14分)(1).2分由题意 的解集是,即 的两根分别是.4分将代入方程得. .6分(2) 由题意知 恒成立,即恒成立.8分设,则.令,得.当时,;当时,10分当时,取得最大值,.因此的取值范围是.12分(文)(1).2分由题意 的解集是,即 的两根分别是.4分将代入方程得. .6分(2) 由题意知 时恒成立,即,所以.由于,于是,得 9分而,所以为所求. 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