湖南省长沙市高三高考模拟题二理科数学试题及答案.doc

2014年湖南省长沙市高考模拟题（二）理科数学一、选择题 1. 已知复数满足（为虚数单位），则z的值为 ( A )Ai B. i C. 1 D. 12. 设随机变量XN(2，32)，若P(Xc)P(Xc)，则c等于 ( C )A0 B1 C2 D33. 二项式的展开式中常数项为 （ B ）A15 B15 C20 D20 4. 设A ，B为两个互不相同的集合，命题P：， 命题q：或，则是的（ B ）A. 充分且必要条件 B. 充分非必要条件 C. 必要非充分条件 D. 非充分且非必要条件5. 已知集合， 若，使得成立，则实数的取值范围是 （ B ）A. B. xABPyOC. D. 6. 函数的部分图象如图所示，设是图象的最高点，是图象与轴的交点，若，则的值为 （ C ）A B C D7. 设变量x，y满足约束条件，则zx3y的最大值为( B ) A B C D8. 如图，正方形的边长为3，为的中点，与相交于，则的值是（ C ）A B C D9 若两条异面直线所成的角为，则称这对异面直线为“黄金异面直线对”，在连接正方体各顶点的所有直线中，“黄金异面直线对”共有 ( C ) A12对 B18对 C24 对 D30对10. 已知函数在区间内任取两个实数，且，不等式恒成立，则实数的取值范围为 （ A ）A B C D【解析】，表示点与点连线的斜率，因为，所以，即函数图象在区间内任意两点连线的斜率大于1，即在内恒成立.由定义域可知，所以，即，所以成立.设，则，当时，函数的最大值为15，所以.故选A.11.如图，是圆的切线，切点为，交圆于两点， ，则=_.由切割线定理知，所以，连接OA，在中求得，所以，故答案是：.12. （选修43：不等式证明）不等式有实数解的充要条件是【解析】13.（选修44：坐标系与参数方程）已知直角坐标系中，直线l的参数方程为. 以直角坐标系xOy中的原点O为极点，x轴的非负半轴为极轴，圆C的极坐标方程为，则圆心C到直线l距离为.14. 设点是双曲线与圆在第一象限的交点，其中分别是双曲线的左、右焦点，且，则双曲线的离心率为.15. 【解析】由，可以推得，当时，当时，所以输出.16. 若三个非零且互不相等的实数、满足，则称、是调和的；若满足，则称、是等差的. 若集合中元素、既是调和的，又是等差的，则称集合为“好集”.若集合，集合.则（1）“好集”中的元素最大值为 2012 ；（2）“好集”的个数为 1006 .【解析】若、既是调和的，又是等差的，则，.即“好集”为形如（）的集合.（1）“好集”中的元素最大值为，又因为，所以最大值为2012.（2）由“好集”是集合的三元子集知，且. ，且.符合条件的可取1006个值. “好集”的个数为1006.三、解答题: 17. 已知函数 .（）求函数f (x)的最小正周期；（）在ABC中，角A，B，C的对边分别是，且满足，求的取值范围【解析】（）f (x)=1+sin2x+(1cos2x) =1+2sin f (x)的最小正周期为 （）由可得，即， ，得， 所以 ， 故，从而2sin， 因此f (x)的值域为. 12分18. 在如图所示的几何体中，平面，是的中点，（）证明：平面；（）求二面角的大小的余弦值【解析】（）因为，所以平面故以为原点，建立如图所示的空间直角坐标系，则相关各点的坐标分别是， 所以，因为平面的一个法向量为，所以，又因为平面，所以平面6分（）由（）知，设是平面的一个法向量，由 得即取，则 8分设是平面的一个法向量，由 得即取，则 10分设二面角的大小为，则故二面角的大小的余弦是 12分19. （本小题满分12分）某联欢晚会举行抽奖活动，举办方设置了甲、乙两种抽奖方案，方案甲的中奖率为，中奖可以获得2分；方案乙的中奖率为，中奖可以获得3分；未中奖则不得分每人有且只有一次抽奖机会，每次抽奖中奖与否互不影响，晚会结束后凭分数兑换奖品（）张三选择方案甲抽奖，李四选择方案乙抽奖，记他们的累计得分为X，若X3的概率为，求；（）若张三、李四两人都选择方案甲或都选择方案乙进行抽奖，问：他们选择何种方案抽奖，累计得分的数学期望较大？【解析】（）由已知得，张三中奖的概率为，李四中奖的概率为，且两人中奖与否互不影响记“这2人的累计得分X3”的事件为A，则事件A的对立事件为“X5”，因为P(X5)，所以P(A)1P(X5)1=，所以.6分（）设张三、李四都选择方案甲抽奖中奖次数为X1，都选择方案乙抽奖中奖次数为X2，则这两人选择方案甲抽奖累计得分的数学期望为E(2X1)，选择方案乙抽奖累计得分的数学期望为E(3X2)由已知可得，X1B，X2B，所以E(X1)2，E(X2)2，从而E(2X1)2E(X1)，E(3X2)3E(X2)6.若E(2X1)E(3X2)，则.若E(2X1)E(3X2)，则.若E(2X1)=E(3X2)，则.综上所述，当时，他们都选择方案甲进行抽奖时，累计得分的数学期望较大；当时，他们都选择方案乙进行抽奖时，累计得分的数学期望较大；当时，他们都选择方案甲或都选择方案乙进行抽奖时，累计得分的数学期望相等12分20 【解析】（）由题设：投放的药剂质量为，渔场的水质达到有效净化 或 或，即：，所以如果投放的药剂质量为，自来水达到有效净化一共可持续8天.6分（）由题设：， ，且， 且， ，投放的药剂质量的取值范围为. 13分21【解析】（）设，（）.易知斜率存在，设为，则方程为.由得， 由直线与抛物线相切，知.于是，方程为.同理，方程为.联立、方程可得点坐标为 ， ，方程为，过抛物线的焦点. ，. ，点在定直线上. 或解：设，则方程为，方程为. 点，坐标满足方程. 直线方程为.由直线过点，知. .点在定直线上. 6分（）由（）知，、的坐标分别为、. . . 设（），由知，当且仅当时等号成立. .设，则. 时，；时，.在区间上为减函数；在区间上为增函数. 时，取最小值. 当，即，时，面积取最小值. 13分22. 【解析】（），当时，由知或， 当时，则，时，在上单调递减，所以当时，时，时，在处取得最大值，即综上所述， （） 4分（）当时，欲证 ，只需证明 ，所以，.所以，当时，都有成立 （III）当时，结论显然成立；当时，由（II）知所以，对任意正整数，都有成立 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