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1、2013 年长春市高中毕业班第一次调研测试数学试题卷(理科)第卷（选择题，共60 分）一、选择题（本大题包括12 小题，每小题5分，共 60 分，每小题给出的四个选项中，只有一项是符合题目要求的，请将正确选项填写在答题纸上）1.已知集合2|20Ax xx，|ln(1|)Bx yx，则()ABIReA.(1,2)B.1,2)C.(1,1)D.(1,22.已知复数1zai()aR（i是虚数单位），3455ziz，则aA.2B.2C.2D.123.如图的程序框图，如果输入三个实数a，b，c，要求输出这三个数中最大的数，那么在空白的判断框中，应该填入下面四个选项中的A.cx?B.xc?C.cb?D.b
2、c?4.一个几何体的三视图如图所示，则这个几何体的体积为A.(8)36B.(82)36C.(6)36D.(92)365.设1130axdx，11201bx dx，130cx dx，则a、b、c的大小关系为A.abcB.bacC.acbD.bca6.在正项等比数列na中，已知1234a a a，45612a a a，11324nnnaa a，则nA.11 B.12 C.14 D.16 7.直线1l与2l相交于点A，动点B、C分别在直线1l与2l上且异于点A，若ABuuu r与ACuuu r的夹角为60o，2 3BCuuu r，则ABC的外接圆的面积为A.2B.4C.8D.128.给定命题p：函数
3、sin(2)4yx和函数3cos(2)4yx的图像关于原点对称；命题q：当2xk()kZ时，函数2(sin 2cos2)yxx取得极小值.下列说法正确的是A.pq是假命题B.pq是假命题xa开始输入abc，?bxxbxc输出x结束是是否否3122正视图侧视图俯视图C.pq是真命题D.pq是真命题9.若两个正实数,x y满足211xy，并且222xymm恒成立，则实数m的取值范围是A.(,24,)UB.(,42,)UC.(2,4)D.(4,2)10.已知直线0 xyk(0)k与圆224xy交于不同的两点A、B，O是坐标原点，且有3|3OAOBABuuu ruuu ru uu r，那么k的取值范围
4、是A.(3,)B.2,)C.2,2 2)D.3,22)11.如图，等腰梯形ABCD中，/ABCD且2ABAD，设DAB，(0,)2，以A、B为焦点，且过点D的双曲线的离心率为1e；以C、D为焦点，且过点A的椭圆的离心率为2e，则A.当增大时，1e增大，12e e为定值B.当增大时，1e减小，12e e为定值C.当增大时，1e增大，12e e增大D.当增大时，1e减小，12e e减小12.对于非空实数集A，记*|,AyxA yx.设非空实数集合M、P满足：MP，且若1x，则xP.现给出以下命题：对于任意给定符合题设条件的集合M、P，必有*PM；对于任意给定符合题设条件的集合M、P，必有*MPI；
5、对于任意给定符合题设条件的集合M、P，必有*MPI；对于任意给定符合题设条件的集合M、P，必存在常数a，使得对任意的*bM，恒有*abP，其中正确的命题是A.B.C.D.第卷（非选择题，共90 分）本卷包括必考题和选考题两部分，第 13 题 21 题为必考题，每个试题考生都必须作答，第 22 题24 题为选考题，考生根据要求作答.二、填空题(本大题包括 4 小题，每小题 5 分，共 20 分，把正确答案填在答题纸中的横线上).13.若实数,x y满足11211xyxyx，则1yx的取值范围是 _.14.ABC中，a、b、c分别是角A、B、C的对边，若222acb，且sin6cossinBAC，
6、则b的值为 _.ABDC文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8
7、I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5
8、文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:C
9、H3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9
10、V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10
11、HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9
12、G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y5文档编码:CH3T6M9V1A10 HA10R9G1S1F1 ZD6N8I2P6Y515.若一个正四面体的表面积为1S，其内切球的表面积为2S，则12SS_.16.定义在R上的函数()f x满足()(5)16f xf x，当(1
13、,4x时，2()2xf xx，则函数()fx在0,2013上的零点个数是_.三、解答题（本大题包括6 小题，共 70 分，解答应写出文字说明，证明过程或演算步骤）.17.（本小题满分12 分）函数()sin()(0,0,)()22f xAxAxR的部分图像如图所示.求函数()yf x 的解析式；当,6x时，求()f x 的取值范围.18.（本小题满分12 分）数列na的前n项和是nS，且112nnSa.求数列na的通项公式；记23log4nnab，数列21nnbb的前n项和为nT，证明：316nT.19.（本小题满分12 分）如图，在三棱柱111ABCA B C中，侧面11AAC C底面ABC
14、，112AAACAC，ABBC，ABBC，O为AC中点 证明：1AO平面ABC；求直线1AC与平面1A AB所成角的正弦值；在1BC上是否存在一点E，使得/OE平面1A AB?若存在，确定点E的位置；若不存在，说明理由.20.（本小题满分12 分）已知椭圆 C:22221(0)xyabab的离心率为22，其左、右焦点分别为1F、2F，点P是坐标平面内一点，且7|2OPuuu r，1234PFPFu uu r u uu u r，其中O为坐标原点.求椭圆 C的方程；如图，过点1(0,)3S，且斜率为k的动直线l交椭圆于A、B两点，在y轴上是否存在定点M，使以AB为直径的圆恒过这个点？若存在，求出点
15、M的坐标；若不存在，请说明理由21.（本小题满分12 分）已知函数2()(22)xf xeaxx，aR且0a.Oxy6132xyASOBOCBAC1B1A1文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B1
16、0M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3
17、HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2
18、U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2
19、 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2
20、H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7
21、文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7 若曲线()
22、yf x在点(2,(2)Pf处的切线垂直于y轴，求实数a的值；当0a时，求函数(|sin|)fx的最小值；在的条件下，若ykx与()yf x的图像存在三个交点，求k的取值范围.请考生在 22、23、24 三题中任选一题做答，如果多做，则按所做的第一题记分.22.（本小题满分10 分）选修41：几何证明选讲.如图，已知 O 和M 相交于 A、B 两点，AD 为M的直径，直线 BD交O 于点 C，点 G 为?BD中点，连结 AG 分别交 O、BD 于点 E、F，连结 CE 求证：GDCEEFAG；求证：.22CEEFAGGF23.（本小题满分10 分）选修 44：坐标系与参数方程选讲.已知曲线C的
23、极坐标方程为4cos，以极点为原点，极轴为x轴正半轴建立平面直角坐标系，设直线l的参数方程为35212xtyt（t为参数）.求曲线C的直角坐标方程与直线l的普通方程；设曲线C与直线l相交于P、Q两点，以PQ为一条边作曲线C 的内接矩形，求该矩形的面积.24.（本小题满分10 分）选修45：不等式选讲.设函数()|1|2|f xxxa.当5a时，求函数)(xf的定义域；若函数)(xf的定义域为R，试求a的取值范围.2013 年长春市高中毕业班第一次调研测试数学（理科）试题参考答案及评分标准一、选择题（本大题共12 小题，每小题5 分，共 60 分）1.B 2.B 3.A 4.A 5.A 6.C
24、7.B 8.B 9.D 10.C 11.B 12.C 简答与提示：1.B由220 xx可 得12x，又ln(1|)yx中1|0 x，则1|x即11x，则|11 Bx xx或Re-，因此()1,2)ABIRe，故选 B.2.B 由题意可知：2222221(1)1212341(1)(1)11155aiaiaiaaaiiaiaiaiaaa，因此221315aa，化简得225533aa，24a则2a，由22415aa可知0a，仅有2a满足，故选B.3.A由于要取a，b，c中最大项，输出的x应当是a，b，c中的最大者，所以应填AOMGFEDBC文档编码:CM6V1B10M9G3 HE10R2U9S8T2
25、 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2
26、H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7
27、文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:C
28、M6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B1
29、0M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3
30、HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2
31、U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7比较x与c大小的语句cx，故选 A.4.A 该几何体由底半径为1 的半圆锥与底面为边长等于2 正方形的四棱锥组成，且高都为3，因此该几何体体积为28311134 313223323636V，故选 A.5.A由题意可计算得1111
32、1231330003312213xaxdxx；131212002111103332xbx dx；141300144xcx dx，综上abc，故选 A.6.C由3312314a a aa q与312456112a a aa q可得93q，333111324nnnnaaaaq，因此36436813nqq，所以14n，故选 C.7.B由题意ABC中60BAC，2 3BC，由正弦定理可知232sin32BCRA，由此2R，24SR，故选 B.8.Bp命题中3cos(2)cos(2)cos(2)44224yxxxsin(2)4x与sin24yx关于原点对称，故p为真命题；q命题中2 sin 2cos22
33、sin24yxxx取极小值时，2242xk，则38xk()kZ，故q为假命题，则pq为假命题，故选B.9.D2142(2)228yxxyxyxyxy，当且仅当4yxxy，即224yx时等号成立.由222xymm恒成立，则228mm，2280mm，解得42m，故选 D.10.C当3|3OAOBABuuu ruu u ru uu r时，O，A，B三点为等腰三角形的三个顶点，其中OAOB，120AOBo，从而圆心O到直线0 xyk(0)k的距离为1，此时2k；当2k时3|3OAOBABuu u ruuu ru uu r，又直xyAOB文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ
34、2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G
35、2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编
36、码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V
37、1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9
38、G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE1
39、0R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S
40、8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7线与圆224xy存在两交点，故22k，综上k的取值范围为2,2 2，故选C.11.B由题可知：双曲线离心率1|ABeDBDA与椭圆离心率2|CDeBDBC设|ADBCt则|2ABt，|22 cosCDtt，|54cosBDt，1254
41、cos1e，222cos54cos1e，0,2时，当增大，cos减小，导致1e减小.12222cos154cos154cos1e e.故选 B.12.C对于，假设1|02MPxx，则*1|2Mx x，则*MPI，因此错误；对于，假设1|02MPxx，则12M，又*12P，则*MPI，因此也错误，而和都是正确的，故选C.二、填空题（本大题共4 小题，每小题5 分，共 20 分）13.1,514.3 15.6 316.604简答与提示：13.由题可知1(1)0yyxx，即为求区域内的点与(0,1)点连线斜率k的取值范围，由图可知1,5k.14.由正弦定理与余弦定理可知，sin6cossinBAC可
42、化为22262bcabcbc，化简可得22223()bbca，又222acb且0b，可计算得3b.15.设正四面体棱长为a，则正四面体表面积为2213434Saa，其内切球半径为正四面体高的14，即1664312raa，因此内切球表面积为22246aSr，则212236 36SaSa.16.由()(5)16f xf x，可知(5)()16f xf x，则(5)(5)0f xf x，所xyO文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE
43、10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9
44、S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 Z
45、Q2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4
46、G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档
47、编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6
48、V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M
49、9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7文档编码:CM6V1B10M9G3 HE10R2U9S8T2 ZQ2W2H4G2G7以()f x是以 10 为周期的周期函数.在一个周期(1,9上，函数2()2xf xx在(1,4x区间内有3 个零点，在(4,9x区间内无零点，故()f x在一个周期上仅有3 个零点，由于区间(3,2013中包含 201 个周期，又0,3x时也存在一个零点2x，故()f x在0,2013上的零点个数为3 201 1604.三、解答题（本大题必做题5 小题，三选一中任选1 小
50、题，共70 分）17.(本小题满分12 分)【命题意图】本小题主要考查三角函数解析式的求法与三角函数图像与性质的运用，以及三角函数的值域的有关知识.【试题解析】解：(1)由图像得1A，24362T，所以2T，则1；将(,1)6代入得1sin()6，而22，所以3，因此函数()sin()3f xx；(6 分)(2)由于,6x，2336x，所以11sin()32x，所以()f x 的取值范围是1 1,2.(12 分)18.(本小题满分12 分)【命题意图】本小题主要考查运用数列基础知识求解数列的通项公式，其中还包括对数的运算与裂项求和的应用技巧.【试题解析】解：(1)由题11112nnSa112n