2010年普通高等学校招生全国统一考试数学理科试题(全国卷I)测验题精品解析.pdf

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1、1/16 2010 年普通高等学校招生全国统一考试理科数学(必修+选修 II)本试卷分第I 卷(选择题)和第卷(非选择题)两部分。第I 卷 1 至 2 页。第卷3 至 4 页。考试结束后，将本试卷和答题卡一并交回。第 I 卷注意事项：1答题前，考生在答题卡上务必用直径0.5 毫 M黑色墨水签字笔将自己的姓名、准考证号填写清楚，并贴好条形码。请认真核准条形码上的准考证号、姓名和科目。2每小题选出答案后，用 2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号，在试卷卷上作答无效。3第 I 卷共 12 小题，每小题5 分，共 60 分。在每小题给出的四个选项中，只

2、有一项是符合题目要求的。参考公式：如果事件A、B互斥，那么球的表面积公式()()()P ABP AP B24SR如果事件A、B相互独立，那么其中 R表示球的半径()()()P A BP AP B球的体积公式如果事件A在一次实验中发生的概率是p，那么334VRn次独立重复实验中事件A恰好发生k次的概率其中 R表示球的半径()(1)(0,1,2,)kknknnP kC ppkn一选择题(1)复数3223ii(A)i(B)i (C)12-13i (D)12+13i【答案】A【命题意图】本小题主要考查复数的基本运算,重点考查分母实数化的转化技巧.【解读】32(32)(23)694623(23)(23)

3、13iiiiiiiii.(2)记cos(80)k,那么tan100A.21kkB.-21kkC.21kkD.-21kk【答案】B 【命题意图】本小题主要考查诱导公式、同角三角函数关系式等三角函数知识，并突出了弦切互化这一转化思想的应用.2/16【解读】222sin801cos 801cos(80)1k,所以tan100tan802sin801.cos80kk(3)若变量,x y满足约束条件1,0,20,yxyxy则2zxy的最大值为(A)4 (B)3 (C)2 (D)1（4）已知各项均为正数的等比数列na，123a a a=5，789a a a=10，则4 5 6aaa=(A)5 2 (B)7

4、 (C)6 (D)4 2【答案】A【命题意图】本小题主要考查等比数列的性质、指数幂的运算、根式与指数式的互化等知识，着重考查了转化与化归的数学思想.【解读】由 等比 数 列 的 性 质 知31231322()5a a aa aaa，37897988()a a aa aaa10,所以132850a a,所以13336456465528()()(50)5 2a a aa aaaa a(5)353(1 2)(1)xx的展开式中 x 的系数是(A)-4 (B)-2 (C)2 (D)4 文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2

5、HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J

6、4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7

7、 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2

8、H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G

9、5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:

10、CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X

11、7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G5文档编码:CA7H4X7N8S2 HH4U6J4G10O7 ZC6C2H7H10G53/16 A B C D A1 B1 C1 D1 O(6)某校开设 A 类选修课 3 门，B类选择课 4 门，一位同学从中共选3 门，若要求两类课程中各至少选一门，则不同的选法共有(A)30种 (B)35种 (C)42种 (D)48种(7)正方体 ABCD-1111AB C D中，B1

12、B与平面 AC1D所成角的余弦值为A 23 B33 C23 D63【答案】D 【命题意图】本小题主要考查正方体的性质、直线与平面所成的角、点到平面的距离的求法，利用等体积转化求出D 到平面AC1D的距离是解决本题的关键所在,这也是转化思想的具体体现.文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T

13、2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J

14、5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T

15、2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J

16、5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T

17、2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J

18、5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y24/16（8）设 a=3log2,b=In2,c=125,则A abc Bbca C cab D cba【答案

19、】C 【命题意图】本小题以指数、对数为载体，主要考查指数函数与对数函数的性质、实数大小的比较、换底公式、不等式中的倒数法则的应用.【解读】a=3log2=21log 3,b=In2=21log e,而22log 3log1e,所以 ab,c=125=15,而2252log 4log 3,所以 ca,综上 cab.(9)已知1F、2F为双曲线C:221xy的左、右焦点，点p 在 C上，1Fp2F=060，则 P到 x 轴的距离为(A)32 (B)62 (C)3 (D)6【答案】B 【命题意图】本小题主要考查双曲线的几何性质、第二定义、余弦定理，考查转化的数学思想，通过本题可以有效地考查考生的综合

20、运用能力及运算能力.【解 读】不 妨 设 点P00(,)xy在 双 曲 线 的 右 支,由 双 曲 线 的 第 二 定 义 得21000|()12aPFe xaexxc，22000|)21aPFe xexaxc.由余弦定理得cos1FP2F=222121212|2|PFPFF FPFPF,即 cos0602220000(12)(21)(2 2)2(12)(21)xxxx,文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S

21、4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I

22、9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S

23、4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I

24、9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S

25、4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I

26、9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S

27、4J5H6Y25/16 解得2052x,所以2200312yx，故 P到 x 轴的距离为06|2y（10）已知函数F(x)=|lgx|,若 0ab,且 f(a)=f(b),则 a+2b 的取值范围是(A)(2 2,)(B)22,)(C)(3,)(D)3,)（11）已知圆O的半径为1，PA、PB为该圆的两条切线，A、B为俩切点，那么PAPB的最小值为(A)42 (B)32 (C)42 2 (D)32 2【答案】D【命题意图】本小题主要考查向量的数量积运算与圆的切线长定理，着重考查最值的求法判别式法,同时也考查了考生综合运用数学知识解题的能力及运算能力.【解读】如图所示：设PA=PB=x(0)x,

28、APO=,则 APB=2，PO=21x，21sin1x，|cos2PA PBPAPB=22(1 2sin)x=222(1)1xxx=4221xxx，令PA PBy，则4221xxyx，即42(1)0 xy xy，由2x是实数，所以2(1)4 1()0yy，2610yy，解得32 2y或32 2y.故min()32 2PA PB.此时21x.（12）已知在半径为2 的球面上有A、B、C、D四点，若AB=CD=2,则四面体ABCD的体积的最大值为P A B O 文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2

29、ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H

30、7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2

31、ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H

32、7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2

33、ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H

34、7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2

35、ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y26/16(A)2 33 (B)4 33 (C)2 3 (D)8 33【答案】B【命题意图】本小题主要考查几何体的体积的计算、球的性质、异面直线的距离,通过球这个载体考查考生的空间想象能力及推理运算能力.【解读】过CD作平面PCD，使 AB 平面PCD,交 AB 与 P,设点 P 到 CD的距离为h,则有ABCD11222323Vhh四面体,当直径通过AB与 CD的中点时,22max2 212 3h,故max4 33V.绝密启用前2010 年普通高等学校招生全国统一考试理科数学(必修+选修 II

36、)第卷注意事项：1答题前，考生先在答题卡上用直径0.5 毫 M黑色墨水签字笔将自己的姓名、准考证号填写清楚，然后贴好条形码。请认真核准条形码上的准考证号、姓名和科目。2第卷共2 页，请用直径0.5 毫 M黑色墨水签字笔在答题卡上各题的答题区域内作答，在试卷卷上作答无效。3。第卷共l0 小题，共90 分。二填空题：本大题共4 小题，每小题 5 分，共 20 分把答案填在题中横线上 (注意：在试卷卷上作答无效)(13)不等式2211xx的解集是 .【答案】0,2 【命题意图】本小题主要考查根式不等式的解法，利用平方去掉根号是解根式不等式的基本思路，也让转化与化归的数学思想体现得淋漓尽致.解读:原不

37、等式等价于2221(1),10 xxx解得 0 x2.(14)已知为第三象限的角，3cos25,则tan(2)4 .文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档

38、编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1

39、D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档

40、编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1

41、D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档

42、编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1

43、D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y27/16 12xy=1 x y a O 12x414ay2yxxa(15)直线1y与曲线2yxxa有四个交点，则a的取值范围是 .【答案】(1,5)4【命题意图】本小题主要考查函数的图像与性质、不等式的解法,着重考查了数形结合的数学思想.【解

44、读】如 图，在 同 一 直 角 坐 标 系 内 画 出 直 线1y与 曲 线2yxxa,观图可知，a 的取值必须满足1,4114aa解得514a.(16)已知F是椭圆C的一个焦点，B是短轴的一个端点，线段BF的延长线交C于点D，且BF2FDuu ruu r，则C的离心率为 .【答案】23【命题意图】本小题主要考查椭圆的方程与几何性质、第二定义、平面向量知识，考查了数形结合思想、方程思想,本题凸显解读几何的特点：“数研究形，形助数”，利用几何性质可寻求到简化问题的捷径.【解读】如图，22|BFbca,作1DDy轴于点 D1,则由BF2FDuu ruu r，得1|2|3OFBFDDBD,所以133

45、|22DDOFc,即32Dcx,由椭圆的第二定义得2233|()22accFDeacaxOyBF1DD文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H

46、7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2

47、ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H

48、7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2

49、ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H

50、7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2 ZK6S4J5H6Y2文档编码:CS1H7K2I9T2 HE5M1D3A5K2