2014年高考理科数学试题全国大纲卷试题及参考答案.pdf

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2014年高考理科数学试题全国大纲卷试题及参考答案.pdf

全国新课标卷理科数学第1页2014 年高考理科数学试题全国大纲卷注意事项：1.本试卷分第卷选择题和第卷非选择题两部分。第卷1 至 3 页，第卷3 至 5 页。2.答题前，考生务必将自己的、准考证号填写在本试题相应的位置。3.全部答案在答题卡上完成，答在本试题上无效。4.考试结束后，将本试题和答题卡一并交回。第卷一、选择题：本大题共12 个小题,每题 5 分,共 60 分.在每题给出的四个选项中，只有一项是符合题目要求的.1.设103izi，则 z 的共轭复数为A13iB1 3iC13iD1 3i2.设集合2|340Mx xx，|05Nxx，则 MNA(0,4B0,4)C 1,0)D(1,03.设0sin 33a，0cos55b，0tan35c，则A abcB bcaC cbaD cab4.假设向量,a b满足：|1a，()aba，(2)abb，则|bA2 B2C1 D225.有 6 名男医生、5 名女医生，从中选出2 名男医生、1 名女医生组成一个医疗小组，则不同的选法共有A60 种 B70种 C75 种 D150 种6.已知椭圆 C：22221xyab(0)ab的左、右焦点为1F、2F，离心率为33，过2F的直线 l 交C 于 A、B 两点，假设1AF B的周长为4 3，则 C 的方程为A22132xyB2213xyC221128xyD221124xy7.曲线1xyxe在点 1,1处切线的斜率等于A2e Be C2 D18正四棱锥的顶点都在同一球面上，假设该棱锥的高为4，底面边长为 2，则该球的外表积为全国新课标卷理科数学第2页A814B16C 9D2749.已知双曲线 C的离心率为 2，焦点为1F、2F，点 A 在 C 上，假设12|2|F AF A，则21cosAF FA14B13C24D2310.等比数列na中，452,5aa，则数列lgna的前 8 项和等于A6 B5 C4 D311.已知二面角l为060，AB，ABl，A 为垂足，CD，Cl，0135ACD，则异面直线 AB 与 CD 所成角的余弦值为A14B24C34D1212.函数()yfx的图象与函数()yg x的图象关于直线0 xy对称，则()yf x的反函数是A()yg xB()ygxC()yg xD()ygx第卷共 90 分二、填空题每题5分，总分值 20 分，将答案填在答题纸上13.8()xyyx的展开式中22x y的系数为.14.设 x、y 满足约束条件02321xyxyxy，则4zxy的最大值为.15直线1l和2l是圆222xy的两条切线，假设1l与2l的交点为 1,3，则1l与2l的夹角的正切值等于.16.假设函数()cos2sinf xxax在区间(,)62是减函数，则 a的取值范围是.三、解答题本大题共 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.本小题总分值 10 分ABC的内角 A、B、C 的对边分别为 a、b、c，已知 3 cos2 cosaCcA，1tan3A，求 B.18.本小题总分值 12 分等差数列na的前 n 项和为nS，已知110a，2a为整数，且4nSS.文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3全国新课标卷理科数学第3页1求na的通项公式；2设11nnnba a，求数列nb的前 n 项和nT.19.本小题总分值12分如图，三棱柱111ABCA B C中，点1A在平面ABC 内的射影 D 在 AC 上，090ACB，11,2BCACCC.1证明：11ACA B；2设直线1AA与平面11BCC B的距离为3，求二面角1AABC的大小.20.本小题总分值12分设每个工作日甲、乙、丙、丁4 人需使用某种设备的概率分别为0.6 0.5 0.5 0.4、，各人是否需使用设备相互独立.1求同一工作日至少3 人需使用设备的概率；2X 表示同一工作日需使用设备的人数，求X 的数学期望.21.本小题总分值12分已知抛物线 C：22(0)ypx p的焦点为 F，直线4y与 y 轴的交点为 P，与 C 的交点为 Q，且5|4QFPQ.1求 C 的方程；2过 F 的直线 l 与 C 相交于 A、B 两点，假设 AB 的垂直平分线l与 C 相较于 M、N 两点，且 A、M、B、N 四点在同一圆上，求 l 的方程.22.本小题总分值12分函数()ln(1)(1)axf xxaxa.1讨论()f x的单调性；2设111,ln(1)nnaaa，证明：23+22nann.文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3全国新课标卷理科数学第4页2014 年高考理科数学试题全国大纲卷参考答案一、选择题：1.D 2.B 3.C 4.B 5.C 6.A 7.C 8.A 9.A 10.C 11.B 12.D 二、填空题：13.70 14.5 15.4316.(,2三、解答题：17.本小题总分值 10 分解：由题设和正弦定理得3sincos2sincosACCA故3tancos2sinACC因为1tan3A，所以cos2sinCC即1tan2C 6 分所以tantan180()BACtan()ACtantantantan1ACAC 8 分1即135B10分18.本小题总分值 12 分解：由110a，2a为整数知，等差数列na的公差 d 为整数又4nSS，故450,0aa即1030,1040dd解得10532d因此3d数列na的通项公式为133nan6 分1111()(133)(103)3103133nbnnnn 8 分于是12.nnTbbb文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3全国新课标卷理科数学第5页1111111()().()371047103133nn111()3 10310n10(103)nn.12分19.本小题总分值 12 分解法一：因为1A D平面1,ABC A D平面11AAC C，故平面11AAC C平面 ABC，又 BCAC，所以 BC平面11AAC C，3 分连结1AC，因为侧面11AAC C为菱形，故11ACAC由三垂线定理得11ACA B 5 分 BC平面11,AAC C BC平面11BCC B，故平面11AA C C平面11BCC B作11,A ECCE为垂足，则1A E平面11BCC B又直线1/AA平面11BCC B，因而1A E为直线1AA与平面11BCC B的距离，13AE因为1AC为1ACC的平分线，故113ADAE 8 分作,DFAB F为垂足，连结1A F，由三垂线定理得1A FAB，故1A FD为二面角1AABC的平面角由22111ADAAAD得 D 为 AC 中点，1525ACBCDFAB，11tan15A DA FDDF所以二面角1AABC的大小为arctan15 12 分解法二：以 C 为坐标原点，射线CA 为x轴的正半轴，以 CB 的长为单位长，建立如下图的空间直角坐标系Cxyz，由题设知1A D与z轴平行，x轴在平面11AAC C内文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3全国新课标卷理科数学第6页设1(,0,)A ac，由题设有2a，(2,0,0)A，(0,1,0)B，则(2,1,0)AB，(2,0,0)AC，1(2,0,)AAac，11(4,0,)ACACAAac，1(,1,)BAac 2 分由1|2AA得22(2)2ac，即2240aac于是221140ACBAaac，所以11ACA B 5 分设平面11BCC B的法向量(,)mx y z，则1,mCB mBB，即0m CB，10m BB因为(0,1,0)CB，11(2,0,)BBAAac，故0y，且(2)0axcz令xc，则2za，(,0,2)mca，点 A到平面11BCC B的距离为22|2|cos,|(2)CA mcCAm CAcmca又依题设，A到平面11BCC B的距离为3，所以3c代入解得3a舍去或1a8 分于是1(1,0,3)AA设平面1ABA的法向量(,)np q r，则1,nAA nAB，即10n AA，0n AB，30pr且20pq，令3p，则2 3q，1r，(3,2 3,1)n，又(0,0,1)p为平面 ABC 的法向量，故1cos,|4n pn pnp所以二面角1AABC的大小为1arccos4 12 分20.本小题总分值 12 分解：记iA表示事件：同一工作日乙、丙中恰有i 人需使用设备，0,1,2i，文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3全国新课标卷理科数学第7页B 表示事件：甲需使用设备，C 表示事件：丁需使用设备，D 表示事件：同一工作日至少3 人需使用设备122DAB CABAB C22()0.6,()0.4,()0.5,0,1,2iiP BP CP ACi 3 分所以122()()P DP AB CABAB C122()()()P AB CP ABP AB C122()()()()()()()()P AP BP CP AP BP AP BP C0.316 分的可能取值为 0，1，2，3，4，其分布列为0(0)()PP BAC0()()()P BP AP C2(10.6)0.5(10.4)0.06001(1)()PP B ACB ACB A C001()()()()()()()()()P BP AP CP BP AP CP BP AP C2220.60.5(10.4)(10.6)0.50.4(10.6)0.5(10.4)0.25222(4)()()()()0.50.60.40.06PP AB CP AP BP C(3)()(4)0.25PP DP(2)1(0)(1)(3)(4)PPPPP1 0.060.25 0.250.060.3810 分数学期望0(0)1(1)2(2)3(3)4(4)EXPPPPP0.2520.383 0.2540.06212 分21.本小题总分值 12 分解：设0(,4)Q x，代入22ypx得08xp文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY10E8J1B10I8 ZC10Y6F10R9Q3文档编码:CK2R5L10G6W8 HY

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