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1、第1页共 26页2014年浙江省高考数学试卷理科一、选择题每题5 分，共 50 分1 5 分 2014?浙江设全集U=x N|x 2，集合 A=x N|x2 5，则?UA=A?B 2 C 5 D2，5 2 5 分 2014?浙江已知i 是虚数单位，a，b R，则“a=b=1”是“a+bi2=2i”的A充分不必要条件B必 要不充分条件C 充分必要条件D既 不充分也不必要条件35 分 2014?浙江某几何体的三视图 单位：cm 如下图，则此几何体的外表积是A90cm2B129cm2C132cm2D138cm24 5 分 2014?浙江为了得到函数y=sin3x+cos3x 的图象，可以将函数y=c
2、os3x 的图象A向右平移个单位B向左平移个单位C向右平移个单位D向左平移个单位5 5 分 2014?浙江在 1+x61+y4的展开式中，记xmyn项的系数为fm，n，则 f3，0+f2，1+f 1，2+f0，3=A45 B60 C120 D210 6 5 分 2014?浙江已知函数fx=x3+ax2+bx+c，其 0f 1=f 2=f 3 3，则Ac 3 B3c 6 C6c 9 Dc9 7 5 分 2014?浙江在同一直角坐标系中，函数fx=xax 0，gx=logax 的图象可能是第2页共 26页ABCD8 5 分 2014?浙江记maxx，y=，minx，y=，设，为平面向量，则Amin
3、|+|，|min|，|Bmin|+|，|min|，|Cmax|+|2，|2|2+|2Dmax|+|2，|2|2+|295 分2014?浙江已知甲盒中仅有1 个球且为红球，乙盒中有 m 个红球和 n 个蓝球 m 3，n 3，从乙盒中随机抽取ii=1，2个球放入甲盒中a放入 i 个球后，甲盒中含有红球的个数记为ii=1，2；b放入 i 个球后，从甲盒中取1 个球是红球的概率记为pi i=1，2 则Ap1p2，E1 E2Bp1 p2，E 1 E 2C p1p2，E1 E2Dp1 p2，E 1 E 210 5 分 2014?浙江 设函数 f1x=x2，f2x=2xx2，i=0，1，2，99记 Ik=|
4、fka1 fka0|+|fka2 fka1丨+|fka99fka98|，k=1，2，3，则AI1I2 I3BI2I1I3CI1I3I2DI3I2 I1二、填空题11 4 分 2014?浙江在某程序框图如下图，当输入50 时，则该程序运算后输出的结果是文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ
5、8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H
6、10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 H
7、G1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9
8、F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 Z
9、G2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W
10、5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3文
11、档编码:CZ8P6Z1H10Y5 HG1P5X9F2K5 ZG2L10W5F6Q3第3页共 26页12 4 分 2014?浙江随机变量 的取值为0，1，2，假设 P=0=，E=1，则D=13 4 分 2014?浙江当实数x，y 满足时，1 ax+y 4 恒成立，则实数a的取值范围是14 4 分 2014?浙江在8 张奖券中有一、二、三等奖各1 张，其余 5 张无奖将这8张奖券分配给4 个人，每人2 张，不同的获奖情况有种用数字作答 15 4 分 2014?浙江设函数fx=，假设 ffa 2，则实数a的取值范围是文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
12、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2
13、E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
14、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2
15、E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
16、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2
17、E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
18、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5第4页共 26页16 4 分 2014?浙江设直线x3y+m=0m 0与双曲线=1a0，b0的两条渐近线分别交于点A，B假设点P m，0满足|PA|=|PB|，则该双曲线的离心率是17 4 分 2014?浙江 如图，某人在垂直于水平地面ABC 的墙面前的点A 处进行射击训练已知点 A 到墙面的距离为AB，某目标点P 沿墙面上的射线CM 移动，此人为了准确瞄准目标点 P，需计算由点A 观察点 P的仰角 的大小假设 AB=15m，AC=25m，
19、BCM=30 ，则 tan的最大值是 仰角 为直线 AP 与平面 ABC 所成角三、解答题18 14 分 2014?浙江在 ABC 中，内角 A，B，C 所对的边分别为a，b，c已知 a b，c=，cos2Acos2B=sinAcosA sinBcosB 求角 C 的大小；假设 sinA=，求 ABC 的面积19 14 分 2014?浙江已知数列 an 和bn满足 a1a2a3 an=n N*假设 an为等比数列，且a1=2，b3=6+b2 求 an和 bn；设 cn=n N*记数列 cn的前 n 项和为 Sn i求 Sn；ii求正整数k，使得对任意n N*均有 Sk Sn20 15 分 20
20、14?浙江如图，在四棱锥ABCDE 中，平面ABC 平面 BCDE，CDE=BED=90 ，AB=CD=2，DE=BE=1，AC=证明：DE平面 ACD；求二面角B AD E 的大小文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D
21、5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8
22、ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D
23、5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8
24、ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D
25、5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8
26、ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5第5页共 26页21 15 分 2014?浙江如图，设椭圆C：ab0，动直线l 与椭圆 C 只有一个公共点P，且点 P 在第一象限 已知直线l 的斜率为k，用 a，b，k 表示点
27、 P 的坐标；假设过原点O 的直线 l1与 l 垂直，证明：点P到直线 l1的距离的最大值为ab22 14 分 2014?浙江已知函数fx=x3+3|xa|a R 假设 fx在 1，1上的最大值和最小值分别记为Ma，ma，求 Ma ma；设 b R，假设 fx+b2 4 对 x 1，1恒成立，求3a+b 的取值范围文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2
28、E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
29、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2
30、E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
31、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2
32、E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档
33、编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5第6页共 26页2014 年浙江省高考数学试
34、卷理科参考答案与试题解析一、选择题每题5 分，共 50 分1 5 分 2014?浙江设全集U=x N|x 2，集合 A=x N|x2 5，则?UA=A?B 2 C 5 D2，5 考点：补集及其运算专题：集合分析：先化简集合A，结合全集，求得?UA解答：解：全集 U=x N|x 2，集合 A=x N|x2 5=x N|x 3，则?UA=2，故选：B点评：此题主要考查全集、补集的定义，求集合的补集，属于基础题2 5 分 2014?浙江已知i 是虚数单位，a，b R，则“a=b=1”是“a+bi2=2i”的A充分不必要条件B必 要不充分条件C 充分必要条件D既 不充分也不必要条件考点：复数相等的充要
35、条件；充要条件专题：简易逻辑分析：利用复数的运算性质，分别判断“a=b=1”?“a+bi2=2i”与“a=b=1”?“a+bi2=2i”的真假，进而根据充要条件的定义得到结论解答：解：当“a=b=1”时，“a+bi2=1+i2=2i”成立，故“a=b=1”是“a+bi2=2i”的充分条件；当“a+bi2=a2b2+2abi=2i”时，“a=b=1”或“a=b=1”，故“a=b=1”是“a+bi2=2i”的不必要条件；综上所述，“a=b=1”是“a+bi2=2i”的充分不必要条件；故选 A 点评：此题考查的知识点是充要条件的定义，复数的运算，难度不大，属于基础题35 分 2014?浙江某几何体的
36、三视图 单位：cm 如下图，则此几何体的外表积是文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4
37、H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7
38、 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4
39、H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7
40、 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4
41、H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7
42、 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5第7页共 26页A90cm2B129cm2C132cm2D138cm2考点：由三视图求面积、体积专题：立体几何分析：几何体是直三棱柱与直四棱柱的组合体，根据三视图判断直三棱柱的侧棱长与底面的形状及相关几何量的数据，判断四棱柱的高与底面矩形的边长，把数据代入外表积公式计算解答：解：由三视图知：几何体是直三棱柱与直
43、四棱柱的组合体，其中直三棱柱的侧棱长为3，底面是直角边长分别为3、4 的直角三角形，四棱柱的高为6，底面为矩形，矩形的两相邻边长为3 和 4，几何体的外表积S=2 4 6+3 6+3 3+2 3 4+2 3 4+4+5 3=48+18+9+24+12+27=138 cm2 故选：D点评：此题考查了由三视图求几何体的外表积，根据三视图判断几何体的形状及数据所对应的几何量是解题的关键4 5 分 2014?浙江为了得到函数y=sin3x+cos3x 的图象，可以将函数y=cos3x 的图象A向右平移个单位B向左平移个单位C向右平移个单位D向左平移个单位考点：函数 y=Asin x+的图象变换专题：三
44、角函数的图像与性质分析：利用两角和与差的三角函数化简已知函数为一个角的一个三角函数的形式，然后利用平移原则判断选项即可解答：解：函数y=sin3x+cos3x=，故只需将函数y=cos3x 的图象向右平移个单位，得到y=的图象故选：C文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E
45、9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编
46、码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E
47、9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编
48、码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E
49、9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编
50、码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5文档编码:CQ7D5D4W4P7 HG1R2E9M6H8 ZS2F2E4H7A5第8页共 26页点评：此题考查两角和与差的三角函数以及三角函数的平移变换的应用，基本知识的考查5 5 分 2014?浙江在 1