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1、1 福建省福州高三第一学期期末质量检查数学（文）试卷（word 版）第 1 卷(选择题共 60 分)一、选择题(本大题共1 2 小题，每小题5 分，共 60 分在每小题所给的四个答案中有且只有一个答案是正确的把正确的选项涂在答题卡的相应位置上)1复数 i(1+i)(i 为虚数单位)等于A0 B1+iC1-iD-1+i2已知全集U=a，b，c，d，e，M=a，c，d，N=b，d，e，则(M)N 等于Ab Bd Cb，e Db，d，e 3 如图是某次大赛中，7 位评委为某选手打出的分数的茎叶统计图，去掉一个最高分和一个最低分后，所剩数据的平均数为A83 B84 C85 D86 4“x2”是“x2-
2、2xbcBacbCcab Dbc a6若变量x，y 满足约束条件,2/2,yxyxx?3?则 z=x2y 的最小值等于A 2 B3 22C33D127已知 cos(+4p)=23，则 sin(4p)的值等于A23B23C53D 538直线y=x 与椭圆2222xyab+=1 的交点在x 轴上的射影恰好是椭圆的焦点，则椭圆C的离心率为A132-+B152+C352-D129已知函数f(x)=2sin(x+)(0)的部分图象如图所示，则函数 f(x)的一个单调递增区间是A（75,12 12pp-）B（7,1 21 2pp-）C（3,36p p-）D（1 1 7,1 2 1 2pp）10若直线x+m
3、y=2+m 与圆 x2+y22x 2y+1=0 相交，则实数m 的取值范围为|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 1 页，共 8 页2 A(，+)B(，0)C(0，+)D(，0)U(0，+)11 如 图，已 知 点O 是 边 长 为1 的 等 边 ABC的 中 心，则(OAOB+)(OAOC+)等于A19B19-C16D16-12已知数列 an中，a1=45，a n+1=12,0,2121,1,2nnnnaaaa?-?=?+?f(x)=x2一 x 一 1+lnx中，属于M 的有(写出所有符合的函数序号)三、解答题(本大题共6 小题，共7
4、4 分，解答应写出文字说明、证明过程或演算过程)l7(本小题满分l 2 分)已知 an 是等比数列，a1=2，且 a1，a3+1，a4成等差数列|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 2 页，共 8 页文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编
5、码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R
6、10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8
7、P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4
8、H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9
9、K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4
10、C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A1
11、0L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L33(I)求数列 a n的通项公式；()若 bn=log2an，求数列 bn的前 n 项和 Sn18(本小题满分12 分)已知 A、B、C 三个箱子中各装有2 个完全相同的球，每个箱子里的球，有一个球标着号码 1，另一个球标着号码2现从 A、B、C 三个箱子中各摸出1 个球(I)若用数组(x，y，z)中的 x、y、z 分别表示从A、B、C 三个箱子中摸出的球的号码，请写出数组(x，y，z)的所有情形，并回答一共有多少种；()如果请您猜测摸出的这三个球的号码之和，猜中有奖那么猜什么数获奖的可能性最大?请说明理
12、由19(本小题满分1 2 分)设 ABC 的内角 A、B、C 所对的边分别为a、b、c已知 a=3，B=3p,SABC=63(I)求 ABC 的周长；()求 sin2A 的值20(本小题满分l 2 分)某书商为提高某套丛书的销量，准备举办一场展销会据市场调查，当每套丛书售价定为 x 元时，销售量可达到15 一 O.1x 万套现出版社为配合该书商的活动，决定进行价格改革，将每套丛书的供货价格分成固定价格和浮动价格两部分，其中固定价格为30 元，浮动价格(单位：元)与销售量(单位：万套)成反比，比例系数为l0假设不计其它成本，即销售每套丛书的利润=售价一 供货价格问：(I)每套丛书定价为100 元
13、时，书商能获得的总利润是多少万元?()每套丛书定价为多少元时，单套丛书的利润最大?21(本小题满分1 2 分)在平面直角坐标系xOy 中，已知点 A(一 1，1)，P 是动点，且三角形 POA 的三边所在直线的斜率满足kOP+kOA=kPA(I)求点 P 的轨迹 C 的方程；()若 Q 是轨迹 C 上异于点P 的一个点，且PQOAl=，直线 OP 与 QA 交于点 M，试探究：点 M 的横坐标是否为定值?并说明理由22(本小题满分1 4 分)已知 m，t R，函数 f(x)=(x-t)3+m(I)当 t=1 时，(i)若 f(1)=1，求函数f(x)的单调区间；(ii)若关于 x 的不等式f(
14、x)x31 在区间 1，2上有解，求m 的取值范围；()已知曲线y=f(x)在其图象上的两点A(x1，f(x1)，B(x2，f(x2)(x1 x2)处的切线分别为 l1、l2若直线 l1与 l2平行，试探究点A 与点 B 的关系，并证明你的结论|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 3 页，共 8 页文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5
15、A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文
16、档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD
17、1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3
18、D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 H
19、N4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7
20、L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 Z
21、W4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L34 福州市 2011-2012 学年第一学期期末高三质量检查数学(文科)试卷参考答案一、选择题（本大题共12 小题，每小题5 分，共 60 分.在每小题所给的四个答案中有且只有一个答案是正确的.）1D 2C 3C 4B 5A 6 B 7A 8A 9D 10D 11D 12C 二、填空题（本大题共4 小题，每小题4 分，共 16 分）1343yx；141320；15sin()sincoscossin；16三、
22、解答题（本大题共6 小题，共 74 分，解答应写出文字说明、证明过程或演算过程）17（本小题满分12 分）解：（）设数列na的公比为q，则22312aaqq，33412aaqq，2 分134,1,aaa成等差数列，1432(1)aaa，即32222(21)qq 4分整理得2(2)0qq，0q，2q，6 分1222nnna（*Nn）8 分（）22loglog 2nnnban，10 分12(1)122nnn nSbbbn 12 分18.（本小题满分12 分）解：（）数组(,)x y z的所有情形为：（1，1，1），（1，1，2），（1，2，1），（1，2，2），（2，1，1），（2，1，2），（2
23、，2，1），（2，2，2），共 8 种答：一共有8 种 5 分注：列出 5、6、7 种情形，得2 分；列出所有情形，得4 分；写出所有情形共8 种，得1 分（）记“所摸出的三个球号码之和为i”为事件iA（i=3，4，5，6），6 分易知，事件3A包含有1 个基本事件，事件4A包含有 3 个基本事件，事件5A包含有3|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 4 页，共 8 页文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文
24、档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD
25、1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3
26、D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 H
27、N4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7
28、L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 Z
29、W4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5
30、A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L35 个基本事件，事件6A包含有 1 个基本事件，所以，31()8P A，43()8P A，53()8P A，61()8P A10 分故所摸出的两球号码之和为4、为 5 的概率相等且最大答：猜 4 或 5 获奖的可能性最大 12 分19（本小题满分12 分）解：（）6 3ABCS，113sin36 3222acBc，8c，2 分由余弦定理
31、得，2222212cos3823 8492bacacB，7b，5 分ABC的周长为38718abc 6 分（）由正弦定理得，sinsinabAB，333 3sinsin7214aABb，8 分ab，AB，故角A为锐角，9 分213cos1sin14AA，10 分3 31339 3sin 22sincos2141498AAA 12 分20（本小题满分12 分）解：（）每套丛书定价为100 元时，销售量为150.1 1005万套，此时每套供货价格为1030325元，3 分 书商所获得的总利润为5(10032)340万元 4 分（）每套丛书售价定为x 元时，由150.10,0,xx得，0150 x，
32、5 分依题意，单套丛书利润10100(30)30150.1150Pxxxx 7 分|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 5 页，共 8 页文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C
33、3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10
34、L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码
35、:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R1
36、0M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P
37、2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H
38、4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K
39、6 ZW4C3G5A10L36 100(150)120150Pxx，0150 x，1500 x，由100100(150)2(150)2 1020150150 xxxx，10 分当且仅当100150150 xx，即140 x时等号成立，此时max20120100P答：（）当每套丛书售价定为100 元时，书商能获得总利润为340 万元；（）每套丛书售价定为140 元时，单套利润取得最大值100 元 12 分（说明：学生未求出最大值不扣分）21（本小题满分12 分）解：（）设点(,)P x y为所求轨迹上的任意一点，则由OPOAPAkkk得1111yyxx，2 分整理得轨迹C的方程为2yx（0 x且
40、1x），4 分（）（方法一）设22112200(,),(,),(,)P x xQ xxM xy，由PQOA可知直线/PQ OA，则PQOAkk，故2221211010 xxxx，即211xx，6分由OMP、三点共线可知，00(,)OMxy与211(,)OPx x共线，201100 x xx y，由（）知10 x，故001yx x，8 分同理，由00(1,1)AMxy与222(1,1)AQxx共线，20220(1)(1)(1)(1)0 xxxy，即2020(1)(1)(1)(1)0 xxxy，由（）知21x，故020(1)(1)(1)0 xxy，10 分将001yx x，211xx代入上式得01
41、01(1)(2)(1)0 xxx x，整理得0112(1)1xxx，由11x得012x，即点 M 的横坐标为定值12 12 分（方法二）设221122(,),(,),P xxQ xx|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 6 页，共 8 页文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L
42、9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW
43、4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A
44、10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档
45、编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1
46、R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D
47、8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN
48、4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L37 由PQOA可知直线/PQ OA，则PQOAkk，故2221211010 xxxx，即211xx，6 分直线OP方程为：1yx x；8 分直线QA的斜率为：2111(1)1211xxx，直线QA方程为：11(2)(1)yxx，即11(2)1yxxx；10 分联立，得12x，点M的横坐标为定值12 12 分22（本小题满分14 分）解：（）(i)因为(1)1f，所以1m，1 分则332()1133f xxxxx，而22()3633(1)0fxxxx恒成立，所以函数()
49、f x 的单调递增区间为(,)4 分（ii）不等式3()1f xx在区间 1,2上有解，即不等式2330 xxm在区间 1,2上有解，即不等式233mxx在区间 1,2上有解，等价于233mxx不小于在区间 1,2上的最小值，6 分因为1,2x时，2213333()0,624xxx，所以m的取值范围是0,)9 分（）因为3()f xx的对称中心为(0,0)，而3()()f xxtm可以由3()f xx经平移得到，所以3()()f xxtm的对称中心为(,)t m,故合情猜测，若直线1l 与2l 平行，则点A与点B关于点(,)t m 对称 10 分对猜想证明如下：因为33223()()33f x
50、xtmxtxt xtm所以222()3633()fxxtxtxt所以，1l，2l 的斜率分别为2113()kxt，2223()kxt又直线1l 与2l 平行，所以12kk，即2212()()xtxt，因为12xx，所以，12()xtxt，12 分|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 7 页，共 8 页文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G5A10L3文档编码:CD1R10M3D8P2 HN4H4C7L9K6 ZW4C3G