本试卷分第1卷《选环题)和第|〔非选择题)同部
全共150分,考试时间120分钟考生作答时,将答家答在答眍卡上,在本试上答
无效考试结束后,将本试卷和答题卡一井交回
1.签前,考生先将自己的姓名、准考证号码填写清,将条形码准诵粘貼在条
区域内
2造择题必须使用B铅笔填途;非送择题必须使用05毫米黑色字迹的签字笔书
宇体工整、笔迹清楚
3请拉照题号序在各目的答题区内
答题区域书写的答案无效;在
秧辄,试卷上答题元效
5.保持卡面清洁,不娶折叠、不要弄酸、弄皱,不准使用涂改液、修正带、刮纸刀
一.单选题:本大题共8小,每小题5分,共40分左每小给出的四个选项中,只有
項是符合题目求的
直线l倾斜知为
2,数
为数单位),則在复平面内x对应的位于
B.二葱限
D.笫四象限
C.若cM月,ma,则m
扫描全能王创建
东方效学书中这样一个效列{F}
这个数列斯是奢名的“波那契数列
则此数列的前10乓和为
5如图所示,在平行六面体ABCD-4BCD中,AB=AD=AA1=1,∠BAD=60°
∠DAA=∠BAA1=90°,则AC1
6.如图所示,将绘有函数∫(x)=2sn(ox+-){>0)部分医像的纸片沿x轴折成
面角,若A、B之间的空间范离为2√5,则f(v3
I2+y2=4.过点P(2)的直线将区C的面积分割成两个部分,若使得
这两部分的画积之若最大,刘直线的方型为
高二学试2页其8
扫描全能王创建
8.如图所示,已知F、F2是相圆C
>b>0)的左,右焦点,A为拓圆的
顶点,B在x轴上,∠BAF2=90°,且F是BF2的中点,O为坐标原点,若点O到直线AB
内距高为3,则相国C的方程为
二,多选题:本大题共4小题,每小题5分,共20分,在每小题给出的选项中,有多项
合题目买求,全部选对的得5分,有迭错的得0分,部分选对得2分
若a⊥b,则m=2
c构成空间的一个茶腐,则{+ba+b+c,也可以构成空间的一个基
D.“直线4与2互相平行”是“直线与4的斜率相等"的充分不必要条件
钉尖分记作A、B、C、D
接这四个顶点构成的几何体为正四面体,组成
则下列结论正确的是
AB⊥CD
C. BC=l
D.四画你ABCD的外接琮表而积为丌
高二敖学试第 页共8
彐描全能王创建
希腊著名数学家阿波罗尼斯发现“若A、B为平面上相异的两点,则所有满足
PA=(>0,且2≠D的点P的轨迹是圜”,后来人们称这个圆为阿波罗尼斯圆在平
面直角坐标系xOy中,A(-2,0),B(40),若=,则下列关于动点P的结论正确的是
A.点P的轨迹方程为x2+y2+8x=0
B,△APB面积的最大值为6
C.在x轴上必存在异于A、B的两定点M、N,使得
D.若点Q(-3),则2|PA+PQ的最小值为5√2
抛物线y2=2px(P>0的焦点为F,点M(2),A(n),B(xy2)都在抛物线
上,且FA+FB+FM=0,则下列结论正确的是
C.FAl+ FMI+IFBF
第Ⅱ卷
三.填空题:本大题共4小题,每小题5分,共20分,把正确答案写在答题卡相应题的横
13已知邳、F分别是椭圆的左、右焦点,B为椭圆的上顶点,且∠FBF2=1200,则椭
14.四叶草也叫运草,四片叶子分划怠征若:
平安
雅宝公司在设计四叶草
吊坠的时候,利用了曲线方程C
y2=2|x|+2|y(如围所示
进行图案绘制试求前线C图成的封闭图形前而积_
扫描全能王创建高二数学试卷参考答案及评分标准
一. 单选题：本大题共 8 小题，每小题 5 分，共 40 分.在每小题给出的四个选项中，只有一项是符合题目要求的.
题号 1 2 3 4 5 6 7 8
答案 B C D C A A C D
二. 多选题：本大题共 4 小题，每小题 5 分，共 20 分.在每小题给出的选项中，有多项符合题目要求，全部选
对的得 5 分，有选错的得 0 分，部分选对得 2 分.
题号 9 10 11 12
答案 AB AB ACD ABD
三. 填空题：本大题共 4 小题，每小题 5 分，共 20 分. 把正确答案写在答题卡相应题的横线上.
3
13、 14、8 4 15、 1 16 x 2 2n、 （注：第一空 2 分；第二空 3 分）
2 n 1
四. 解答题：共 70 分，解答应写出文字说明、解答过程或演算步骤.
17.（10 分）
解： （1）【解法一】
2
直线 l的斜率为 kl ··················································································································································1分3
l //m 2 直线m的斜率为 km ·························································································································2分3
2
直线m的方程为 y 3 (x 2) ······················································································································3分
3
即 2x 3y 13 0 ····························································································································································4分
【解法二】
l //m 直线m的方程为 2x 3y C 0 ············································································································1分
直线m过点 (2,3) 2 2 3 3 C 0 C 13 ························································································3分
直线m的方程为 2x 3y 13 0 ··························································································································4分
（2）易知圆心C(1,2)，半径 r 1 ······························································································································5分
C l d | 2 1 3 2 6 | 2 2 13 圆心 到直线 的距离 ······································································7分
22 32 13 13
| AB | 2 r 2 d 2 2 1 4 6 13 ··················································································································9分
13 13
6 13
线段 AB的长为 ···············································································································································10分
13
高二数学试卷参考答案及评分标准 第 1 页（共 4页）
18.（12 分）
a2 2a5 12 3a1 9d 12 a1 1
解：（1）设等差数列{an}的公差为 d ·····································3 分
3a4 2a1 10 a1 9d 10 d 1
an 1 (n 1) 1 n 等差数列{an}的通项公式为 an n ···········································································5 分
（2） bn n 2
n ··································································································································································6 分
S 1n 1 2 2 2
2 3 23 (n 1) 2n 1 n 2n ①·····················································································7 分
2Sn 1 2
2 2 23 3 24 (n 1) 2n n 2n 1 ②·····················································································8 分
①－② S 2 22 23 2n n 2n 1得： n ·······································································································9 分
2(1 2n )
Sn n 2
n 1 ············································································································································10 分
1 2
数列{bn}的前 n项和 Sn (n 1) 2
n 1 2 ············································································································12 分
19.（12 分）
3csin B bcos(C 解：（1） ) ，由正弦定理可得： 3 sinC sin B sin Bcos(C ) ····················1分
3 3
0 B sin B 0 3 sinC 1 cosC 3 sinC ·····················································································3分
2 2
tanC 3 ··········································································································································································5分
3
0 C C 角C的大小为 ·····················································································································6分
6 6
（2） c2 a2 b2 2abcosC c2 12 b2 6b ·······························································································7分
3c2 3b2 2a2 c2 8 b2 8 b2 12 b2 6b b2 3b 2 0 b 1，或b 2 ···················9分
①当b 1 c 7 S 1 3 时， ABC absinC ·····························································································10分2 2
②当b 2时， c 2 S 1 ABC absinC 3 ································································································11分2
ABC的面积为 3 3或 ·········································································································································12 分
2
高二数学试卷参考答案及评分标准 第 2 页（共 4页）
20.（12 分）
证明：（1）四边形 ABCD是等腰梯形， AD // BC OD 1,OB 2 ··················1 分
EO PE DO 1连接 ， EO // PD ··································································2 分
PB DB 3
EO 平面 AEC， PD 平面 AEC ·····································································3 分
PD //平面 AEC ·············································································································4 分
（2） PO 平面 ABCD， AC 平面 ABCD PO AC
tan PAC 2 OP 2 OP 2 ·························································································································5 分
OA
OB OC 以O为坐标原点，分别以OB，OC，OP所在直线为 x，y，z轴，建立空间直角坐标系
Oxyz O(0,0,0)，P(0,0,2)， A(0, 1,0)，C(0,2,0)，B(2,0,0) ······························································7 分
PA (0, 1, 2)， PC (0,2, 2)，PB (2,0, 2)，···························································································8 分
n PC 2y 2z 0
设平面 PBC 的法向量为 n (x, y, z)
n PB 2x 2z 0
令 x 1 y 1， z 1 n (1,1,1) ···························································································································10 分
PA PBC sin | PA n | 3 15设 与平面 所成角为 | cos PA ,n | ··························11 分
| PA || n | 5 3 5
PA 与平面PBC 15所成角的正弦值为 ···············································································································12 分
5
21.（12 分）
b 3
解： （1）由题，可知 a 2 ········································································································································2 分
2b 2 3
a 2
解得 ·······································································································································································3 分
b 3
2
C x y
2
双曲线 的方程为 1.·······························································································································4 分
4 3
x2 y2
1 2 2
（2） 4 3 ,消 y得: (3 4k )x 8kx 16 0 ·······················································································6 分
y kx 1
高二数学试卷参考答案及评分标准 第 3 页（共 4页）
3 4k 2 0 3 1 k 1，且k ··············································································7分
2 ( 8k) 64(3 4k
2) 0 2
设 A(x1, y1) B(x , y ) x
8k 16
， 2 2 1 x2 2，x1x2 2 ··············································································8分3 4k 3 4k
OA OB x1x2 y1y2 x1x2 (kx 1)(kx 1) (1 k
2
1 2 )x1x2 k(x1 x2) 1 1 ·······························9分
16(1 k 2) 8k 2 10 16k 2
2 2 2 0 2 0 3
3 3
4k 2 0 k ，或k ··················11分
3 4k 3 4k 3 4k 2 2
k的取值范围为 1 k 3 3 ，或 k 1 ································································································12 分
2 2
22.（12 分）
y log3(xn 1) 解：由题可知 n (xn 0，n N )，·········································································································1分xn 1
n 2 n x 1（1） ，且 N * , xn 3xn 1 2 xn 1 3(xn 1 1)， n 3(常数) ··································2分xn 1 1
x1 1 3 {xn 1}是首项为3，公比为3的等比数列································································································3分
xn 1 3 3
n 1 3n x nn 3 1 数列{x
n
n}的通项公式为 xn 3 1 ···························································4分
n
（2） y log 3 n y n 1n 3n n 0
n 1 1 yn 1 yn 数列{yn}单调递减··········································5分3 3 yn 3n
y 1n 最大值为 y1 · 9t
2 18mt 0恒成立···········································································································6分
3
9t 2 18t 0 m [ 1,1] t 2，或t 2
9t
2 18t 0
t的取值范围为 ( ， 2) (2， ) ···························································································································7分
(y y )(x x ) 4n 1
（3）四边形 PQ Q n n 1 n 1 nn n n 1Pn 1的面积是Tn ································································8分2 3
3 3 3(1 1 ) ·························································································································10分
n(4n 1) n(n 1) n n 1
1 1 1 3[(1 1) (1 1) (1 1 1 1 1 ) ( )] 3(1 ) ·······················11分
T1 2T2 nTn 2 2 3 3 4 n n 1 n 1
n N * 1 3 3(1 ) 3 3
n 1 n 1
1 1 1
3 .············································································································································12分
T1 2T2 nTn
如有其他解法，保证各问分值不变的前提下，酌情给分.
高二数学试卷参考答案及评分标准 第 4 页（共 4页）
