齐齐哈尔市2021-2022学年高二上学期期末考试
数 学 试 卷
本试卷分第I卷（选择题）和第II卷（非选择题）两部分.
全卷共150分,考试时间120分钟.考生作答时，将答案答在答题卡上，在本试卷上答题无效.考试结束后，将本试卷和答题卡一并交回.
注意事项：
1.答题前，考生先将自己的姓名、准考证号码填写清楚，将条形码准确粘贴在条形码区域内.
2.选择题必须使用2B铅笔填涂；非选择题必须使用0.5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚.
3.请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效.
4.作图可先使用2B铅笔填涂；非选择题必须用黑色字迹的签字笔描黑.
5.保持卡面清洁，不要折叠、不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀.
第I卷
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.
1. 直线的方程为，则直线的倾斜角为
A．30° B．45° C．60° D．90°
2. 复数(为虚数单位)，则在复平面内对应的点位于
A．第一象限 B．第二象限 C．第三象限 D．第四象限
已知互不重合的直线，互不重合的平面，下列命题正确的是
A．若，则 B．若 ，则
C．若，则 D．若，则
4.1202年，意大利数学家斐波那契(Leonardo Fibonacci，约1170-约1250)出版了他的《算
盘全书》(Liber Abaci)，在书中他向欧洲人介绍了东方数学.书中有这样一个数列:
，，且，这个数列就是著名的“斐波那契数列”，则此数列的前项和为
A．10 B．88 C．143 D．232
5.如图所示，在平行六面体中，，，，则
A． B．
C． D．
6. 如图所示，将绘有函数部分图像的纸片沿轴折成直二面角，若之间的空间距离为，则
A． B． C． D．
7.已知圆,过点的直线将圆的面积分割成两个部分，若使得这两部分的面积之差最大，则直线的方程为
A． B． C． D．
8. 如图所示，已知是椭圆的左、右焦点，为椭圆的上
顶点，在轴上，，且是的中点，为坐标原点，若点到直线的距离为，则椭圆的方程为
A． B．
C． D．
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.
9. 下列叙述正确的是
A．集合，，，则“任取，使得”的概率为
B．向量，，若，则
C．若构成空间的一个基底，则也可以构成空间的一个基底
D．“直线与互相平行”是“直线与的斜率相等”的充分不必要条件
10. 扎马钉(图1)，是古代军事战争中的一种暗器.
如图2所示，四个钉尖分别记作、、、，
连接这四个顶点构成的几何体为正四面体，组成
该“钉”的四条等长的线段公共点为，设，
则下列结论正确的是
A．
B．为正四面体的中心
C．
D．四面体的外接球表面积为
11. 古希腊著名数学家阿波罗尼斯发现“若为平面上相异的两点，则所有满足：且的点的轨迹是圆”，后来人们称这个圆为阿波罗尼斯圆.在平面直角坐标系中，，若，则下列关于动点的结论正确的是
A. 点的轨迹方程为
B. 面积的最大值为
C. 在轴上必存在异于的两定点，使得
D. 若点，则的最小值为
12．抛物线的焦点为，点，，都在抛物线上，且，则下列结论正确的是
A．抛物线方程为 B．是的重心
C． D．
第II卷
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.
13.已知分别是椭圆的左、右焦点，为椭圆的上顶点，且，则椭圆的离心率
14.四叶草也叫幸运草，四片叶子分别象征着:成功、幸福、平安、
健康,表达了人们对美好生活的向往.梵克雅宝公司在设计四叶草
吊坠的时候，利用了曲线方程(如图所示)
进行图案绘制.试求曲线围成的封闭图形的面积
15.如图，棱长为的正方体中，为线段上的动点，为线段上的动点，则长度的最小值为
16.已知一元二次函数满足：，且恒成立，则
若，则数列的前项和为
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.本小题满分10分
已知直线
（1）求过点，且与直线平行的直线的方程；
（2）直线与圆相交于两点，求线段的长.
18.本小题满分12分
已知等差数列满足：，
（1）求等差数列的通项公式；
（2）若，求数列的前项和.
19.本小题满分12分
在中，角所对边分别为，且.
（1）求角的大小；
（2）若，，求的面积.
20.本小题满分12分
如图，某种风筝的骨架模型是四棱锥，四边形是等腰梯形，,,平面，，,
，在上.
（1）为保证风筝飞行稳定，需要在处引一尾绳，使得，
求证：直线平面；
（2）实验表明：当时，风筝表现最好，求此时直线与平面
所成角的正弦值.
21.本小题满分12分
已知双曲线的渐近线方程为，
且虚轴长为.
（1）求双曲线的方程；
（2）若直线与双曲线相交于不同的两点，且满足，
求的取值范围.
22.本小题满分12分
已知函数的图象上有一点列，
点在轴上的射影是，且，且，．
（1）求证：是等比数列，并求数列的通项公式；
（2）对任意的正整数，当时，不等式恒成立，
求实数的取值范围；
（3）设四边形的面积是，求证：.
高二数学试卷 第 1 页 共8页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）
