2023一2024学年度第一学期教学质量检查
高二数学
一、单项选择题：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.请把正确选项在答题卡中的相应位置涂黑.
1.在等比数列{an}中，4=2，a2a4=16,则a5=
A.4
B.8
C.10
D.12
2.若直线1的一个方向向量是L,-V3),则直线1的倾斜角是
A.30°
B.60°
C.120°
D.150°
3.已知平面C的一个法向量为m=(2,3,一1)，平面B的一个法向量为n=(4,k,2),若a上B,
则k=
A.-2
B.2
C.6
D.-6
4.已知点P在抛物线y2=4x上，且点P与点A(3,0)的距离和点P到直线x=-1的距离相等，
则PA=
A.1
B.2
C.3
D.4
5.若{a,,武构成空间的一个基底，则下列各组中不能构成空间的一个基底的是
A.a+i,i+c,c+武
B.fa,b,a+b+c
C.fa+b,b+c,c}
D.a-i,-c,c-欧
6.东莞鸿福路大桥是一座系杆拱桥，其圆拱结构可近似看作圆的一部分，经查询资料知该拱桥（如
下图)的跨度AB约为126米，拱高OP约为9米，该拱桥每隔约7米用一根吊杆连接圆拱与系杆，
则与OP相距35米的吊杆MN的高度约为
(参考数据：√494≈22.23)
A.7.3米
B.6.3米
C.5.3米
D.4.3米
7.已知双曲线二-X=@>06>0的左，石焦点分别为，E,且孩双曲线与圆x+y2=a2+b2在
第二象限的交点为点P,若m∠PFF=2,则双曲线的离心率为
A.3
B.2
c.5
D.√6
8.在数列的每相邻两项之间插入此两项的和，形成新的数列，再把所得新数列按照同样的方法进
行构造，可以不断形成新的数列.现对数列1,2进行构造，第1次得到数列1,3,2；第2次得
高二数学第1页（共4页）
到数列1,4,3,5,2；…依次构造，记第nn∈N)次得到的数列的所有项之和为T,则T,=
A.1095
B.3282
C.6294
D.9843
二、多项选择题：本大题共4小题，每小题5分，共20分.在每小题给出的四个选项中，有多项
符合题目要求，全部选对的得5分，有选错的得0分，部分选对的得2分.请把正确选项在答题
卡中的相应位置涂黑，
9.已知数列{a.}的前n项和Sn=-n2+9n，则下列说法正确的是
81
A.Sn的最大值为
B.
巴}是等差数列
C.{an}是递减数列
D.as+as+as+a+as =0
10.已知圆C1:x2+y2-4x+2my+m2=0和圆C2:x2+y2-4y-12=0,则下列说法正确的是
A、若m=0,则圆C和圆C2相离
B.若m=0,则圆C和圆C2的公共弦所在直线的方程是x一y一3=0
C.若圆C和圆C2外切，则m=42-2
D.若圆C和圆C2内切，则m=-2
11.已知曲线C:xx+4yy=4,则
A.曲线C在第一象限为椭圆的一部分
B.曲线C在第二象限为双曲线的一部分
C.直线y=
方1与由线C有阿个交点D.当线y=怎41与瓮c有三个交点
12.在如图所示的试验装置中，ABCD和ABBF均为边长为1正方形框架，且它们所在的平面互
相垂直活动弹子M,N分别在对角线AC,BF上移动，且CM=2CA,=2FB(0<九<1).则
下列结论正确的是
A.32∈(0，)，MN⊥AC
B.Me0,D,Mw≥
2
C.V2∈(0，)，MN/平面CEF
D.3入∈(0,1)，平面MWB⊥平面MAF
三、填空题：本大题共4小题，每小题5分，共20分.请把答案填在答题卡的相应位置上
13.已知空间两点A(1,2,3),B(2,4,5),则与AB方向相同的单位向量的坐标是
14.数列{a}满足an=
+3+2a∈N),则数列{a,}的前10项和为
15.一条光线从点A(0,1)射出，经直线y=x反射后与圆C:(x-2)2+(y-4)=1相切，则反
射光线所在直线的方程可以为
(写出满足条件的一条直线方程即可)
高二数学第2页（共4页）2023－2024学年度第一学期教学质量检查
高二数学 参考答案
一、单项选择题
题号 1 2 3 4 5 6 7 8
答案 B C A C D B C B
二、多项选择题（全部选对的得 5分，选对但不全的得 2分，有选错的得 0分）
题号 9 10 11 12
答案 BC BD ABC ACD
三、填空题（16题第一空 2分，第二空 3分）
(1 , 2 213. , ) 514. 15. x 1或15x 8y 15 0（一个方程即给满分） 16. 3
3 3 3 12
四、解答题
17. 解：(1)因为 BD1 BA AD DD1 a b c .·················································3分
(2) A1D AD AA1 b c，············································································· 4分
因为 a c b c 2 2 cos60 2，a b 2 2 cos90 0 2 2， c b 4，············ 5分
BD AD a b c b c 2 2所以 1 1 a b a c b b c b c c 0 2 4 4 2 ，
······················································································································· 6分
2 2 2 2 BD1 a b c a b c 2a b 2a c 2b c 4 4 4 0 4 4 2 3 ，
······················································································································· 7分
2 2 2
A1D b c b c 2b c 4 4 4 2，············································8分
所以 cos BD , A
BD1 A1D 2 3
1 1D 2 3 2 6 ，···············································9分BD1 A1D
3
所以异面直线 BD1与 A1D夹角的余弦值为 ．·····················································10分
6
18. 解：（1）由题意可得 2b 2 3，b 3，···························································1分
e c 1 1 ，即 c a，························································································2分
a 2 2
高二数学参考答案 第 1 页 (共 6 页)
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因为 a2 b2 c2，得 a 2，··················································································3分
x2 y2
所以，椭圆C 的方程为 1；······································································ 4分
4 3
（2）椭圆C 的右焦点 F2 (1, 0) ，············································································ 5分
直线 l的倾斜角为 45 ，则直线 l的斜率为 1，··························································· 6分
所以直线 l的方程为 y x 1，············································································· 7分
y x 1
由 x2 y2 可得7x2 8x 8 0，·······································································8分
1
4 3
设交点M (x1, y1),N (x2 , y2 )
x
8
1 x 2
7
则 ，······························································································· 9分
x1x
8
2 7
|MN | (1 1)[(x1 x
2
2 ) 4x1x2 ]，··································································· 10分
(1 1)[(8)2 4( 8 24 )] ，·········································································11分
7 7 7
24
所以线段M N 的长为 ····················································································12分
7
19. 解（1）：设等差数列{an}的公差为 d ，正项等比数列{bn}的公比为 q(q 0)，
由题意 a1 1， a4 7和 a4 a1 3d ，得7 1 3d ，则 d 2，·······························1分
故数列{an}的通项公式为 an a1 (n 1)d 1 2(n 1) 2n 1．·····························2分
2b1 a1 a3
由题意b1是 a1和 a3的等差中项， a5是b1和b3的等比中项，得 2 ，···············3分
a5 b1b3
2b 6 b 3
由 a1 1， a3 2 3 1 5， a5 2 5 1 9
1 1
，得 ，则 ，··············4分
81 b1b3 b3 27
2 b1 3
由b3 b1q 且q 0，得 ，·········································································5分
q 3
{b } b b qn 1 n 1 n故数列 n 的通项公式为 n 1 3 3 3 ．················································ 6分
高二数学参考答案 第 2 页 (共 6 页)
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（2）由题意和（1），得 a2， a4， ， a2n构成了首项为 a2 3，公差为 2d 4的等差数列，
······················································································································· 7分
b1，b3， ，b2n 1构成了首项为b1 3，公比为 q2 9的等比数列，··························8分
Sn (a2 b1) (a4 b3) (a2n b2n 1)，·························································· 9分
(a2 a4 a2n ) (b1 b3 b2n 1)，····························································10分
(3 4n 1)n 3(1 9n )
，···············································································11分
2 1 9
(2n 1)n 3(9
n 1)
.······················································································ 12分
8
20. 解：（1）如图，以D为原点，分别以DA，DC，DP为 x轴， y轴， z轴，建立空间直角
坐标系，则D 0,0,0 ， A 3,0,0 ，C 0,6,0 ， B 3,6,0 ， P 0,0,6 ，····················1分
因为 E是 PC的中点，则 E 0,3,3 ，····································································· 2分
因为 PB 3PF，得 F 1,2, 4 ，············································································ 3分
所以DF 1,2,4 ，DE 0,3,3 ， AC 3,6,0
n DF x 2y 4z 0
设平面DEF 的法向量为 n x, y, z ，则 ，
n DE 3y 3z 0
令 y 1则 z 1， x 2，所以n 2,1, 1 ，·······················································6分
所以 AC n 3 2 6 1 0 1 0，···························································7分
所以 AC n，所以 AC / /平面DEF .··································································· 8分
(2)因为DA 3,0,0 , 2所以DA n 3 2 0 0 6 , n 22 12 1 6 ,········· 10分
DA n 6
所以点 A到平面DEF的距离为 6 ··················································12分
n 6
高二数学参考答案 第 3 页 (共 6 页)
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21. 解：（1）当 n 1时， a1 S1 2S1 1，得 S1 1，故 S1 1 2，·························1分
由 na1 (n 1)a2 an 2Sn n(n N
)①，
得 (n 1)a1 na2 ... 2an an 1 2Sn 1 (n 1)②，·············································2分
② ①，得 a1 a2 ... an an 1 2Sn 1 2Sn 1，
得 Sn 1 2Sn 1 2Sn 1，即 Sn 1 2Sn 1，························································3分
Sn 1 1
所以 S 2n 1 1 2(Sn 1)，即 S 1 ，··························································· 4分n
所以，数列{Sn 1}是以 2为首项， 2为公比的等比数列.··········································· 5分
n n
（2）由（1）知 Sn 1 2 ，故 Sn 2 1，
n 1 n 1
当n 2， an Sn Sn 1 2 ，显然 a1 1也适合上式，故 an 2 (n N )，·········· 6分
设T
1 2 n
n a1 a2 a
，
n
1 2 3
则Tn 0 1 2
n 1 n
2 2 2 2n 2
2n 1
③，
1T 1 2 3 n 1 nn 1 2 3 n 1 n ④，································································7分2 2 2 2 2 2
1 n
1 1 1 1 1 n
1 ( ) n 1
③ ④得 Tn 0 1 2
2 2 (n 2) ( )n，········· 8分
2 2 2 2 2n 1 2n 1 1 2
n 2
2
T 4 n 2所以 n n 1 ，···························································································9分2
高二数学参考答案 第 4 页 (共 6 页)
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1 2 n n n n
由 4 a a a 1012 ，即Tn 4 ，得 2
n 1 1012，··················· 10分
1 2 n 1012 n 2
c n 2n 1 c c n 1 n
2
2n 2n 1 n 3n 4 2n 1设 n ，则n 2 n 1 n
0，
n 3 n 2 (n 2)(n 3)
所以，数列{cn}为递增数列，··············································································11分
c 11 210因为 11 1012
12 11
， c 2 1012，
13 12 14
1 2 n n
所以，满足不等式 4 na a a 1012 的 的最小值为12．····························· 12分1 2 n
22. 解：（1）由题可知，圆心C到定点 (1,0)的距离与到直线 x 1的距离相等，则点C的轨迹
为以 (1,0)为焦点， x 1为准线的抛物线，··························································· 2分
所以其轨迹方程为 y2 4x .·················································································· 4分
（2）由题可知直线 PQ的斜率不为 0，设 PQ的方程为 x ty m，
x ty m
设 P(x1, y1),Q(x2 , y2 ) ，联立方程组 2 ，得 y
2 4ty 4m 0，····················· 5分
y 4x
y1 y2 4t
所以 y1y2 4m ，················································································· 6分
16t
2 16m 0
又因为 AP AQ，则 AP AQ 0，
即 (x1 1)(x2 1) (y1 2)(y2 2) 0 ，································································ 7分
化简得 x1x2 (x1 x2 ) y1y2 2(y1 y2 ) 5 0 ，
y2 y2
消元得 1 2 (ty m ty m) y y 2(y y ) 5 0
4 4 1 2 1 2 1 2
(y1y )
2
即 2 y1y2 (t 2) (y1 y2 ) 2m 5 016
所以m2 4t2 6m 8t 5 0，···········································································8分
高二数学参考答案 第 5 页 (共 6 页)
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因式分解得 (m 2t 5)(m 2t 1) 0，
当m 2t 1 0时，直线 PQ经过点 A，不合题意，舍去·········································· 9分
当m 2t 5 0时，m 2t 5，直线PQ： x ty 2t 5·····································10分
d | 4t 4 |
2
A PQ 4
1 2t t
2 4 1
2
1 4 2此时点 到直线 距离 1 t 2 1 t ，···············11分t
t
当且仅当 t 1时取等号，此时 16t2 16m 0满足要求，
所以点 A到直线 PQ距离的最大值 4 2 .································································12分
高二数学参考答案 第 6 页 (共 6 页)
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