福建省福州市福清市高中联合体2023-2024学年高二上学期期末考试质量检测数学试题2023—2024 学年第一学期高二年期末考试质量检测
数学参考答案及评分细则
评分说明：
1．本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考
查内容比照评分标准制定相应的评分细则.
2．对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难
度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数的一半；如果后继
部分的解答有较严重的错误,就不再给分.
3．解答右端所注分数,表示考生正确做到这一步应得的累加分数.
4．只给整数分数.选择题和填空题不给中间分.
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．
1.D 2.C 3.C 4.B
5.A 6.A 7.C 8.B
二、多项选择题：本题共 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
是符合题目要求的．全部选对的得 5分，部分选对的得 2分，有选错的得 0分．
9.BC 10.ABC 11.ABC 12.BD
二、填空题：每小题 5分,满分 20分．
10
13. x 2y 3 0 14. 14 15. 21 16. 1,
2 7 2
三、解答题：本大题共 6小题，共 70分．解答应写出文字说明、证明过程或演算步骤．
17．解：（1）因为点O(0,0) , N (4,0) ,
所以线段ON的中点为 E(2,0) ·············································································· 2分
因为直线ON的斜率为 k 0 ,所以ON垂直平分线的斜率不存在.·································· 4分
所以ON垂直平分线的方程为 x 2；··································································· 5分
（评分说明：若直接写出 x 2 ,也给 5分）
（2）解法一：因为 A关于直线 y x对称,则可设 A的方程为 x a 2 y a 2 r2 ,······7分
(0 a)2 (0 a)2 r2
又因为点O(0,0) , N (4,0)
在 A上,所以 2 2 2 ,········································ 8分
(4 a) (0 a) r
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a 2
解得 ,·································································································· 9分
r 2 2
所以 A的标准方程为 (x 2)2 (y 2)2 8．·························································· 10分
解法二：因为直线ON与直线 y x的交点为圆心，
y x x 2
由 x 2，解得 y 2，····················································································· 7分
故圆心 A 2,2 .································································································· 8分
又因为 r (2 0)2 (2 0)2 2 2 .·········································································9分
所以 A的标准方程为 (x 2)2 (y 2)2 8．·························································· 10分
18.解：(1)设 N (x, y) ,Q x0 , y0 ···············································································1分
x x0 1 2 x0 2x 1
由题意知 y ,则 y 2y ,········································································· 3分 y 0 0
2
又点Q在 M 2上,所以 2x 1 1 2y 3 2 4,························································· 5分
2 3 2
所以 的方程为 x 1 y
1．···································································6分
2
(2)因为 O的直径为 2,
故圆心为O 0,0 ,半径 r1 1 .················································································· 8分
由（1）可知 的圆心 P(1,
3) ,半径 r2 12 .································································· 9分
3 13
所以 OP (0 1)2 (0 )2 ,······································································· 10分
2 2
13
又因为 r1 r2 0 , r1 r2 2 , 0 2 ,即 r1 r2 OP r1 r2 ,········································· 11分2
所以点 N的轨迹与 O的位置关系是相交．························································· 12分
19.解：（1）因为 E , F分别为棱 AB , PB中点,所以 EF AP ,
因为 EF 平面 PAD , AP 平面 PAD ,所以 EF 平面 PAD ,··········································· 1分
因为 AB CD ,CD 2 , AB 4 , E为棱 AB中点,所以 AE CD .
所以四边形 ADCE为平行四边形,故 AD CE .·························································· 2分
因为CE 平面 PAD , AD 平面 PAD ,所以CE 平面 PAD ,···········································3分
因为 EF CE E , EF ,CE 平面CEF ,所以平面CEF 平面DAP；································4分
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（2）因为平面 PAB 平面 ABCD ,平面 PAB 平面 ABCD AB , AD 平面 ABCD , AD AB ,所以 AD
平面 PAB .因为 AD CE ,所以CE 平面 PAB，·························································5分
因为 PA PB ,所以 EP AB .以 E为原点,分别以 EP , EB , EC所在直线为 x轴、y轴、z轴,建立如图所
示的空间直角坐标系.························································································6分
可得 D 0, 2,2 , P 2,0,0 , B 0,2,0 ,C 0,0,2 .
所以 DP 2,2, 2 , BP 2, 2,0 , BC 0, 2,2 .···························7分
n BP 0, 2x 2y 0,
设平面 PBC的法向量 n x, y, z ,则 所以 ···· 8分
n BC 0. 2y 2z 0.
取 x 1 ,则 n 1,1,1 .···························································· 9分
cos DP,n DP n 2 2 2 1则 DP n 2 3 3 3 .······································································ 10分
设直线 DP与平面 PBC所成角为 ,则 sin
1
cos DP,n
3 .········································11分
因为 [0， ] ,所以 cos 1 sin2 1 2 2 1 ( )2 2 .3 3
2 2
故直线 DP与平面 PBC所成角的余弦值为 .······················································ 12分
3
p
20.解：（1）因为 MF 2 ,根据抛物线的定义,知点M到 的准线 y 的距离为2 2 ,········ 1分
p
因为点M到 y轴的距离为 2 p ,所以点M的坐标为 2 p，2 2 ,································· 3分
2
2 p
因为点M 在 : x 2py p 0 上,所以 2p 2p 2 ,··································· 4分
2
即 p p 2 0 ,因为 p 0 ,所以 p 2 ,···································································5分
所以 的方程为 x2 4y ···················································································· 6分
（2）由（1）可知 的焦点 F (0,1) , l : x 2 y 2 0经过 的焦点,······························· 7分
x 2y 2 0
由 2
x2
,得 x 2x 4 0．
4y
设 A(x1, y1) ,B(x2 , y2 ) ,则 x1 x2 2 , x1x2 4．························································ 8分
所以 | x1 x2 | (x1 x2)
2 4x1x2 2
2 4 4 2 5 ,···················································· 10分
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S 1 | OF | | x x | 1因此△ABO的面积 △ABO 2 1
2 2 5 5．2 ·········································· 12分
21. 解法一：（1）在正方形 ABCD中, AB BM , AD DN .
所以在四棱锥 P AMCN 中, AP PM , AP PN .························································1分
因为平面 APM 平面 APN＝AP , PM 平面 APM , PN 平面 APN ,
所以 MPN 为二面角M AP N 的平面角.····························································· 2分
2 2 2
因为在正方形 ABCD中, CM CN MN ,
所以在四棱锥 P AMCN 2 2 2中, PM PN MN .·······················································3分
所以 PM PN ,即二面角M AP N 为直二面角.所以平面 PAM 平面 PAN .··················· 4分
（2）由（1）得, PA,PM ,PN两两垂直.以 P为原点,分别以 PN , PM , PA所在直线为 x轴, y轴, z轴,建
立如图所示的空间直角坐标系.······························ 5分
所以M (0,2,0) , N (2,0,0) ,T (0,0,1) , A 0,0,4 .
可得MN (2, 2,0) ,MA 0, 2,4 ,MT 0, 2,1 .··········· 6分
设平面MNA的法向量m x1, y1, z1 ,
m MN 0, 2x 2y 0,
则
1 1
所以
m MA 0. 2y 4z 0.
···························· 7分
1 1
取 z1 1 ,则m 2,2,1 .························································································ 8分
n MN 0, 2x 2y 0,
设平面MNT
的法向量 n x2 , y
2 2
2 , z2 ,则 所以 2y z 0.······························9分 n MT 0. 2 2
取 y2 1 ,则 n 1,1,2 .······················································································· 10分
则 cos m,n
m n 6 6
m n 3 6 3 .·········································································· 11分
所以平面MNA与平面MNT 6夹角的余弦值为 .·····················································12分
3
解法二：（1）在正方形 ABCD中,M , N分别是 BC ,CD的中点,
所以在四棱锥 P AMCN 中, PM CM , PN CN .
所以△PMN △CMN ,所以 MPN MCN 90 ,即 PM PN .······································· 1分
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又在正方形 ABCD中, AB BM ,所以在四棱锥 P AMCN 中, AP PM .····························2分
由于 AP PN P , AP , PN 平面 APN ,所以 PM 平面 APN .········································· 3分
因为 PM 平面 APM ,所以平面 APM 平面 APN .·····················································4分
（2）依题, AP PM , AP PN , PM PN P ,
所以 AP 平面PMN .所以 AP MN .
设 AC MN E ,可得 AP PE .
由 AC MN , AP AC A ,得MN 平面 PAC .
因为MN 平面 AMN ,所以平面 AMN 平面PAC .
过 P作 PO垂直 AC ,垂足O .以O为原点,过O作MN 的
平行线为 x轴,OC ,OP所在直线为 y轴、 z轴,建立如图所示的空间直角坐标系.·············· 5分
4
在 Rt△APE中,OE 2 ,OA 8 2 ,OP .
3 3 3
M ( 2, 2
4
可得 ,0) , N ( 2, 2 ,0) , A(0, 8 2 ,0) , P 0,0, 3 .············································· 6分3 3 3
8 2 4 3 8 2 4
所以 AP (0, , ) , AT (0, , ) (0,2 2,1) ,
3 3 4 3 3
2 2
可得T (0, ,1) ,MT ( 2, 2,1) , NM 2 2,0,0 .··················································7分
3
m MT 0, 2x 2y z 0,
设平面MNT 的法向量m x, y, z ,则 所以 ························ 8分
m NM 0. 2 2x 0.
取 y 1 ,则m 0,1, 2 .······················································································· 9分
取平面 AMN的一个法向量 n 0,0,1 ,·································································· 10分
cos m,n m n 2 6则 m n 3 3 .··············································································11分
6
所以平面MNA与平面MNT夹角的余弦值为 .·····················································12分
3
22.解：（1）设点 P的坐标为 x，y ,因为点 A的坐标为 2，0 ,
y
所以直线 AP的斜率 kAP x 2 x 2 ,····································································1分
y
同理,直线 BP的斜率 kBP x 2 x 2 ,····································································· 2分
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y y 1
由已知,有 x 2 x 2 x 2 4 ,········································································· 3分
2
化简, x得 的方程为 y2 1 x 2 .·····································································4分
4
（评分说明：没写出 x 2 ,只扣 1分,不重复扣分.）
（2）点M位于定直线 x 4上．·········································································· 5分
理由如下：
设C(x1, y1) ,D(x2 , y2 ) ,
x ty 1 0
由 x2 2 2
y2
得 (t 4)y 2ty 3 0 ,
1 4
2t 3
所以 4t2+12(t2 4) 16 t2+3 0 ,且 y1 y2 2 ，y1·y2 2 ，······························ 6分t 4 t 4
因为 A , B两点的坐标分别为 2，0 , 2，0 ,
y1 y
直线 AC方程为 y (x 2) 2x 2 ,直线 BD方程为
y (x 2)
x 2 ,···································· 8分1 2
y y 1 (x 2) x1 2 2(x1y2 x2 y1 2y2 2y1)
由 ,得 x y x y x y 2y 2y ,······················································ 9分 y 2 (x 2) 1 2 2 1 1 2
x2 2
x1 ty1 1 2(2ty1y2 3y2 y1 )
又 x ty 1 ,代入得
x
2 2 3y y
,·····························································10分
2 1
2t y y 2t
由 y1 y2 2 , y y
3
1 21 2 得 t 4 2 , t 4 y1y2 3
即 2ty1y2 3(y1 y2 ) ,
2[3(y1 yx 2
) 3y2 y1 ]
所以 43y y ,··········································································11分2 1
所以点M在定直线 x 4上．············································································· 12分
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