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桂林市 2025-2026学年度下学期非毕业年级日常考试题库卷
高二年级数学参考答案及评分标准
一、选择题：本题共 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 B C C D B A A
二、选择题：本题共 3小题，每小题 6分，共 18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 6分，部分选对的得部分分，有选错的得 0分．
题号 9 10 11
答案 ABC ABD AC
三、填空题：本题共 3小题，每小题 5分，共 15分．
5 1
12．1 13．6.6 14．
2
四、解答题：本题共 5小题，共 77分，解答应写出文字说明、证明过程及演算步骤.
15．(本小题满分 13分)
解析：（1）设等差数列 an 的首项为a1，公差为 d，
2 a1+d + a1+4d =3a1+6d=21由题意得 6a1+15d=51
，解得a1=1,d=3，······································ 4分
所以an=1+ n 1 ×3=3n 2.················································································6分
（2）由（1）知an=3n 2，则an+1=3n+1······························································7分
b = 1所以 n =
1 = 1 1 1 ···························································· 11分
anan+1 3n 2 3n+1 3 3n 2 3n+1
T = 1 1 1 + 1 1 + + 1 1得 n =
1 1 1 = n .··········································· 13分
3 4 4 7 3n 2 3n+1 3 3n+1 3n+1
16．(本小题满分 15分)
证明：(1)因为 PD 底面 ABCD， BC 底面 ABCD
所以 BC PD ······································1分
又因为 BC CD ·································· 2分
PD CD D，且CD,PD 面 PCD
所以 BC 面 PCD ·······················································································3分
又DE 面PCD，所以 BC DE ··································································4分
又因为 PD AB DC， E是 PC的中点
所以DE PC ···························································································· 5分
又 PC BC C ， PC,BC 面 PBC
所以DE 面 PBC ·······················································································6分
(2)不妨设 AB 1
因为 PD 底面 ABCD，且底面 ABCD是正方形，所以DP,DA,DC 三条直线两两垂直，
以D为坐标原点，建立如图所示空间直角坐标系················································ 7分
则 P(0,0,1),B(1,1,0)， E(0, 1 , 1) ···································································8分
2 2

所以DB (1,1,0),DP (0,0,1) ······································································9分

设平面 PBD的一个法向量为m (x, y, z)

m DP z 0
则 ·················································································10分
m DB x y 0
令 x 1，则 y 1

所以m (1, 1,0) ······················································································· 11分

由(1)得DE (0, 1 , 1)是平面 PBC 的法向量··················································· 13分
2 2
1 1
所以 cos m,DE
m DE 1
2 ··············································14分
|m | | DE | 1 22
2
平面 PBC 与平面 PBD 1夹角的余弦值为 ······················································· 15分
2
17. (本小题满分 15分)
2
解析：(1)设M (x, y) y y y 1，由题意有 kAM kBM 2 ·················2分x 2 x 2 x 2 4
x2 4y2 2 x
2
化简得： ，即 2y2 1，x 2
2
2
故所求动点 M
x
的轨迹方程为 2y2 1 (x 2) ············································4分
2

(2)设直线 TP的倾斜角为 ， (0， )
2
由题意得 2 PTQ ，又 PTQ ，所以 2 ，得
2 2 4
故 kTP 1，kTQ 1 ······················································································· 5分
因为T (1 1 1， )，所以直线 TP： y x 1 3，即 y x ········································· 6分
2 2 2
3
y x
(i) 2 x
2
联立 ，得 2(x
3
)2 1 ························································ 8分
x2 2 2 2 2y 1
2
化简得5x2 12x 7 7 0，解得 x 或x 1 (舍去)
5
P(7 1故 ， ) ······························································································10分
5 10
1
(ii) 直线 TQ： y (x 1)，即 y x 1
2 2
y 1 x
2 x2 1
联立 ，得 2(x )2 12
x 2 2 2y2 1
2
化简得5x2 4x 1 1 0，解得 x 或x 1 (舍去)
5
1 7
故Q( ， ) ······························································································12分
5 10
|TP | ( 7 1)2 ( 1 1)2 2 2 2 2所以 ( )2 ( )2 ，
5 10 2 5 5 5
|TQ | ( 1 1)2 ( 7 1)2 ( 6)2 ( 6)2 6 2 ，
5 10 2 5 5 5
因为 PTQ 1 1 2 2 6 2 12 ，所以 s
2 TPQ
|TP | |TQ | ··························· 14分
2 2 5 5 25
12
故△PTQ的面积是 ·················································································· 15分
25
18．(本小题满分 17分)
解析：(1)当 a 1时， f (x) 1 x 2ln x ， x (0, )，此时切点坐标为 (1,0)，
x
················································································································· 1分
f (x) 1+ 1 2 x
2 2x 1
2 2 ，······························································· 2分x x x
f (1) 0，即切线斜率为 0········································································· 3分
所以切线方程为 y=0.·····················································································4分
f (x) 1+ 1 2a x
2 2ax 1
(2) 2 x x x2
,
令 g(x) x2 2ax 1 ,则 =4a2 4 ································································ 5分
① 当 0即 1 a 1时, g(x) 0恒成立，即 f (x) 0恒成立，
f (x)在（0，+ ）上单调递增··········································································· 6分
② 当 0即 a 1或a 1时, 2 2令 g(x)=0得 x1 a a 1， x1 a+ a 1
若 a 1，则0 x1 x2 ，
当 x (0, x1)时, g(x) 0 ,即 f (x) 0 , f (x)单调递增
当 x (x1，x2 )时, g(x) 0 ,即 f (x) 0 , f (x)单调递减
当 x (x2，+ )时, g(x) 0 ,即 f (x) 0， f (x)单调递增·································· 8分
若 a 1，则 x1 x2 0，由 g(0)=1>0知 x （0，+ ）时 g(x) 0恒成立，
即 f (x) 0恒成立， f (x)在（0，+ ）上单调递增···············································9分
综上: 当 a 1时， f (x)在（0，+ ）上单调递增, 无单调递减区间
a 1 f (x) (0,a a2当 时， 在 1)， (a+ a2 1，+ )上单调递增，
(a a2在 1，a a2 1)上单调递减·························································10分
1 1
-
(3)由 nem = me n 得， lnm ln n 1 1 ln m m n ，即 ······························ 11分
m n n mn
由于m,n m n> 0，所以 0 m，所以 1，
mn n
t m 1 m tn ln t tn n t 1 n t 1 t 1不妨令 ，则 ， 2 ，所以 ，m n tn tn t ln t ln t
m n t 1 t 1 t
2 1
所以 ··································································· 13分
ln t t ln t t ln t
t 2 1 1 1
方法一：要证m- n > 2，即证 2，即证 ln t (t )， t (1, ) ·········· 14分
t ln t 2 t
2
令 k(t) ln t- 1 (t 1) k (t) 1- 1 (1 1 ) (t 1) ，则 2 2 ，显然 k (t) 0 ············ 15分2 t t 2 t 2t
k(t) t (1, ) k(x) k(1)=0 ln t 1 1所以 在 上单调递减，所以 ，即 (t )恒成立，
2 t
即m- n > 2成立························································································· 17分
2
方法二: 要证m- n > 2 t 1，即证 2 2，即证 t 1 2t ln t， t (1, ) ············14分
t ln t
令 k(t) t 2 1 2t ln t，则 k (t) 2t 2(1 ln t) 2(t 1 ln t) ，
令 p(t) t 1 ln t ，则 p (t) 1 1 t 1 ，当 t 1时，p (t) 0恒成立，p(t)在（1，+ ）
t t
上单调递增，又 p(1)=0，所以 p(t) 0在 (1, )上恒成立，
所以 k (t) 0在 (1, )上恒成立， k(t)在（1，+ ）上单调递增··························· 15分
又 k(1)=0，所以 k(t) 0在 t (1, )恒成立，即 t 2 1 2t ln t在 (1, )上恒成立，
即m- n > 2成立.························································································ 17分
19．(本小题满分 17分)
解析：(1)X的取值为 2，4，6·········································································· 1分
P(X 2) 2 2 4 ···········································································2分
5 5 25
P(X 4) 2 3 3 2 16 ································································ 3分
5 5 5 3 25
P(X 6) 3 1 1 ············································································· 4分
5 3 5
所以 X的分布列为：
X 2 4 6
4 16 1
P
25 25 5
···········································································································5分
EX 2 4 4 16 6 1 102 ··························································· 6分
25 25 5 25
(2)(i)第 4次摸球后，游戏结束可以分为以下三种情况：
2 3 1 1 2
①第 1次摸到红球，第 2、3、4次摸到黑球，其概率为 ······· 7分
5 5 3 7 175
3 2 1 1 2
②第 1次摸到黑球，第 2次摸到红球，第 3、4次摸到黑球，其概率为
5 3 3 7 105
···········································································································8分
3 1 6 1 6
③第 1、2次摸到黑球，第 3次摸到红球，第 4次摸到黑球，其概率为
5 3 7 7 245
···········································································································9分
P 2 2 6 202所以 4 ·························································· 10分175 105 245 3675
(ii)依题意知，第 n+1次摸球后，游戏结束的概率为 Pn 1
第 n+1次摸球后，游戏结束可以分为以下两种情况：
2
①第 1次摸到红球，游戏结束的概率为 Pn ···············································12分5
②第 1次摸到黑球，从第 2次到第 n次摸球中只摸到一次黑球，第 n+1次摸到黑球，
游戏结束的概率为：
3 [1 (6)n 2 (2) 1 (6)n 3 (2)2 1 (6)n 4 (2)n 2 1] 1 ········14分
5 3 7 3 3 7 3 3 7 3 3 7
7
(6)n 9 (2 )n ·········································································15分
40 7 40 3
P 2 P 7 (6 n 9 2由全概率公式得 n 1 n ) ( )
n ·····································16分
5 40 7 40 3
P 2 P 7 6故 n 1 n ( )
n 9 2 ( )n ······················································· 17分
5 40 7 40 3桂林市2025-2026学年度下学期非毕业年级日常考试题库卷
7.已知地物线y2=4h的焦点为F,若点A(,-2]在谈抛数线上，MF-
A.2
B.2v2
C.4
D.5
高二年级数学
8.若关干￥的不等式”一lr 网+1恒成立，则实效a的欧盆范阅是
A.【-第，1
B.《-01
C.（-s.-1
(奇该周时120分外，满分50分)
二、选择预：本丽共3小丽，每小贸6分，共18分.在每小圆给出的选网中，有多项符合翻目要求。
注意事项：
全部选对的得6分，部分选对的得部分分，有选带的得0分.
【,荟卷首，背生异必用黑色字诚创花我签字笔将白己的校名灶名、非奴，学号和准考证号
9.己知直就ar+y+3=0,图M:+y之-2x-y=0,则
班牙在答是十上。齐条形码情贴在茶题卡的“条特码林粘处”。
A,回的半轻为r-V5
2.作答这样罚叶，迹白斗小题签常后，用2B船笔花答题卡上对应随日遂项的器案信岳点杂
黑：如君政功，局樟应擦干净后，再这涂其他答常，云常不能答在试卷上。
B.图心坐标为(12)
3,非滋补通必频用黑色李连阳笔发签李笔作爹，多策必领写在答题卡各斑日指完区盛内和
C.当战1平分四时c"-5
应位五上：如需放动，先划掉原来的茶米，然后再写上新的器紧：不难使闲铅笔和涂改液，
D,当直线！与圆制相切时a=2
不按烈上安求作器的答案无姓。
I0.已知致列{a.的前r项和为S,下列说法正确的是
4.芳生必顷保持答卡的整法。考扰菇束后，将答期卡文四。
一、选择题：本题共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
A.若3.=n',则{a是每接数列
一

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