山东省青岛市2024-2025学年高三上学期期初调研检测数学试题（PDF版，含答案）

山东省青岛市2024-2025学年高三上学
期期初调研检测数学试题
2024年高三年级期初调研检测
数学参考答案及评分标准
一、单项选择题：本题共 8小题，每小题 5分，共 40分．
1--8：BABD BCAC
二、多项选择题：本题共 3小题，每小题 6分，共 18分．
9．BC 10．ABC 11．ABD
三、填空题：本题共 3个小题，每小题 5分，共 15分．
12． 1 3 9 2； 13． ； 14． ．
2 8
四、解答题：本题共 5小题，共 77分．解答应写出文字说明，证明过程或演算步骤．
15．（13分）
解：（1）设事件 A表示在一次猜谜活动中有一方获胜，······································· 1分
事件 A包含两种情况：甲猜对乙猜错或甲猜错乙猜对，······································· 2分
2 1 2 1 1
则 P(A) (1 ) (1 ) ，·························································· 3分
3 2 3 2 2
1
所以在一次活动中,有一方获胜的概率为 ·························································4分
2
（2）由题意知，猜谜次数 X 可能取值为1, 2,3 ···················································5分
P(X 1) P(A) 1 ···················································································7分
2
P(X 2) P(AA) (1 1 ) 1 1 ·······························································9分
2 2 4
P(X 3) P(AA) 1 1 (1 )2 ·································································11分
2 4
所以 X 的分布列为
X 1 2 3
1 1 1
P
2 4 4
X E(X ) 1 1 2 1 3 1 7期望为 ······················································ 13分
2 4 4 4
16．（15分）
解：（1）因为 2cos A(c cos B b cosC) 2a，
由正弦定理得 2cos A(sinC cos B sin B cosC) 2 sin A，····························· 2分
即 2cos Asin(B C) 2 sin A，··································································4分
即 cos A 2 π ， A ················································································ 7分
2 4
数学答案 第 1页（共 5页）
（2）作CD AB于D，因为CD AD 1 AB，所以D在线段 AB上··············· 9分
3
BD 2所以 AB ························································································ 11分
3
所以 sin BCD BD 2 5 ， cos CD 5 BCD ·································· 13分
BC 5 BC 5
sinC sin( ACD BCD) 2 5 2 2 5 3 10所以 .·················· 15分
2 5 2 5 10
17．（15分）
解：（1）因为 PD 平面 ABCD,BC ABCD ，所以 BC PD，
又 BC CD,CD PD D ，所以 BC 平面 PCD ·········································· 2分
又DE 平面PCD，所以DE BC ，
又DE PC,PC BC C，所以DE 平面 PBC ，······································· 4分
又 PB 平面 PBC ，所以DE PB，
又 EF PB， EF DE E ，所以 PB 平面DEF ········································ 6分
（2）以D为原点，DA,DC,DP所在直线分别为 x轴， y轴， z 轴，
建立空间直角坐标系，·················································································· 7分
设DC 1，则 P(0,0,1),D(0,0,0), A(1,0,0),B(1,1,0),C(0,1,0),E(0, 1 , 1)，
2 2
1 1 EP (0, , ),PA (1,0, 1)，设 PG PB ( , , ), [0,1]，则
2 2

EG EP PG ( , 1 1 , )，
2 2

| EG PA | | 2
1
|
所以 cos 45 1 2 ，解得： ························ 10分
| EG | | PA | 2 22 3 2 1
2

DG (1 , 1 , 1故 )，
2 2 2
n DE 0 y z 0
设平面DEG的一个法向量为m (x1, y1, z
1 1
1)，则 ，即 ，
n DG 0 x1 y1 z1 0
令 y1 1，得m (0,1, 1) ··········································································· 12分
取平面DEF 的法向量为 n BP ( 1, 1,1) ···················································· 14分
设平面DEG 与 平 面DEF 的夹角为 ，
cos | m n | 6则 ·············································································15分
|m | | n | 3
数学答案 第 2页（共 5页）
18．（17分）
解:（1）由题知： 4 12 1 m，解得m 3 ···················································· 1分
设直线 P1Q1 : y x，记Q1( x2 , y2 ),P2 (x2 , y2 ),Q2 ( x3 , y3),P3(x3 , y3) ，
由C和直线 P1Q1均关于原点对称得：Q1( 1, 1)，即 P2坐标为 (1, 1) ···················· 3分
所以直线 P2Q2 : y 1 x 1，即直线 P2Q2 : y x 2，
y x 2
联立方程 2 ，得3x 4x 7 0 x 1
7 7
；解得 ，即 x ，
4x
2 y2 3 3 3 3 3
13 7 13
又因为 y3 x3 2 ，所以P3坐标为 ( , ) ········································· 5分3 3 3
（2）（法 1）设直线 PnQn : y x xn yn ，代入 4x
2 y2 3得：
3x2 2(x y )x (x y )2n n n n 3 0 ，··························································· 6分
x 2 2 5 2所以 n xn 1 (xn yn )，解得 xn 1 xn (xn yn ) xn yn ，3 3 3 3
又因为 yn 1 x x y
8 x 5n 1 n n n yn ，··················································· 9分3 3
因为 2x1 y
2x y 6x 3y
1 1 0，所以 n 1 n 1 n n 3，2xn yn 2xn yn
数列{an}是公比为3的等比数列·····································································10分
yn 1 yn
1, (1)
xn 1 xn
（法 2）因为 4x2n y
2
n 3, (2) ，································································ 6分
4x2 n 1 y
2
n 1 3, (3)

y y 4(x x ) 2(yn 1 yn ) 2(x x )因为 1 n 1 n n 1 n n 1 n ，
x n 1 xn yn 1 yn 4(xn 1 xn ) (yn 1 yn )
所以 4xn 1 2yn 1 (4xn 2yn ) 2xn 1 yn 1 2xn yn
所以 2xn 1 yn 1 3(2xn yn ) ，又因为 2x1 y1 1 0，
2x y
所以 n 1 n 1 3，数列{a
2x y n
}是公比为3的等比数列；···································10分
n n
n 1
2x y (2x y )3n 1 3n 1
2x
n
yn 3
（3）由（2）知： n n 1 1 ，即 ，
2 4xn y
2
n 3 2x y
3

n n

3n 1
数学答案 第 3页（共 5页）
x 1 (3n 1 3 n n 1 )
解得 4 3 ·············································································12分
y 1 ( 3 3n 1)
n 2 3n 1
x x y y 5
因为OG ( n n 2 , n n 2 ) ( (3n 1 1 ), 5 ( 1 n 1
2 2 4 3 n 2 3 n
3 )) ，

又因为OP (x , y ) 3 1 3 1 ( (3n 1 ), ( 3n 1n 1 n 1 n 1 n n ))，4 3 2 3
5 5
所以OG OPn 1，同理得：OH OPn 2 ，··············································· 16分3 3
S 9
所以 1 ·····························································································17分
S2 25
19．（17分）
解：（1）（ⅰ）由题知，当a 0时， f (x) 2ln(1 x) 2x，
2 2x
所以 f (x) 2 ······································································· 1分
x 1 x 1
当 x ( 1,0)时， f (x) 0，所以 f (x)在 ( 1,0)上单调递增；
当 x (0, )时， f (x) 0，所以 f (x)在 (0, )上单调递减；··················· 2分
所以， f (x) f (0)，所以0为 f (x)在 ( 1, )上的最大“ 点”·························· 3分
（ⅱ）由（ⅰ）知： ln(1 x) x ········································································4分
若 a 0，则 f (x) 2ln(1 x) 2x ax ln(1 x) 0 ；
此时， f (x) f (0)，不可能存在“ 点”，符合题意····································· 5分
a 0 2 ax 2 a若 ，则 f (x) a ln(1 x) 2 a ln(1 x) a 2 ，
1 x 1 x
2
g(x) a ln(1 x) 2 a a 2 a
a[x ( 2)]
令 a 2 ，则 g (x) a2 2 ····· 6分1 x 1 x (1 x) (1 x)
若 a 1 2，则 2 0， g (x) 0， g(x)在 (0,1)上单调递增，且 g(0) 0；
a
所以，当 x (0,1)时， f (x) 0， f (x)在 (0,1)上单调递增；
所以1为 f (x)的“ 点”，不合题意···························································· 7分
0 a 1 2若 ，则 2 0，
a
2
当 x (0, 2)时， g (x) 0，所以 g(x)在 (0, 2 2)上单调递减；
a a
当 x 2 2 ( 2, )时， g (x) 0，所以 g(x)在 ( 2, )上单调递增；
a a
数学答案 第 4页（共 5页）
所以 g(x) g( 2 2)且 g(2 2) g(0) 0，
a a
2 2 2 2 2
又因为 ln(1 x) x，1 x e x，所以 1 e a ，故 2 e a 3 e a 1，
a a
2 2
因此 g(ea 1) a ln(1 ea 1) a 2 a 0 ，
2
所以，存在 (2 2,e a 1)，使得 g( ) 0，
a
当 x (0, )时， f (x) 0， f (x)在 (0, )单调递减；
当 x ( , )时， f (x) 0， f (x)在 ( , )单调递增；
因为 f (x)不存在“ 点”，所以 0 a 1f (1) 0，
2
又当 a 1时， f (x)在 (0,1)上单调递增，得 f (1) 3ln 2 2 0，即 ln 2 ，
3
2 2 1 2因此 ，解得0 a 2，
ln 2 ln 2
2
综上， a的取值范围为 ( , 2] ······························································10分
ln 2
（2）假设 f (x)的“ 点”有 k个，从小到大依次为 a1,a2 , ,ak，由“ 点”定义可知：
因为 f (a1) f (1)且 f (1) f (a1 1)，所以 f (a1) f (1) f (a1) f (a1 1) 1 ·····11分
对于1 i k ，因为 f (ai ) f (ai 1)且 f (ai 1) f (ai 1)，
所以 f (ai ) f (ai 1) f (ai ) f (ai 1) 1 ······················································15分
这样就有：
f (a1) f (1) 1，
f (a2 ) f (a1) 1，
……
f (ai ) f (ai 1) 1，
……
f (ak ) f (ak 1) 1，
迭加得： f (a1) f (0) f (a2 ) f (a1) f (a k) f (a k 1) f (a k) k ，
又因为 f (m) f (ak )，所以 f (m) k ，
综上， f (x) 在D上的“ 点”个数不小于 f (m) ················································ 17分
数学答案 第 5页（共 5页）