高二数学答案第 1 页 共 4 页（数学是有生命的，题目是有经典的）
2019-2020 学年度第一学期期末学业水平检测高二数学参考答案
一、单项选择题：本大题共 8 小题，每小题 5 分，共 40 分。
1-8: CBCD CADA
二、多项选择题：本大题共 4 小题，每小题 5 分，共 20 分。
9.CD; 10. AD; 11. CD; 12. ABD;
三、填空题：本大题共 4 个小题，每小题 5 分，共 20 分。
13. 32 ； 14. 4
3
； 15. 9759.0 ； 16. 3 ；
四、解答题：共 70 分。解答应写出文字说明，证明过程或演算步骤。
17.（10 分）
解：（1）由题知： 10, 120x y ··································································· 2 分
所以 1
22
1
n
i i
i
n
i
i
x y nxy
b
x nx
5120 6000 880 8610 500 110
············································ 4 分
所以 200108120 a ，所以线性回归方程为： 2008 xy ·······················5 分
所以估计该国年度国际足联排名为1时注册足球运动员的人数约为
19220018 y （万人）······································································ 6 分
（2）由题知： (10,0.1)X B ········································································ 7 分
所以 10 0.1 1EX ···················································································8 分
10 1 9
10( 2) 1 ( 0) ( 1) 1 0.9 (0.9) (0.1) 0.26P X P x P x C ············10 分
注：若仅仅结果算错只扣一分！
18.（12 分）
解：（1）因为{ }na 为等比数列，所以 3
1 2 3 2
1( ) 64a a a a ，所以 2
1
4a ··················· 2 分
又因为 2 3 2
1 12 (2 1) (2 1)4 2a a a q q ，所以 1
2q ···································4 分
所以 1
1
2a ， 1
1
1( )2
n n
na a q ····································································· 5 分
（2）由题知： 1
2
logn nb a n ······································································ 6 分
所以 1( )2
n
n n nc a b n ·················································································· 7 分
所以 2 31 1 1 11 2 ( ) 3 ( ) ...... ( )2 2 2 2
n
nT n ··········································· 8 分
2 3 4 11 1 1 1 1 1 1 ( ) 2 ( ) 3 ( ) ...... ( 1) ( ) ( )2 2 2 2 2 2
n n
nT n n ········9 分
所以 2 3 11 1 1 1 1 1( ) ( ) ...... ( ) ( )2 2 2 2 2 2
n n
nT n ··········································10 分
所以 1 1
1 1(1 ( ) )1 1 1 1 12 2 ( ) 1 ( ) ( ) 1 ( 1)( )12 2 2 2 2 21 2
n
n n n n
n
nT n n
··········· 11 分
所以 12 ( 2)( )2
n
nT n ·············································································· 12 分高二数学答案第 2 页 共 4 页（数学是有生命的，题目是有经典的）
19.（12 分）
解:（1）设 ( , )M x y ，由题知： 1 2
1 1,+2 2
y yk kx x
··········································2 分
所以 2 1 2 2
1 1 ( 1)( 2 2) 4 4
2 2 4 4
y y y x x yk k x x x x
································4 分
又因为 yx 42 ，所以 2 1
4 4 14 4
yk k y
························································6 分
（2）设 ( , )M x y ，因为 MO AO ，故可得 2 2 2MO AO MA
所以 2 2 24= 4 4= 2MA x y y y y ················································ 9 分
又因为 M 到直线 02 y 的距离 2d y MA ··········································11 分
所以⊙ M 与直线 02 y 相切····································································· 12 分
20.（12 分）
解: （1）设频率分布直方图中 6 组数据的频率分别为 1 2 3 4 5 6, , , , ,P P P P P P
4 0.04 5 0.2P ，所以 8 400.2N ···························································3 分
由题意 1 2 3 4 5 6 1P P P P P P
而 2 3 1 4 5 61 ( ) 1 5(0.01 0.04 0.02 0.01) 0.6P P P P P P ················ 5 分
所以 20 30 之间的志愿者人数 1 2 340 ( ) 40 0.6 24N P P ························6 分
（2）35 45 之间共有 5 (0.01 0.02) 40 6 人，其中 4 名女教师，2 名男教师
从中选取三人，则女教师的数量为 的可能取值为1,2,3······································ 7 分
所以
1 2
4 2
3
6
1( 1) 5
C CP C
；
2 1
4 2
3
6
3( 2) 5
C CP C
；
3
4
3
6
1( 3) 5
CP C
所以 的分布列为：
1 2 3
( )P k 1
5
3
5
1
5
·····································································10 分
所以数学期望为 1 3 11 2 3 25 5 5E ···················································12 分高二数学答案第 3 页 共 4 页（数学是有生命的，题目是有经典的）
21.（12 分）
解：（1）设 6( , ), ( , )2QP x y Q x ，由题知： 2 2
1 1 1
OP OQ
，
所以 2 2
2
1 1 13
2Q
x y x
·············································································· 1 分
又因为OP OQ ，所以 6 60, ( 0)2 2Q Q
y yxx x xx
·································· 2 分
所以得:
2
2
22 2
2
1 1 1, 13 3 3
2 2
x yyx y
x
····················································4 分
所以轨迹W 的方程为：
2
2 1( 0)3
x y x ························································5 分
（2）由题意:设 ),(),,( 2211 yxByxA ， AB 的中点为 ),( 00 yxE , ),3( mD
由
13
)1(
2
2
yx
xky
得： 0336)31( 2222 kxkxk ········································ 6 分
由韦达定理得： 2
2
21 31
6
k
kxx
，所以 2002
2
0 31,31
3
k
kkkxyk
kx
，
即点 )31,31
3( 22
2
k
k
k
kE
················································································8 分
因为
kkOE 3
1 ，所以射线
k
xyOE 3: ························································ 9 分
由
3
3
x
k
xy
得：
kyD
1 ,即点 )1,3( kD ；····················································· 10 分
由
13
3
2
2
yx
k
xy
得： 2
2
31
1
kyG ·································································· 11 分
又因为 2 2
1 1
1 3 1 3D E
ky y k k k
，所以 EDG yyy 2 ，即 2OG OD ∙ OE ·······12 分高二数学答案第 4 页 共 4 页（数学是有生命的，题目是有经典的）
22．（12分）
解：（1）
2
2 100 (10 30 20 40) 4 25 4 3.84150 50 30 70 21K ．····························2分
故有95%的把握认为男、女顾客对该商场服务的评价有差异··································· 3分
（2）由题知： X 的可能取值为： 1,0,1
( 1) (1 ) 0.1P X ， ( 0) (1 )(1 ) 0.5P X ， ( 1) (1 ) 0.4P X
因此 X 的分布列为：
X 1 0 1
P 0.1 0.5 0.4
·················································································································6 分
由题知： 0 100, 1P P , 1 10.1 0.5 0.4i i i iP P P P ，········································ 7 分
1
1 1
1
10.5 0.1 0.4 , 4
i i
i i i
i i
P PP P P P P
·······························································8 分
所以 1 ( 0,1,2,...9)i iP P i 为等比数列··························································9 分
（ⅱ）由（ⅰ）知： 1 1 0
1( )( )4
i
i iP P P P
所以 1
1 1 1 1 1
1 1 1( ) ( ) ( )4 4 4
i i i
i i iP P P P P P
0 1 1 1
1 1 4 1(1 ) (1 )4 4 3 4i iP P P ············································ 10 分
又因为 10 1 10
4 1(1 ) 13 4P P ，所以 5 1 5
4 1(1 ),3 4P P
所以 5
5
10
5
1 1280
1 12811 4
PP P
·······································································11 分
从概率看甲最终更流畅的概率 5P 的接近1，由于 0.8 0.5 ，
此时得到正确结论的概率很高，因此该实验方案合理！······································ 12 分
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