2024届福建省三明市普通高中高三毕业班5月质量检测数学参考答案_2024年5月_01按日期_8号_2024届福建省三明市高三下学期三模_福建省三明市普通高中2024届高三毕业班5月质量检测数学试题

三明市 2024 年普通高中高三毕业班质量检测
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、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分．
1．C 2．C 3．D 4．A 5．A 6．B 7．B 8．C
二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，
部分选对的得部分分，有选错的得0分．
9．BC 10．ACD 11．BCD
三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．
1 
12． 6 13．  ,3  14．  6,7,8,9  , 21 （第一空2分，第二空3分）
e 
四、解答题：本大题共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.解法一：（1）证明：取BD的中点M ，连接PM、MC,·······················1分
∵△BPD和△BCD均为等边三角形，
∴BDPM，BD CM .··································································2分
又PM CM M ，
∴BD平面CPM ，·········································································3分
又CP平面CPM ，
∴BDCP.····················································································4分
 
（2）以M 为原点，MB,MC所在直线为x,y轴，过M 作平面BCD的垂线所在直
线为z轴，如图所示建立空间直角坐标系，···········································5分
∵平面ABD平面PBD，平面ABD平面PBDBD，PM 平面PBD，PM  BD
∴PM 平面ABD.
∵△PBD和△CBD均为等边三角形，
∴PM MC  PC  3，PMC 60，
第 1 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}∴P  0, 3 , 3 ，C  0, 3,0  ，B  1,0,0 ，··············································6分
 
 2 2
  3 3      3 3
∴BP 1, , ，BC  1, 3,0 .MP 0, , 
   
 2 2  2 2
设平面PBC 的法向量为m (x,y,z)
  3 3
 mBP 0, x y z 0,
∴  即 2 2
mBC 0 
x 3y 0
 
取z 1,则m  3, 3,1 ，···································································8分
  3 3
平面ABD的法向量MP 0, , ，·················································10分
 
 2 2
设平面ABD与平面PBC 的夹角为，

 MPn 3 39
∴cos cos MP,n     ··································12分
MP n 3 13 13
39
∴平面ABD与平面PBC 夹角的余弦值为 .····································13分
13
解法二：（1）同解法一······································································4分
（2）如图，取MC 的中点E为原点，连接PE，过点E作EF //MB,交BC于点F ,
由（1）知CM  BD,EF  MC,
又由（1）知BD平面CPM ，又PE平面CPM ，∴BDPE，
∵△PBD和△CBD均为等边三角形且棱长为2，
∴PM MC  PC  3，PE  MC,
BDMC  M PE 平面CBD
  
以E为原点，EF,EC,EP所在直线为x,y,z轴，
建立空间直角坐标系，如图所示··························································5分
∵平面ABD平面PBD，平面ABD平面PBDBD，PM 平面PBD，PM  BD
∴PM 平面ABD，
第 2 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}  3 3
平面ABD的法向量MP 0, , ···················································7分
 
 2 2
 3  3   3 
∴P0,0, ，C

0, ,0

，B

1, ,0

·············································8分
 2  2   2 
     3 3
∴CB  1, 3,0 ，CP 0, , ，
 
 2 2
设平面PBC 的法向量为m  x,y,z ，
  x 3y 0
 mCP 0   
∴  ,即 3 3 ，取z 1,则m  3, 3,1 ，················· 10分
mCB 0  y z 0
 2 2
设平面ABD与平面PBC 的夹角为，

 MPm 3 39
∴cos cos MP,m     ，······························12分
MP m 3 13 13
39
∴平面ABD与平面PBC 夹角的余弦值为 .····································13分
13
16.解法一：
 1 3 
（1）由题意 f(x)sinxcos(x ) sinx cosxsin(x )
6 2 2 3
·····································································································2分
π
因为 f  x 图象的两条相邻对称轴间的距离为 ，
2
所以周期T  2π 2 π ，故2，所以 f  x sin  2x π ，·····················4分
 2  3
π π π
当x 0,m 时，2x   ,2m ，·················································5分
3 3 3
π π 3π
因为 f  x 在区间 0,m 上有最大值无最小值，所以 2m  ，·········6分
2 3 2
π 7π  π 7π
解得 m ，所以m的取值范围为 , .···································7分
12 12 12 12
第 3 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}
（2）将函数 f  x 图象向右平移 个单位长度，
6
  
得到 y sin

2(x )

sin2x的图象，············································8分
 6 3
再将图象上所有点的横坐标变为原来的2倍（纵坐标不变），
得到g(x)sinx的图象，···································································9分
1 1
所以函数h(x) xsinx，所以h(x) cosx，································10分
2 2
1
令h(x)0得cosx ，
2
因为x(2,)，
4
所以当x(2, )时，h(x)0，h(x)单调递增，····························11分
3
4 2
当x( , )时，h(x)0，h(x)单调递减，································ 12分
3 3
2 2
当x( , )时，h(x)0，h(x)单调递增，·································· 13分
3 3
2
当x( ,)时，h(x)0，h(x)单调递减.·········································14分
3
4 2
所以函数h(x)的极大值点为 和 .··············································15分
3 3
解法二：（1）同解法一.····································································· 7分

（2）将函数 f  x 图象向右平移 个单位长度，
6
  
得到 y sin

2(x )

sin2x的图象，············································8分
 6 3
再将图象上所有点的横坐标变为原来的2倍（纵坐标不变），
得到g(x)sinx的图象，···································································9分
第 4 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}1 1
所以函数h(x) xsinx，所以h(x) cosx，································10分
2 2
1
令h(x)0得cosx ，
2
2 2
当 2k x 2k时，h(x)0，h(x)单调递增，
3 3
因为x(2,)
4
所以k 1时，2 x ，h(x)单调递增，··································11分
3
2 2
k 1时，  x h(x)单调递增·················································12分
3 3
2 4
当 2k x 2k时，h(x)0，h(x)单调递减，
3 3
因为x(2,)
2
k 0时，  x，h(x)单调递减，·············································· 13分
3
4 2
k 1时，  x ，h(x)单调递减，······································ 14分
3 3
4 2
所以函数h(x)的极大值点为 和 .··············································15分
3 3
解法三：（1）同解法一.····································································· 7分

（2）将函数 f  x 图象向右平移 个单位长度，
6
  
得到 y sin

2(x )

sin2x的图象，············································8分
 6 3
再将图象上所有点的横坐标变为原来的2倍（纵坐标不变），
得到g(x)sinx的图象，···································································9分
1 1
所以函数h(x) xsinx，所以h(x) cosx，································10分
2 2
第 5 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}1
令h(x)0得cosx
2
因为x(2,)，所以x,h(x),h(x)的变化情况如下：
4 4 4 2 2 2 2 2 2
x (2, )  ( , )  ( , ) ( ,)
3 3 3 3 3 3 3 3 3
h(x)  0  0  0 
极大 极小 极大
h(x) 单调递增 单调递减 单调递增 单调递减
值 值 值
···································································································14分
4 2
所以函数h(x)的极大值点为 和 .··············································15分
3 3
17.解: (1)记随机任选1题为家政、 园艺、 民族工艺试题分别为事件A(i 1,2,3)，
i
记随机任选1题，甲答对为事件B，··············································1分
3
则P(B)P(A)P(B| A)P(A)P(B| A)P(A )P(B| A )P(A )P(B| A )
i i 1 1 2 2 3 3
i1
······························································································ 2分
1 2 1 2 1 4 3
       ,·······························································4分
4 5 4 5 2 5 5
3
所以随机任选1题，甲答对的概率为 ；···········································5分
5
(2) 乙答对记为事件C,则
1 1 1 1 1 1 1
P(C) P(A)P(C| A)P(A )P(C| A )P(A )P(C| A )      
1 1 2 2 3 3 4 2 4 2 2 2 2
·····································································································7分
设每一轮比赛中甲得分为X ,
3  1 3
则P(X 1) P(BC) P(B)P(C)  1   ，································· 8分
5  2 10
3 1  3  1 1
P(X 0) P(BCBC) P(BC)P(BC)   1   1   ，········9分
5 2  5  2 2
第 6 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#} 3 1 1
P(X 1) P(BC) 1    .····················································10分
 5 2 5
三轮比赛后，设甲总得分为Y ,
3
 3  27
则P(Y 3)
 
 ，······························································11分
10 1000
2
 3  1 27
P(Y 2)C2     ，··························································12分
3 10 2 200
2 2
3 1  3  1 279
P(Y 1)C1    C2     ，··································· 13分
3 10 2 3 10 5 1000
所以甲最终获得奖品的概率为
27 27 279 441
P  P(Y 3)P(Y 2)P(Y 1)    .····················15分
1000 200 1000 1000
18.（1）因为a a a a ( 2)n2n①
1 2 n1 n
所以当n2,a a a a ( 2)(n1)2n1②，·············································1分
1 2 n n1
②
由 得a 2n··················································································2分
① n
因为n1时a 2也符合上式，···························································· 3分
1
所以数列a 是以2为首项，2为公比的等比数列，
n
所以a 2n,nN*.············································································· 4分
n
2
 12n
（2）由（1）知， S   2n12 ，···············································5分
n 12
因为不等式(1)ntS 14S 2对任意的 nN恒成立，又S 0且S 单调递增，
n n n n
·····································································································6分
14
所以(1)ntS
n

S
对任意的 nN恒成立，··········································· 7分
n
因为S =2，S 6，S =14，S =30，·························································· 8分
1 2 3 4
14  14
所以当n为偶数时，原式化简为tS
n

S
对任意的 nN恒成立，即t

S
n

S


n n min
25
因为S 6 14，所以当n2时，t ，············································10分
2 3
第 7 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}14  14
当n为奇数时，原式化简为tS
n

S
对任意的 nN恒成立，即t

S
n

S


n n min
因为S 2 14S 14，所以当n1时，t9，所以t9，··················12分
1 3
25
综上可知，9t .······································································ 13分
3
1 1
（3）因为b   ，······························································14分
n log a 2 2n
2 n
所以b 是单调递减数列，
n
所以 b  b ···············································································15分
n1 n
b b 2(b b ) 2(b b )
所以 n n1  n n1  n n1 2( b  b )
b 2 b b  b n n1
n n n n1
b b b b b b
1 2  2 3 + n n1 2( b  b  b  b  b  b )
1 2 2 3 n n1
b b b
1 2 n
1
2  2
2
原不等式得证.················································································17分
 1
1 y 
19.解法一：（1）由题意可知双曲线y 的实轴为y  x，联立 x ，
x  y  x
x1 x 1 1
解得 或 ，即双曲线y 的两顶点为(1,1),(1,1)，
y 1 y 1 x
故实轴长2a   11 2  11 2 2 2，即 a 2····································2分
1 
函数y 的图象绕原点O顺时针旋转 后渐近线为y x，····················3分
x 4
1
所以 ab 2 ，c2， 所以，双曲线y 的离心率 e 2.·················· 4分
x
1 
（2）由（1）知函数y 的图象绕原点O顺时针旋转 得到双曲线x2  y2  2的图象，
x 4
 1
所以，双曲线x2  y2  2的图象绕原点O逆时针旋转 得到函数y 的图象，
4 x
·····································································································5分
第 8 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}1
其渐近线分别为x轴和 y轴，此时设P(x , )，
0 x
0
1
因为y ，················································································6分
x2
1 1 1
所以P处切线斜率k  ，P处切线方程为y  (xx ) ，·········7分
x2 x x2 0
0 0 0
2
过P作切线分别交两条渐近线于A(0, ),B(2x ,0)·································· 8分
x 0
0
1 1 2
所以S  OAOB   2x 2，
AOB 2 2 x 0
0
所以AOB的面积为定值2，即AOB的面积为定值2.·····························9分
3 3 3
（3）依题意知，函数 y  x 的两条渐近线分别为y  x和y轴，
3 2x 3
则实轴为 y  3x
 3 3  x  3
y  x 
2
由 3 2x ,联立解得 ，所以双曲线实半轴长a  3，
  3
 y  3x y 

2
3 3  x2
将函数 y  x 的图象绕原点O顺时针旋转 得到曲线的方程为  y2 1，
3 2x 3 3
···································································································10分

将F(1, 3),直线l:x 3y30绕原点O顺时针旋转 得到
3
3
F(2,0)，直线l:x  ·····································································11分
2
则过F(2,0)直线交曲线右支于M,N两点，
3 3
设M(x ,y )，N(x ,y )，则C( ,y )，D( ,y )
1 1 2 2 2 1 2 2
因为直线MN斜率不为0，所以设直线MN方程为x my2，
x my2,

由x2 得(m2 3)y2 4my10，

 y2 1
 3
第 9 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#} 4m
y  y  ,
m2 30,   1 2 m2 3
 且 ··························································12分
  0  1
y y 
 1 2 m2 3
因为 y y 0，所以m2 3，·····························································13分
1 2
y  y y y 3
k  2 1 y y  2 1 (x )
因为 NC 3 ，所以NC方程为 1 3 2 ，
x  x 
2 2 2 2
3 1
y (x  ) my y  y
令 y 0得 1 2 2 3 1 2 2 1 3
x    
y  y 2 y  y 2
2 1 2 1
y  y
由韦达定理可得my y  1 2 ，
1 2 4
y  y 1
1 2  y
所以 4 2 1 3 1 3 7
x     
y  y 2 4 2 4
2 1
7 7
所以直线NC过定点( ,0)，由图象的对称性可知，MD过定点( ,0)，
4 4
7
即直线NC与直线MD交点H( ,0)，··············································15分
4
1 1 1 3 m2 1
那么S  HF y  y    (y  y )2 4y y  
MNH 2 1 2 2 4 1 2 1 2 4 (m2 3)2
···································································································16分
令t m2 1，则1t 4，
m2 1 t 1
 
则 (m2 3)2 (t4)2 16
t 8
t
16
因为函数 y t 8在 1,4 上单调递减
t
3 1 3
所以当t 1，即m2 0时，MNH面积取最小值，最小值为   ，
4 9 12
3 1 3
即MNH面积最小值为   .················································ 17分
4 9 12
解法二：
第 10 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}（1）同解法一.················································································4分
（2）当曲线E在点P处切线斜率不存在时，切线方程为 x  2,此时AOB面积为2，
·····································································································5分
当曲线E在点P处切线斜率存在时，设在点P处切线方程为 y kxm，
y kxm,
联立 得(1k2)x2 2kmxm2 20，
x2  y2 2
1k2 0,
则 ,化简得m2 2k2 2,k2 1···················· 6分
 4k2m2 4(1k2)(m2 2)0
y kxm, m m
由 得A( , )，
 y  x 1k 1k
y kxm, m m
由 得B( , ).····························································7分
 y x 1k 1k
m m m m 4m2(1k2) 4 1k2
则 AB  (  )2 (  )2   ，
1k 1k 1k 1k 1k2 m
m
因为原点O到直线y kxm的距离d  ，···································8分
1k2
1 1 4 1k2 m
所以S  AB d  2，
AOB
2 2 m 1k2
所以AOB面积为定值2.····································································9分
（3）同解法一.·············································································· 17分
第 11 页 共 11 页
{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}