2026届漳州三检数学答案_2024-2027高三（6-6月题库）_2026年05月高三试卷_260520福建2026届各地市五月质检_2026届漳州三检_2026届漳州三检数学试题+答案

2026 届高三毕业班教学质量检测
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．B 3．A 4．A 5．D 6．A
7．C 8．C
二、多项选择题：本大题共 3小题，每小题 6分，共 18分。在每小题给出的四个选项中，
有多个选项符合题目要求，全部选对的得6分，部分选对的得部分分，有选错的得0分。
9．ABC 10．ABC 11．ACD
三、填空题：本大题共3题，每小题5分，共15分。
12．7π 13．x y10 14．50
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。
2c|FF |4,
1 2

4 6
  1,

15．解法一：（1）由题意，得 a2 9b2 ·················································· 3分

c a2 b2,

a b0,
a 6,


解得b 2, ······················································································· 5分

c2,

x2 y2
所以C的方程为  1． ································································· 6分
6 2
（2）设Q(x ,y )，x 0，y 0，由题意，得F(2,0)，F (2,0)．
0 0 0 0 1 2
因为点Q在C上且在第一象限，△QFF 为直角三角形，
1 2
所以QF F 90，或FQF 90． ··················································· 7分
2 1 1 2
6
若QF F 90，则Q(2, )； ···························································· 8分
2 1 3
 
若FQF 90，则QF QF (2x ,y )(2x ,y ) x2 4 y2 0，
1 2 1 2 0 0 0 0 0 0
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第1页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}········································································································· 11分
x2 y2
又Q在C上，所以 0  0 1， ····························································· 12分
6 2
x  3,
所以 0 即Q( 3,1)．
 y 1,
0
6
综上，点Q的坐标为(2, )，或( 3,1)． ················································ 13分
3
6
解法二：（1）由题意，得椭圆C的左、右焦点分别为F(2,0),F (2,0)，点P(2, )
1 2 3
在C上， ······························································································ 1分
6 6
所以c2，2a|PF ||PF |  16 2 6， ····························· 3分
1 2 3 9
所以a 6，b a2 c2  2， ·························································· 5分
x2 y2
所以C的方程为  1． ································································· 6分
6 2
（2）因为点Q在C上且在第一象限，△QFF 为直角三角形，
1 2
所以QF F 90，或FQF 90． ··················································· 7分
2 1 1 2
6
若QF F 90，则Q(2, )； ···························································· 8分
2 1 3
2 2 2
若FQF 90，则 QF  QF  FF 16，
1 2 1 2 1 2
因为( QF  QF )2 2a2 24，即 QF 2 2 QF  QF + QF 2 24，
1 2 1 1 2 2
解得 QF  QF 4．
1 2
设Qx ,y  ， x 0，y 0，
0 0 0 0
1 1
所以S  FF y  QF  QF ，即2y 2，所以y 1． ··········· 11分
△F 1 QF 2 2 1 2 0 2 1 2 0 0
x2 y2
又Q在C上，所以 0  0 1， ····························································· 12分
6 2
x  3,
所以 0 即Q( 3,1)．
 y 1,
0
6
综上，点Q的坐标为(2, )，或( 3,1)． ················································ 13分
3
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第2页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}a cosC
16．解法一：（1）由  得，acosAccosC，所以sinAcosAsinCcosC， ....... 2分
c cosA
1 1
即 sin2A sin2C，即sin2Asin2C． ................................................................... 3分
2 2
因为A，C0，π，所以2A，2C0，2π
，
2A2C π 2A2C 3π
所以根据y sinx的图象可得2A2C ，或  ，或  ， ... 5分
2 2 2 2
π 3π
所以AC，或AC  ，或AC  ， ................................................................. 6分
2 2
π
又AC(0,π)，ABC  π，所以AC，或B  ，
2
所以△ABC是等腰三角形或直角三角形．. .................................................................... 7分
（2）①若AC，则a c2，则ac4b，不满足三角形三边关系，舍去；
............................................................................................................................................... 8分
②若B π ，则BC a b2 c2 2 3． ................................................................. 9分
2
以B为原点，分别以BC,BA为x，y轴建立平面直角坐标系，
则B0,0，A0,2，C  2 3,0  ，M  3,0  ，N  3,1  ，
 
   
所以AM  3,2 ，BN  3,1 ， ........................................................................ 12分
y
   
所以cosMPN cos PM,PN cos AM,BN A
N
     
3,2  3,1 P
AM BN 7
     ．. ..... B ............... M ............ C ...... x ..... 15分
AM  BN  3 2 +22   3 2 +12 14
解法二：（1）同解法一； . ................................................................................................ 7分
（2）①若AC，则a c2，则ac4b，不满足三角形三边关系，舍去；
............................................................................................................................................... 8分
π
②若B ，则BC a b2 c2 2 3， ................................................................. 9分
2
AM  BM2  AB2  ( 3)2 22  7，
1
BN  AC 2． ............................................................................................................ 10分
2
因为BC，AC边上的两条中线AM ，BN 相交于点P，所以P为△ABC的重心，
1 7 1 2 1
所以PM  AM  ，PN  BN  ，MN  AB1， ............................ 13分
3 3 3 3 2
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第3页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}PM2 PN2 MN2 7
所以cosMPN   ．. ........................................................ 15分
2PM PN 14
a cosC b2 c2 a2 a2 b2 c2
解法三：（1）由  ，得a c ， ............................ 2分
c cosA 2bc 2ab
   
即a2 b2 c2 a2 c2 a2 b2 c2 ，
  
化简，得 a2 c2 b2 a2 c2 0， ................................................................................. 5分
从而ac，或b2a2 c2， .............................................................................................. 6分
所以△ABC是等腰三角形或直角三角形．. .................................................................... 7分
（2）
①若ac2，则ac4b，不满足三角形三边关系，舍去； ............................... 8分
π
②若b2a2 c2，则B ，所以BC a b2 c2 2 3， .................................. 9分
2
a 3 π
cosC   ，所以C  ．
b 2 6
1 π
又BN  AC CN，所以CBN C  ． .......................................................... 11分
2 6
2 3
在Rt△ABM 中，AM  AB2 BM2  7 ，sinAMB ，cosAMB ，
7 7
............................................................................................................................................. 13分
π
所以cosMPN cos(CBN AMB)cos( AMB)
6
3 1 7
 cosAMB sinAMB ． . .................................... 15分
2 2 14
17．解：（1）由题意，机器人无需进行第二轮巡检的概率为
2 2 2 20
PC2( )2(1 )C3( )3  ； ······························································· 3分
3 3 3 3 3 27
（2）由题意得，X 3,5,6， ···································································· 4分
且P(X 3)C2p2(1 p)C3p3 3p2(1 p) p3， ········································· 5分
3 3
P(X 5)C1p(1 p)2 3p(1 p)2， ··························································· 6分
3
P(X 6)C0(1 p)3 (1 p)3， ································································ 7分
3
所以E(X)3[3p2(1 p) p3]53p(1 p)26(1 p)3=3p33p23p6，
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第4页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}2 
所以E(X)3p33p23p6，p

,1． ················································· 9分
3 
（3）应选方案一，理由如下： ································································ 10分
记Y 为机器人巡检的检测总费用，Z 为人工巡检的检测总费用，
由题意得，E(Z)Z 303=90， ···························································· 11分
2 
令 f p3p33p23p6，p

,1，
3 
则 f(p)9p26p33(3p22p1)33p1p1，
2  2 
因为p

,1，所以 f(p)0，即 f(p)在

,1上单调递减，
3  3 
2 32
所以E(X) f(p)≤f( ) ， ···························································· 14分
3 9
32
所以E(Y)E(18X)18E(X)≤18 64＜90E(Z)，
9
故选用智能机器人巡检的检测平均总费用更低，应选方案一． ························ 15分
18．（1）证明：在△SBH 中，SB4，HB2 3，SBO30，
所以SH  SB2 HB2 2SBHBcosSBO 2，
所以SH2 HB2  SB2，所以SH OB． ················································· 1分
又因为平面SOB平面OABC，平面SOB平面OABC OB，SH 平面SOB，
所以SH 平面OABC， ········································································ 2分
又AC 平面OABC，所以SH  AC ． ··················································· 3分
（2）取HA的中点E，SH 的中点F ，连接EF ，则EF 即为直线l．
·········································································································· 4分
证明如下：连接DE，DF，因为D，F 分别是BH 和SH 的中点，所以DF//BS ，
又DF 平面SAB，BS 平面SAB，所以DF//平面SAB， ······················ 5分
同理，DE//平面SAB， ········································································· 6分
又DEDF  D，DE,DF 平面DEF ，所以平面DEF//平面SAB， ········ 7分
又过D且与平面SAB平行的平面有且只有一个，D平面DEF ，
所以平面DEF 即平面，又平面DEF平面SAB EF，所以EF 即为直线l．
·········································································································· 8分
（3）以O为坐标原点，OB为x轴建立如图所示的空间直角坐标系Oxyz，其中y轴
在平面OABC内，z轴平面OABC，则S( 3,0,2)，B(3 3,0,0)，
 
设A(x,y,0)，y 0，则SA(x 3,y,2)，SB (2 3,0,2)，
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第5页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}因为ASB60，
 
 
SASB 2 3x2 1
所以cos SA,SB     ，
|SA||SB| 4 (x 3)2  y2 4 2
x2 y2 3
化简得  1(x )，又y 0，所以x 3，
3 6 3
x2 y2
所以点A在xOy面内的曲线:  1(x 3)上，
3 6
同理，点C也在上． ·········································································· 10分
在平面直角坐标系xOy内，D(2 3,0)，
x2 y2
设直线AC的方程为xmy2 3 ，代入:  1(x 3)，
3 6
得(2m2 1)y2 8 3my180， z
S
设A(x ,y ),C(x ,y )，则
A A C C
2m2 10,

(8 3m)2 72(2m2 1)0,
F

8 3m
 C
y  y  ,
 A C 2m2 1 y

18
y y  0, O H E D B x
 A C 2m2 1
A
2 2
解得 m ． ·········································································· 11分
2 2
| AC| (x x )2 (y  y )2  1m2 (y  y )2 4y y
A C A C A C A C
8 3m 18 (m2 1)(2m2 3)
 1m2 ( )2 4 2 6 ，
2m2 1 2m2 1 (2m2 1)2
········································································································· 12分
t1
( 1)(t13)
令t 2m2 1，则m2  t1 ，| AC|2 6 2 2 6 t2 7t12
2 t2 2t2
7 12 1 7 7
2 3 1  2 3 12(  )2 12( )2 1，
t t2 t 24 24
2 2 1 7
因为 m ，所以t 2m2 1[1,0)， ≤1 ，
2 2 t 24
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第6页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}1
所以当 1，即m0时，| AC| 6 2， ·········································· 13分
t min
此时在空间直角坐标系Oxyz中，A(2 3,3 2,0)，C(2 3,3 2,0)，
  
BA( 3,3 2,0)，BS (2 3,0,2)，BC ( 3,3 2,0)，

设n (x ,y ,z )是平面SAB的法向量，
1 1 1 1
 
 n BA0,   3x 3 2y 0, 
1  即 1 1 取n ( 6,1,3 2)， ························ 14分
1
 n BS 0,  2 3x 2z 0,
1 1 1

设n (x ,y ,z )是平面SBC的法向量，
2 2 2 2
 
 n BS 0,  2 3x 2z 0, 
2  即 2 2 取n ( 6,1,3 2)， ························· 15分
2
 n BC 0,   3x 3 2y 0,
2 2 2
设二面角ASBC的大小为，
 
 
|n n | 6118 23
则|cos||cosn ,n | 1 2   ， ······························ 16分
1 2 |n ||n | 55 25
1 2
4 6
又[0,π]，所以二面角ASBC的正弦值为sin 1cos2 ．
25
········································································································· 17分
19．解法一：（1）因为u(x) f(x)g(x)ex axb1，所以u'(x)ex a，
············································································································ 1分
①若a≤0，则u'(x)ex a0，所以u(x)在(,)上单调递增； ············· 2分
②若a 0，则由u'(x)ex a0，得xlna，由u'(x)ex a0，得xlna，
所以u(x)在(,lna)上单调递减，在(lna,)上单调递增． ························· 4分
（2）由u(1)≥0，即eab1≥0，得ab≤e1． ·································· 5分
当ae，b1时，abe1，下面证明此时u(x)≥0成立， ······················· 6分
此时u(x)ex ex，由（1）知u(x)在(,1)上单调递减，在(1,)上单调递增，
所以u(x)≥u(1)0成立． ········································································ 7分
综上，ab的最大值为e1． ··································································· 8分
（3）若h(x)有零点，设零点为x (x 0)，
0 0
则h(x ) x f(x2)e(lnx 1)g2(x )0， ············································· 9分
0 0 0 0 0
即g(x ) x f(x2)e(lnx 1)  x ex 0 2 e(lnx 1)，
0 0 0 0 0 0
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第7页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}即ax b1 x ex 0 2 e(lnx 1) 0， ···················································· 10分
0 0 0
这说明点P(a,b)在直线l:x x y1 x ex 0 2 e(lnx 1) 0上， ················· 11分
0 0 0
设点A(0,1)到直线l的距离为d ，
| x ex 0 2 e(lnx 1)|
则|PA|≥d ，即 a2 (b1)2≥ 0 0 ， ··························· 13分
x2 1
0
由（2）知，ex≥ex，仅当x1时，“=”成立， ·········································· 14分
elnx 0 x 0 2 e(lnx 1) e(lnx x2)e(lnx 1)
所以 a2 (b1)2≥ 0 ≥ 0 0 0
x2 1 x2 1
0 0
e(x2 1)
 0  e 0， ········································································· 16分
x2 1
0
所以a2 (b1)2≥e． ············································································· 17分
解法二：（1）同解法一； ··········································································· 4分
（2）①若a0，则
当x0时，u(x)ex axb11axb1axb2，
b2
当x 时，axb20，
a
b2
所以当x0且x 时，u(x)0，不合题意． ······································· 5分
a
②若a 0，则u(x)ex b1的值域为(b1,)，
所以b1≥0，b≤1，所以ab≤1e1． ·············································· 6分
③若a 0，则结合（1）得，[u(x)] u(lna)≥0，
min
即aalnab1≥0，即b≤aalna1，所以ab≤2aalna1，
令 p(a)2aalna1，则 p'(a)1lna，
当0ae时， p'(a)0， p(a)单调递增；当ae时， p'(a)0， p(a)单调递减，
所以ab≤p(a)≤p(e)e1，
当ae，baalna11时，abe1． ·········································· 7分
综上，ab的最大值为e1． ··································································· 8分
（3）若h(x)有零点，设零点为x (x 0)，
0 0
则h(x ) x f(x2)e(lnx 1)g2(x )0， ············································· 9分
0 0 0 0 0
即g(x ) x f(x2)e(lnx 1)  x ex 0 2 e(lnx 1)，
0 0 0 0 0 0
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第8页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}即ax b1 x ex 0 2 e(lnx 1) 0， ···················································· 10分
0 0 0
设a rcos,b1rsin，r≥0，则a2 (b1)2 r2，
rx cosrsin x ex 0 2 e(lnx 1) 0，
0 0 0
1
即r x 0 2 1sin() x 0 ex 0 2 e(lnx 0 1) 0，其中tan x ，
0
所以 x ex 0 2 e(lnx 1)≤r x2 1，
0 0 0
x ex 0 2 e(lnx 1)
所以 a2 (b1)2 r≥ 0 0 ， ··········································· 13分
x2 1
0
由（2）知，ex≥ex，仅当x1时，“=”成立， ·········································· 14分
elnx 0 x 0 2 e(lnx 1) e(lnx x2)e(lnx 1)
所以 a2 (b1)2≥ 0 ≥ 0 0 0
x2 1 x2 1
0 0
e(x2 1)
 0  e 0， ········································································· 16分
x2 1
0
所以a2 (b1)2≥e． ············································································· 17分
2026届高三毕业班教学质量检测 数学参考答案及评分细则 第9页（共9页）
{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}