高一数学答案_2026年01月高一试卷_260118黑龙江省哈尔滨市2025-2026学年高一上学期期末学业质量检测（全）_黑龙江省哈尔滨市2025-2026学年高一上学期期末学业质量检测数学含答案

哈尔滨市 2025 级高一学年学业质量检测试卷
数学答案
1 2 3 4 5 6 7 8
C B D B B C A A
9 10 11
BC ACD ABD
1 
12.1 13.

,
2 
14.2025
答案详解：
2 1 2
7.当角的终边在第二象时，sin ，cos ,则5sin 0；
5 5 cos
2 1 2
当角的终边在第四象时，sin ，cos ，则5sin 0.
5 5 cos
8.令t 3x 0,则只需关于t的方程t2 ta 0有两个不相等的正实数根，
 0
 1
即x x 0得:0a
1 2
 4
x x 0
1 2
2   2 2    2 π
10.T  4 3, ;   2k, 0,  f x  2sin x  ,B错；
  4 3 3 4 2 3 3 3
 2 π π  3
f 02sin  3 ,A对； x  2kπ，解得x  k,kZ，C对；
3 3 3 2 4 2
 2    5  
 2k x  2k,x

 3k, 3k

,kZ ，D对. 选ACD。
2 3 3 2  4 4 
11.对于A：若Mx的图象是一条直线，则xx2 3xm在xR上恒成立，即x24xm0在xR上
恒成立，显然不成立，故A正确；
对于B：若Mx的图象是一条抛物线，则x2 3xmx在xR上恒成立，即x24xm0在xR上恒
成立，只需164m 0，即m4即可，故B正确；
42 6
对于C：当m2时，f(x)x23x2 ，令 f(x) g(x)，即x23x2 x，解得x 2 6，
2
     
此时M 62  f 62 g 6 2  6 2 1，故C错误；
对 于 D ： ① 由 选 项 B 可 知 ， 当 m4 时 ， f(x)g(x) 在 xR 上 恒 成 立 ， 故
3 9
M(x) f(x)x2 3xm(x )2 m 0在xR上恒成立，显然不符题意；
2 4
②当m4时，当x0时，M(x)gxx0，即当x0时，M(x)0无解；
数学答案 第 1 页 共 5 页
{#{QQABJQCEogigAJBAARhCQwXyCgKYkBACAIgGhBAYoAAAwRFABCA=}#}f(x )0
若x 1且x Z，使得M(x )0，则 0 ，因为 f(x)x23xm 在x(0，)单调递减，
0 0 0 g(x )0
0
f(1) f(x )0
故 0 ，则M(1)0，此时x1也是满足题意的整数解，与题意不符.因此该唯一整数解只能
g(1)10
为1，
M(1)0 f 14m 0
即 ，所以 ，解得4m10，
M(2)0 f 210m0
综上所述，实数m的取值范围为4,10 ，故D正确.故选：ABD.
14.y  f(x1)1 f x对称中心为(1,1),
f(1011) f(1010) f(1009) f(1012) f 101321012 f(1)2025
15.
3sincos 1 3
方法一：（1）   tan ···································································3分
sin3cos 3 4
sin2sincoscos2 tan2tan1 19
sin2sincoscos2   ···········8分
sin2cos2 tan21 25
     
1cos 1cos 1cos 1cos 1cos 1cos
（2）为第三象限角，   
     
1cos 1cos 1cos 1cos 1cos 1cos
1cos 1cos 2 8
    ·············································································13分
sin sin tan 3
3sincos 1
方法二：  4sin3cos································································3分
sin3cos 3
 3
sin
4sin3cos   5
为第三象限角，  ························································7分
sin2cos21

4
cos

 5
2 2
 3 4 3  4 19
（1）sin2sincoscos2       ······································10分
 5 5 5  5 25
1cos 1cos 8
（2）   ·················································································13分
1cos 1cos 3
1x
16.（1）因为 0所以1 x1， f(x)的定义域为：(1,1)·············································2分
1x
1x 1x
因为log ( )log ( )log 10所以则 f(x) f(x)0，所以 f(x)为奇函数.···············5分
2 1x 2 1x 2
数学答案 第 2 页 共 5 页
{#{QQABJQCEogigAJBAARhCQwXyCgKYkBACAIgGhBAYoAAAwRFABCA=}#}1x 1x 1x 3x1
（2）由（1）可知 f(x)log ( )，所以， f(x)log ( )1 2 0········ 6分
2 1x 2 1x 1x 1x
1
所以，(3x1)(x1)0，即：  x1·············································································8分
3
1
所以，不等式 f(x)1的解集为：{x|  x1}.·································································10分
3
1x 1x
（3）对于函数 f(x)log ( )，令g(x) ，由反比例函数性质可知，g(x)在(1,1)内单调递增，
2 1x 1x
故 f(x)在(1,1)内单调递增，··························································································12分
由 f(2m1) f(m1)0可得 f(2m1)f(m1)，
因为 f(x)是奇函数，故 f(2m1) f(1m)·········································································13分
2m11m
 2
12m11，解得m(0, )························································································15分
 3
11m1
17.（1）总收入：12x·····································································································1分
当0 x6时，F  x  12x8  2x2 2x  2x2 10x8················································3分
 128  128
 
当x6时，F x 12x814x 652x 57·······································5分
 x   x 
2x2 10x8,0 x6

 
所以，2025年总利润为：F x   128 ··················································7分

2x 57,x6
  x 
2
 5 9
（2）当0 x6时，F  x  2x2 10x82x  
 2 2
5 9
当x 时，利润最大，最大为 万元.··············································································10分
2 2
 128 128
当x6时，F  x  2x 572 2x 5725
 x  x
128
当且仅当2x ，即：x8时，利润最大，最大为25万元.··············································13分
x
9
因为25 ，所以年产量为8万件时，利润最大，最大为25万元.··········································15分
2
1
18.（1） f(1)g(1)2， f(1)g(1)f(1)g(1) ，·············································2分
2
3 5
解得 f(1) ，g(1) ································································································ 3分
4 4
数学答案 第 3 页 共 5 页
{#{QQABJQCEogigAJBAARhCQwXyCgKYkBACAIgGhBAYoAAAwRFABCA=}#}（2）因为 f(x)是奇函数，g(x)是偶函数，且 f(x)g(x)2x①，
则 f(x)g(x)2x，即 f(x)g(x)2x②，···························································4分
2x 2x 2x 2x
联立①②可得 f(x) ，g(x) ································································6分
2 2
2x 2x
又因为 f(x) 2x12x1，而2x1在R上为增函数，2x1在R上为减函数，············7分
2
2x 2x
则 f(x) 在R上为单调递增函数··········································································8分
2
（2）由（1）可知， f(x)在R上为单调递增函数，
 
则sincossincos1在

0,

恒成立，·····················································10分
 4
 
即sincossincos10③在

0,

恒成立
 4
t2 1
令sin cos t，则sincos ，
2
t2 1
则③式变为t 10，即(t2 1)2(t1)，························································ 12分
2
    
又因为

0,

，sin cos 2sin( )  t 1，2 ······································14分
 4 4
所以当t 1时，R····································································································15分
  2 2
当t 1, 2 时， ，即( ) 2( 21)······················································16分
t1 t1 min
 
综上，实数的取值范围 ，2( 21) ···········································································17分
19.（1） f(x) x2 axa3·························································································4分
a a2
（2） f(x)(x )2 a3， f(x) 0
min
2 4
①当a4时， f(x)  f(2)a70,解得7a4················································5分
min
a a2
②当4a4时， f(x)  f( ) a30 ,
min 2 4
解得6a2，故4a2························································································6分
数学答案 第 4 页 共 5 页
{#{QQABJQCEogigAJBAARhCQwXyCgKYkBACAIgGhBAYoAAAwRFABCA=}#}③当a4时， f(x)  f(2)73a 0 ,无解·······························································7分
min
综上，7a2··········································································································8分
（3）x ，x 是一元二次方程x2axa30的两个不相等的实数根，
1 2
故Δa24a120，解得：a6或a2，
x x a
由韦达定理得： 1 2 ，························································································10分
xx a3
1 2
x 1 2  x 2 2  x 1 3 x 2 3  (x 1 x 2 )  (x 1 x 2 )2 3x 1 x 2    a(a2 3(a3))  a3 3a2 9a ，··················12分
x x x x x x  a3 a3
2 1 1 2 1 2
27 27
a26a9 a32 .·················································································14分
a3 a3
27
此式需为整数，因a为整数，(a3)2为整数，故 需为整数.
a3
27
又aN*，所以a3，又 Z，
a3
所以27 的因数：-27，-9，-3，-1，1，3，9，27，
故a4,6,12,30.·············································································································17分
数学答案 第 5 页 共 5 页
{#{QQABJQCEogigAJBAARhCQwXyCgKYkBACAIgGhBAYoAAAwRFABCA=}#}