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=}#}