2026届广西壮族自治区高一数学月考质量检测QS01黑白可打印原创仿真卷B1第068套（含答案详解、评分标准与作答空间）

高一数学月考质量检测·黑白打印版第页2026届广西壮族自治区高一数学月考质量检测QS01黑白可打印原创仿真卷B1第068套（含答案详解、评分标准与作答空间）高一数学月考质量检测仿真卷B1第068套适用对象高一年级月考复习与课堂检测考试时间150分钟试卷总分150分内容范围集合、常用逻辑表述、函数初步、不等式难度比例基础约60%，中档约30%，综合约10%呈现形式黑白可打印，含作答空间作答要求选择题涂写答案栏，填空题写最终结果，解答题写步骤批改口径按参考答案与评分标准执行注意事项1.本卷共22题，满分150分，考试时间150分钟；请使用黑色字迹签字笔在规定区域内作答。2.选择题必须把答案填入答案栏；填空题只写最终结果，必要时注明取值范围。3.解答题应写出必要的文字说明、演算步骤或推理过程；仅写最后答案但缺少关键步骤的，按评分标准酌情给分。4.全卷围绕集合、函数与不等式展开，侧重基础概念、符号表达、运算转化和综合应用。5.作答空间按A4黑白打印预留，若空间不足，可在对应题号旁续写并保持步骤清晰。试卷结构与考点分布题型题号分值主要考点能力侧重单项选择题1—840分集合运算、命题关系、函数定义域、函数性质、绝对值不等式、二次函数、复合函数、反比例型函数识记、辨析、快速运算多项选择题9—1220分集合关系、二次不等式、绝对值函数、基本不等式思想多结论判断、反例意识填空题13—1620分一次函数参数、分式不等式、集合包含、闭区间最值准确表达、端点处理解答题17—2158分集合含参、分式不等式、二次函数最值、函数模型、区间关系规范推理、步骤书写综合题2212分函数比较、恒成立问题、二次方程根的分布转化思想、综合分析答题规范提示：集合题要写清区间端点的开闭；函数题要先确认定义域；不等式题要说明变形依据，涉及分式时必须排除分母为零的点；含参数问题要把条件转化为端点、区间或最值关系后再求参数范围。学生信息与客观题答案栏学校班级姓名座号总分阅卷人选择题答案栏：题号123456789101112答案备注多选题全部选对得5分，少选且正确得2分，有错选得0分。填空题答案栏：题号13141516答案一、单项选择题（本大题共8小题，每小题5分，共40分。每小题只有一个选项符合题意。）1.已知集合A={x|-1≤x<4}，B={x|x²-3x≤0}，则A∩B=A.[0,3]B.[-1,4)C.(-∞,0]∪[3,+∞)D.[3,4)2.命题“若x>2，则x²>4”的逆命题是A.若x²>4，则x>2B.若x≤2，则x²≤4C.若x²≤4，则x≤2D.若x>2，则x²≤43.函数f(x)=√(2-x)+1/(x+1)的定义域为A.(-∞,2]B.(-∞,2]且x≠-1C.[-1,2]D.(-1,2]4.下列函数中，在其定义域上为奇函数的是A.f(x)=x/(x²+1)B.f(x)=x²+1C.f(x)=|x|D.f(x)=x+15.不等式|2x-1|<3的解集是A.(-1,2)B.[-1,2]C.(-2,1)D.(1,2)6.二次函数y=x²-4x+1的最小值是A.-4B.-3C.1D.47.已知f(x)=2x-1，g(x)=x²+1，则(g∘f)(2)=A.5B.8C.10D.178.函数y=1/(x-2)的单调性表述正确的是A.在(-∞,2)和(2,+∞)上均为减函数B.在定义域上为减函数C.在(-∞,2)上增、在(2,+∞)上减D.在(-∞,2)上减、在(2,+∞)上增二、多项选择题（本大题共4小题，每小题5分，共20分。全部选对得5分，少选且正确得2分，有错选得0分。）9.设全集U={1,2,3,4,5,6,7,8}，A={2,4,6,8}，B={1,2,3,4}，下列说法正确的是A.A∩B={2,4}B.A∪B={1,2,3,4,6,8}C.∁UA={1,3,5,7}D.B⊆A10.关于不等式x²-5x+6≤0，下列结论正确的是A.解集为[2,3]B.其对应二次函数开口向上C.x=1满足该不等式D.x=2和x=3均使等号成立11.函数f(x)=|x-1|+2具有的性质是A.最小值为2B.图象关于直线x=1对称C.在(-∞,1]上单调递减D.值域为(-∞,2]12.若a>0，b>0，且a+b=4，则下列判断正确的是A.ab≤4B.1/a+1/b≥1C.a²+b²≥8D.a,b均大于2三、填空题（本大题共4小题，每小题5分，共20分。）13.若一次函数f(x)=kx+3满足f(2)=7，则k=答：____________________14.不等式1/(x-1)≥0的解集为答：____________________15.已知A={x|x²-5x+6=0}，B={x|x<a}，若A⊆B，则整数a的最小值为答：____________________16.函数f(x)=x²-2x在区间[0,3]上的值域为答：____________________

四、解答题（本大题共6小题，共70分。解答应写出必要步骤、推理过程与结论。）17.（10分）已知集合A={x|x²-5x+6≤0}，B={x|m<x<5}。（1）当m=1时，求A、B以及A∪B；（2）若A⊆B，求实数m的取值范围。17题作答区____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

18.（12分）解不等式：(x-1)(x+2)/(x-3)≥0，并用数轴或符号表说明理由。18题作答区______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

19.（12分）已知函数f(x)=x²-2x-3。（1）求函数的零点；（2）解不等式f(x)≤0；（3）写出函数的单调区间，并求f(x)在[-1,4]上的最小值与最大值。19题作答区________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

20.（12分）某校印制月考练习卷。方案甲的费用由固定制版费180元和每份0.15元组成；方案乙无固定制版费，但每份0.30元。设印制份数为n（n为正整数）。（1）分别写出甲、乙两种方案的费用函数C₁(n)、C₂(n)；（2）若预算不超过600元，按方案甲最多可印多少份？（3）从节约费用角度看，印制份数达到多少时选择方案甲更合适？20题作答区________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

21.（12分）已知A={x|x²-2x-3<0}，B={x|a-1≤x≤a+2}。（1）求集合A；（2）若B⊆A，求实数a的取值范围；（3）若A∩B≠∅，求实数a的取值范围。21题作答区________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

22.（12分）已知f(x)=x²-2x-3，g(x)=kx+1，k为实数。（1）当k=1时，解不等式f(x)≤g(x)；（2）若对任意x∈[0,2]，都有f(x)≤g(x)，求k的取值范围；（3）若方程f(x)=g(x)的两个实数根都在区间(-2,5)内，求k的取值范围。22题作答区__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案、详解与评分标准一、客观题答案表题号123456789101112答案AABAABCAABCABDABCABC得分555555555555二、填空题答案表题号13141516答案2(1,+∞)4[-1,3]三、逐题解析、评分标准与易错提醒1.【答案】A解析：由x²-3x≤0得x(x-3)≤0，故B=[0,3]。又A=[-1,4)，所以A∩B=[0,3]。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：先求B再交集，不能把二次不等式解成外侧区间。2.【答案】A解析：把原命题的条件和结论互换得到逆命题，即“若x²>4，则x>2”。该逆命题不一定为真，但题目只问形式。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：逆命题与否命题、逆否命题不同，不需要判断真假后再改写。3.【答案】B解析：根式要求2-x≥0，即x≤2；分式要求x+1≠0，即x≠-1。综合得定义域为(-∞,2]且x≠-1。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：定义域要同时满足所有限制，分母为零的点必须剔除。4.【答案】A解析：f(-x)=(-x)/[(-x)²+1]=-x/(x²+1)=-f(x)，定义域关于原点对称，所以为奇函数。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：判断奇偶性前要看定义域，再验证f(-x)与f(x)的关系。5.【答案】A解析：|2x-1|<3等价于-3<2x-1<3，解得-1<x<2。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：严格不等号不会包含端点。6.【答案】B解析：y=x²-4x+1=(x-2)²-3，故最小值为-3。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：顶点纵坐标才是最小值，不是顶点横坐标。7.【答案】C解析：f(2)=3，g(f(2))=g(3)=3²+1=10。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：复合函数按由内到外计算，不能写成f(g(2))。8.【答案】A解析：反比例型函数y=1/(x-2)在x<2与x>2两个区间上分别为减函数，但不能说在整个定义域上为减函数。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：函数在断开的定义域上讨论单调性时，要分区间表述。9.【答案】ABC解析：A∩B={2,4}，A∪B={1,2,3,4,6,8}，∁UA={1,3,5,7}均正确；B中有1、3不属于A，因此B⊆A错误。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：补集必须在给定全集U内求。10.【答案】ABD解析：x²-5x+6=(x-2)(x-3)，开口向上，≤0时取两根之间，解集为[2,3]；x=1时代入为2>0，不满足。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：二次不等式的开口方向决定取内侧还是外侧。11.【答案】ABC解析：f(x)=|x-1|+2的图象是顶点为(1,2)的V形，最小值2，关于x=1对称，在(-∞,1]上随x增大而减小，值域为[2,+∞)。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：绝对值函数向上平移后下界随之改变。12.【答案】ABC解析：由(a-b)²≥0得a²+b²≥2ab；又(a+b)²=a²+2ab+b²≥4ab，所以ab≤4。1/a+1/b=(a+b)/ab=4/ab≥1。a²+b²=(a+b)²-2ab≥16-8=8。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：正数条件很关键，不能推出a、b都大于2。13.【答案】2解析：由f(2)=2k+3=7，得2k=4，k=2。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：代入自变量2后再解参数，不能把f(2)误写为2f。14.【答案】(1,+∞)解析：分母x-1不能为0。分式1/(x-1)的分子为正，因此分式非负等价于x-1>0，解得x>1。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：由于分子不是0，等号不可能成立。15.【答案】4解析：方程x²-5x+6=0的两根为2、3，故A={2,3}。若A⊆B={x|x<a}，需2<a且3<a，即a>3，整数a的最小值是4。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：集合包含要求A中每一个元素都属于B。16.【答案】[-1,3]解析：f(x)=x²-2x=(x-1)²-1，在[0,3]上最小值为-1；端点值f(0)=0，f(3)=3，最大值为3。评分：选择题或填空题答对得5分；多选题按卷首说明计分。易错提醒：区间最值要比较顶点和端点。17.【答案】（1）A=[2,3]，当m=1时B=(1,5)，A∪B=(1,5)；（2）m<2。解析：先因式分解x²-5x+6=(x-2)(x-3)。由(x-2)(x-3)≤0，得A=[2,3]。当m=1时，B=(1,5)，由于[2,3]完全落在(1,5)内，所以A∪B=(1,5)。若A⊆B，则集合A中的端点2、3都必须属于B。因为3<5恒成立，而2属于B要求m<2，故m的取值范围为m<2。评分点关键内容分值求A正确因式分解并得A=[2,3]3分第（1）问写出m=1时B=(1,5)，并求A∪B=(1,5)3分第（2）问由A⊆B转化为m<2，并说明端点2必须落入B4分易错提醒：B是不含端点的开区间，若写成m≤2会把x=2是否属于B判断错。18.【答案】[-2,