广东省惠州市2025届高三上学期第一次调研考试数学答案

惠州市2025届高三第一次调研考试试题高三数学参考答案与评分细则一、单项选择题：本题共8小题，每小题满分5分，共40分.题号12345678答案DABCACBC22所以等腰三角形A1BC的面积为×2×2=2，22ab，由余弦定理可得cosC=因为，所以C=，故选：C.高三数学答案第1页，共9页,B(x2,8．【解析】f(x)的正零点为与所以f(x)从左到右的零点依次为:,,,,,,.为了使得f(x)在区间(0,π)恰有6个零点,只需解得.故选：C.二、多项选择题：本题共3小题，每小题满分6分，共18分。在每小题给出的四个选项中，有多项符合题目要求。全部选对得6分，部分选对得部分分，有选错的得0分。题号9全部正确选项BDACDABD乙机构测评分数的平均分乙==93，所以选项A错误，对于选项B，甲机构测评分数的方差22222D2对于选项C，乙机构测评分数从小排到大为：91，92，93，94，95，所以乙机构测评分数的中位数为93，所以选项C错误，对于选项D，因为甲机构测评分数中有且仅有2个测评分数超过平均分，由对立事件的定义知，事件M,N互为对立事件，所以选项D正确，故选：BD.高三数学答案第2页，共9页9T所以log9T2a8当且仅当a3=a7时，等号成立．所以D正确．故选:ACD..2.(x1)2+y2.2.(x1)2+y2.(x1)2+y2.2方程不变，所以曲线C关于y轴对称，A选项正确；..y22y2，所以B选项正确；对于C选项：△F1PF2面积为×F1F2×yP=yP，+9则x2+1则2y2y2则△F1PF2面积的最大值为，所以C选项错误；则OP的取值范围为,2，所以D选项正确，故选：ABD.三、填空题：本题共3小题，每小题5分，共15分.与B,sin关于y轴对称，高三数学答案第3页，共9页即θ,θ+关于y轴对称，θ++θ=π+2kπ,k∈Z，则θ=kπ+,k∈Z，当k=0时，可取θ的一个值为.故答案为满足θ=kπ+,k∈Z即可）.(5,2,14．【解析】在f(|x)|=1f(x)中令x=0,得f(0)=0,又f(x)+f(1—x)=1,(5,2,,当0≤x1<x2≤1时f(x1)≤f(x2),可知f(x)是[0,1]上的非递减函数,四、解答题：本题共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本小题满分13分，其中第一小问6分，第二小问7分。）【解析】（1）因为f,(x)=lnx+1+a·······································································2分解得a=3.·······································································································6分令f,(x)>0，得x>e2，令f,(x)<0，得0<x<e2，·········································【注】没有说明极大值扣一分16本小题满分15分，其中第一小问6分，第二小问9分。）【解析】（1）因为学生初试成绩X服从正态分布N(μ,σ2)，其中μ=65，σ2=102，高三数学答案第4页，共9页=0.15865，·······································································································4分所以估计初试成绩不低于的人数为0.15865×1000≈159人.···········································6分故Y的分布列为：Y0P125 25 20 2717本小题满分15分，其中第一小问6分，第二小问9分。） DE2=CE2····························································3分由PC∩CD=C,DE垂直于平面PCD内两条相交直线,················································5分（2）解:由(1)知,ΔCDE为等腰直角三角形,上DCE=,如图,过点D作DF垂直CE于F,易知又已知EB=1,故FB=2.==,故AC=DF=.···········································································8分高三数学答案第5页，共9页------(3)(2,以C为坐标原点,分别以CACB,------(3)(2,则C(0,0,0),,P(0,0,3),A|,0,0|,E(0,2,0),D(1,1,0),·················9分设平面PAD的法向量1=(x1,y1,z1),l2l2x1-y1=0 6【注】写出公式给1分，代入数据给1分，结果计算正确给1分方法二：设平面PAD的法向量1=(x1,y1,z1),[x+[x+y-1l3y1-3z1-x-18本小题满分17分，其中第一小问3分，第二小问5分，第三小问9分。）+y2则抛物线C2的焦点为F(0,1)，··············································································2分故p=2.··········································································································3分（2）若直线MN与y轴重合，则该直线与抛物线C2只有一个公共点，不符合题意，设直线MN的方程为y=kx+1，点M(x1,y1)、N(x2,y2)，高三数学答案第6页，共9页2OM.ON=x1x2+y1y2=x1x2+44=-4+1（3）设直线NO、MO的斜率分别为k1、k2，其中k1>0，k2<0，2点B在第四象限，同理可得xB=·4k+1，······························································10分S△OMNOM.ONx1.x2x1.x2 2214k2+1当且仅当k1=时，等号成立.··············································································16分的取值范围为[2,+∞).···············································································17分（3）方法二：设直线NO、MO的斜率分别为k1、k2，其中k1>0，k2<0，解得x=±4k+1，点解得x=±4k+1，点A在第三象限，则xA=-4k+1，2点B在第四象限，同理可得xB=·4k+1，······························································10分高三数学答案第7页，共9页设点B到直线OA的距离-4x1x2ΔOABΔOAB19本小题满分17分，其中第一小问3分，第二小问6分，第三小问8分。）-an，数列{2n}是“速增数列”.····································3分又an∈Z,故a3-a2≥3·······················································································4分依次递推可得a4-a3≥4,a5-a4≥5, 当k≥3时，ak=2023=(ak-ak-1)+(ak-1-ak-2)+ 2-a1)+a1································6分 高三数学答案