2025届贵州省毕节市高三三诊数学试题及答案

项是符合题目要求的.2.已知函数则f=3.已知a是正项等比数列8，b，2，…的第5项，则logab=4.若命题p：k<2，命题q：直线y=kx−1与抛物线y=x2无公共点，则p是q的5.如图，体积为2τ的圆柱的轴截面ABCD为正方形，点E在底面CDCBABE6.函数f(x)=ex−e−x，若对于x∈R，f(sinx+m)+f(cosx)≤0恒成立，则m的7.省会贵阳已开通的地铁线路如图所示.某人乘图例A.6条B.7条D.图例oo●洛湾oo北京路oo●洛湾oo北京路BOBO贵州大学8.已知双曲线C右焦点为F，过点F作互相垂直的直线l1，l2.l1与C的右支交于M，N两点，NF=3FM，若l2与C的左支交于P点，且P，），二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项 D.若AC在AD上的投影向量为AD，则m=310.已知函数f(x)满足对任意的x,y∈R，都有f(x+y)+f(x−y)=2f(x)f(y)，且f(0)≠0.下列结论正确的是B.f(x)是偶函数C.若f(2)=3，则f(4)=6D.若f(1)=0，则4是f(x)的一个周期A.M1(x1,y1)，M2(x2,y2)是C上两点，x2>x1>0，则y2>y1D.点M到直线x−y=0和x=0的距离之积为定值 13.△ABC的内角A，B，C所对边分别为a，b，c.若B=边上的高为a，,a2,...,an}（n∈N*）的元素个数为Card(A)，子集个数为A(n)，称时，n的最小值为.四、解答题:本题共5小题，共77分,解答应写出f男女抽取3名进行座谈.记这3人中女员工人数为X，求X的分布列和数学期望.α0.001x2在且不为0的直线l经过点F2，且与E交于P，Q两点，△F1PQ周长为8.（2）已知P，Q在直线x=t上的射影分别为M，N.若直线PN与QM的交点T在x轴上，求t的值.（2）已知AB=2，E为线段BC中点.(ⅰ)若P是平面ADE内一点，AP=求线段CP长的最小值；若AD=，BD=2，l是平面ADE内一条直线，求l与平面ACD所成角an,数列{bn}的前n项和为Sn，证明：4Si+Si+1<12345678DCCBBABD9（1）由已知，切点坐标为(0,1).····························又f'(x)=2cos2x+2ax−1，········································所以切线方程为y=x+1；··············································· 因为f'(x)在上具有单调性，则f,,(x)≥0或f,,(x)≤0恒成立，··························································9分即时，有2a≥4sin2x，或2a≤4sin2x (1)零假设H0:喜欢户外运动与性别无关.··················································1分由于4.040＞3.841，··················故零假设H0不成立，所以喜欢户外运动与性别有关.··································7分（2）不喜欢户外运动的员工按分层抽样抽取9人，其中女员工4人.··············8分.····················································X的分布列为X0123P 42204242 42X的期望为.··························因为△F1PQ的周长为4a，·························································椭圆E方程为=1；··························································································································6分2所以y1+y2=，y1.y2=，···················直线PN方程为y−y2=.令y=0，得x=2(1)取线段BC中点E，连接EA，ED.·················································1分因为△ABC是正三角形，有BC丄AE.········································所以BC丄平面ADE，·····················································有BC丄DE，···············································································4分又E为线段BC中点，则BD=CD.············································································（2i）连接EP，AP22.···························································7分AE−AP当且仅当点P在线段AE的中点处，等号成立······8分所以线段CP的最小值为；····························································10···········································································以E为坐标原点，以EC，ED分别作为x轴，y轴正方向建立空间直角坐标系如图所示.·········································································yCC则A(0,，C(1,0,0)，D(0,, 设直线l的方向向量n=(0,m,n)，直线l与平面ACD所成角为θ,于是sinθ=.·································15分2所以sinθ≤.······················································16分故0≤sinθ≤.即直线l与平面ACD所成角的正弦值的取值范围是.·················17分即，··································3分所以数列是首项为2公差为1的等差数列；································5分所以cosan=，sinan.················································6分所以Sn=[log21−log22+log22−log23+…+log2n−log2(n+1)]log2.················································································7分，····························8分cosa0cosa1+cosa1cosa2+…+cosan−1cosan>ln+ln+…+ln即，················