2027届湖北武汉三中高三上学期9月月考数学试题及答案

高三数学1．答题前，务必将自己的个人信息填写在答题卡上，并将条形码置.2．回答选择题时，选出每小题答案后，用铅笔把答题卡对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其他答案标号．回答非选择题时，将答案写在答题卡上.一项是符合题目要求的.63θ为7．在平面直角坐标系xOy中，抛物线C:x2=2py(p>0)的焦点为F，过点F且倾斜角为的直线与C在第一象限的交点为A，与x轴的交点为B，若ABn+2(n*)的展开式中，含xyz项的系数为bn，记数列{}的前n项和为Tn，若不lbnJ2346合题目要求，全部选对的得6分，部分选对的得部分分，有选错的得0分.B．存在实数k，使直线l与曲线E无公共点C．直线l被曲线E截得的弦长的最小值为22），1 -/22 D．点Q的轨迹长度为τ甲胜乙的概率为p(0<p<1)，无平局，设甲本场比赛获胜且乙至少胜一局的概率为f(p)，则当f(p)取最大值时，p=.四、解答题：本题共5小题，共77分．解答应写（1）请完善上述2×2列联表，并分别估计老年人与非老年人患白内障的（2）根据小概率值α=0.005的独立性2n(ad−bc)22n(ad−bc)2αxα（1）求BD;（2）求△BCD的面积.（1）求证：MN//平面PAB；（2）求平面PAD与平面PCD夹角的余弦值.（2）讨论f(x)的单调性；点.（1）求E的方程.（2）设O为坐标原点，过点F的直线l与E的右支交于M，N两点，其中点N在第四象限，直线MO与E的另一个交点为R.(ⅱ)求△MNR面积的取值范围.对的得部分分，有选错的得0分.12四、解答题：本题共5小题，共77分．解答应写································································（2）零假设为H0：患白内障与年龄无关.······················22 11分） :sinA=.·············································································∠ABC=，:A为锐角， ABACsinCsin∠ABCABACsinCsin∠ABC:在△ABC中，由正弦定理得=，75917．解析（1）如图，取PA的中点为Q，连接BQ，MQ.M为PD的中点，:MQ//AD，MQ=1AD.························································································（2分）2:3BN=NC，:BN=:AD//BC，BC=2AD=2，:MQ//BN，MQ=BN，···························································································（4分）:四边形BNMQ为平行四边形，:MN//BQ，·············································································································（5分）:MN丈平面PAB，BQ⊂平面PAB，:MN//平面PAB.····································································································（6分）（2）连接AC.:PB=PA=AB=BC=2，∠ABC=，:△PAB，△ABC均为等边三角形，:平面PAB丄平面ABCD，平面PAB∩平面ABCD=AB，:AB，OP，OC两两互相垂直.········································以OB，OC，OP所在直线分别为x轴、y轴、z轴建立如图所示的空间直角坐标系Oxyz，则P0,,0 −y2取y2=22:cosm,n(3)22 (3)2255 :平面PAD与平面PCD夹角的余弦值为.···························································（15分） (2τ)(2τ),:f(x)在(0,2τ)上单调递减.·····················································································（9分）:H(x)在(|(,τ),|上单调递增，·····················(2,(2,,故h(x)在(0,x0)上单调递减，在(x0,τ)上单调递增，:h(x)在(0,τ)上有且只有一个极小值点x=x0.···························································（17分）19．解析（1）设双曲线E的半焦距为c.a :E的方程为x2−=1.··············由{lx23m2)y2由{lx23m2)y2:FM=5NF，2,2)，:y1=−5y2，······················································································· 45m29:y1y2， :(ⅱ)由双曲线的对称性知，MO=RO，:S△MON=S△NOR，···································································································（12分） :S△MNR=2S△MON=2××OF×y1−y2=OFy1−y2=2y1−y2 (3m−1,3m−1(3m