四川省眉山市2026届高三第二次模拟测试数学+答案

第1页共6页一、选择题每小题5分，共40分）1.D；2.A；3.B；4.A；5.B；6.C；7.C；8.B.二、选择题每小题6分，共18分）9.AD；10.BCD；11.ACD；三、填空题每小题5分，共15分）214.f=sinx或fcosπx或f(x)=|x_2k|,x∈[2k_1,2k+1],k∈Z（答案不唯一四、解答题共77分）因为sin2A_sin(B+C)=0，所以2sinAcosA_sinA=0，所以cosA所以A··································································6分（2）因为AD为BC边上的中线，有两边同时平方可得：222由重要不等式：c2+b2≥2bc（当且仅当b=c时取得等号)得bc≤.···由三角形面积公式SbcsinA当且仅当b=c时取等316.解1）取PA中点F，连接EF,DF.212所以四边形CDFE是平行四边形，所以CE∥DF.················································································3分（2）解：因为三棱锥E_ABC与四棱锥P_ABCD的高之第2页共6页比等于，所以三棱锥E_ABC与四棱锥P_ABCD的体积之比等于解得λ=2······················································································7分取AD中点O，则在等腰直角三角形PAD中，PO丄AD.又因为平面PAD丄平面ABCD，平面PAD平面ABCD=AD，POC平面PAD，取BC中点G，则在梯形ABCD中，OG∥AB.以点O为坐标原点，分别以OD,OG,OP的方向为x轴，y轴，z轴的正方向，建立设平面ACE的法向量为n=(x,y,z)，所以平面ACE与平面PAB的夹角的余弦值为615分9又f(1)=_2，:切点为(1，_2)···························································4分第3页共6页f,(x)<0满足题意····································································································8分2a若即a时，f,(x)≥0恒成立，f(x)在(0，+∞)单调递增，与题意矛盾舍掉.盾舍掉.掉·································································································10分由题f(x)在区间(0,1)上单调递增，在区间(1，+∞)单调递减，x→0时，f(x)→_∞,x→+∞时，f(x)→_∞,又f(x)max=f(1)=_a_1·································12分_a_1>0，即a<_1时，函数f(x)的有两个零点，···································14分所以C是以A(_2,0)，B(2,0)为焦点，6为长轴长的椭圆，·······················4分所以C的方程为.·······························································5分由（1）可知点B为椭圆的右焦点，而=λ,λ>0所以P,Q为过右焦点B的直线与椭圆的交点因为C与直线l2均关于x轴对称，所以若直线QR过定点，该点必在x轴上.····································································································7分第4页共6页当直线PQ的斜率不为零时，设直线PQ的方程为x=my+2，P(x1,y1),Q(x2,y2),2因为t，所以··············································14分__所以xy1y2y1y__ _t4所以当t=时，直线QR恒过定点··············································17分方法二：当PQ的斜率存在且不为零时，若直线QR恒过定点H而所以只需y2=0，xx1x所以只需第5页共6页所以t·······················································································16分19.解：为方便表述，设事件Bn：第n天该学生晨读打卡为无效打卡（1）根据全概率公式，第2天为有效打卡的概率，由第1天有效、无效两种情况拆分：P(A2)=P(A1)P(A2|A1)+P(B1)P(A2|B1)，（2）代入已知条件：P下面推导P(An)与P(An-1)的递推式对任意n≥2，第n天有效打卡仅与第n-1天结果相关，由全概率公式P(An)=P(An-1)P(An|An-1)+P(Bn-1)P(An|Bn-1)，且P(Bn-1)=1-P(An-1)代入条件概率得P····································································································4分故X2的分布列为：X2012P 1613 12····································································································9分所以数学期望E···········································10分 第6页共6页结合（1）中递推式P且P(An-1)=E(Xn-1)-E(Xn-2)(n≥3)联立得E-E-E··············