四川省绵阳市2026届高三上学期第二次诊断性考试数学（A）试题含答案

绵阳市高中2023级第二次诊断性考试一、选择题：本题共8小题，每小题5分，共40分．二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不全的得部分分，有选错的得0分．三、填空题：本题共3个小题，每小题5分，共15分．四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17分，共77分．解答应写出文字说明、证明过程或演算步骤．15．解1）由正弦定理得：3sinAcosB=3sinC—sinB，································2分∴3sinAcosB=3sin(A+B)—sinB，························································4分∴3cosAsinB=sinB，·········································································6分∵△ABC的周长为8，16．解·························································2分 =0，又fln3，··························································4分 ····························8分由f’(x)<0得f(x)的单调递减区间为：(3，6)，······································10分∵fln6，····················································12分∵e44>27，则ffln3>0，····································14分即·······································································4分∴数列{anbn}是首项为2，公比为2的等比数列；···································6分，··························································7分an=4(n≥2)，·································9分2n，··················································13分0n2解得：a2=8，b2=2，········································································2分（2）方法一：设P(x0，y0)，A(x1，y1)，B(x2，y2)，代入椭圆方程可得：2=8，代入上式可得：x1x2+4y1y2=—2，平方可得：x12x22+8x1x2y1y2+16y12y22=4，····6分∴4S2=|y1x2x1y2|2=x12y222x1x2y1y2+x22y12=x12(2)2x1x2y1y2+(84y22)y12+8y··············9分∴△OAB的面积为定值；······························································10分（2）方法二：设P(x0，y0)，A(x1，y1)，B(x2，y2)，代入椭圆方程可得：，代入上式可得：x1x2+4y1y2+2=0，又A(x1，y1)，B(x2，y2)在直线y=kx+t上，+t，则x1x2+4(kx1+t)(1x212)24x1x2···························································8分 （3）设A(x1，y1)，B(x2，y2)，D(x3，y3)，E(x4，y4)，∵直线AB与直线DE平行，则直线AB与直线DE的斜率均为，由，则(m-x1，n-y1)=λ(x3-m，y3-n)，······································································12分同理，由=λ可得： 同理：(x4+x3)+4kCD(y4+y3)=0，则λ(x4+x3)如解图1，取PE中点H，连接HF，HG，则HF⊥PE，GH⊥PE，结合HF，HG二平面HGF，又∵FG二平面HGF，故PA⊥FG；（2）设AC与BD的交点为O，∵PC二平面PAC，平面PAC平面EBD=EO，∵E为PA的中点，故O为AC的中点，如解图2所示，建立空间直角坐标系，设B(xB，yB，0)，D(xD，yD，0)，AC中点O(1，0，0)，21得2n(m21)=2m(n21)，则mn=1，··················································6分m2同理可得n2=(2n，2n2，2n)，·····························································7分······························8分（3）∵BD//FG，且PA⊥FG，故PA⊥BD，结合PC⊥平面ABD，则可得PC⊥BD，因此BD⊥平面PAC，故BD⊥AC，故B，D关于平面PAC对称，设△ABD的外心为S，显然S应在x轴上，故有(x02)2=(x0t)2+4t，整理得x······························12分同时PA在平面PAC中的垂直平分线恰为CE，因此球心T即为过S且垂直于平面ABD的直线与CE的交点，令v=t2，则v>2且v≠0，代入x0及R2表达式，得Rv2+6v且给定该球的半径时，三棱锥P-BCD的体积有3个可能的值，等价于t有3个不同的解，即v有3个不同的解，①当R时，关于w的方程Rw2+6w+20，4·、i2，方程v在区间(‒2，0)有1解，v有唯一解2v2．故共有2组解，不满足题意；···········································································16分 分别有一解.此时关于v的方程w=v在区间(‒2，0)有一解，在(0，+∞)有2解，共3解，符合题意，代入x0的表达式为：x则x02024x0+4，后同解法一的讨论．方法二3）