綿陽市高中2023級(jí)第二次診斷性考試一、選擇題：本題共8小題，每小題5分，共40分．二、選擇題：本大題共3小題，每小題6分，共18分．全部選對(duì)的得6分，選對(duì)但不全的得部分分，有選錯(cuò)的得0分．三、填空題：本題共3個(gè)小題，每小題5分，共15分．四、解答題：本題共5小題，第15題13分，第16、17小題15分，第18、19小題17分，共77分．解答應(yīng)寫出文字說明、證明過程或演算步驟．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的周長為8，16．解·························································2分 =0，又fln3，··························································4分 ····························8分由f’(x)<0得f(x)的單調(diào)遞減區(qū)間為：(3，6)，······································10分∵fln6，····················································12分∵e44>27，則ffln3>0，····································14分即·······································································4分∴數(shù)列{anbn}是首項(xiàng)為2，公比為2的等比數(shù)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：設(shè)P(x0，y0)，A(x1，y1)，B(x2，y2)，代入橢圓方程可得：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的面積為定值；······························································10分（2）方法二：設(shè)P(x0，y0)，A(x1，y1)，B(x2，y2)，代入橢圓方程可得：，代入上式可得：x1x2+4y1y2+2=0，又A(x1，y1)，B(x2，y2)在直線y=kx+t上，+t，則x1x2+4(kx1+t)(1x212)24x1x2···························································8分 （3）設(shè)A(x1，y1)，B(x2，y2)，D(x3，y3)，E(x4，y4)，∵直線AB與直線DE平行，則直線AB與直線DE的斜率均為，由，則(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，則λ(x4+x3)如解圖1，取PE中點(diǎn)H，連接HF，HG，則HF⊥PE，GH⊥PE，結(jié)合HF，HG二平面HGF，又∵FG二平面HGF，故PA⊥FG；（2）設(shè)AC與BD的交點(diǎn)為O，∵PC二平面PAC，平面PAC平面EBD=EO，∵E為PA的中點(diǎn)，故O為AC的中點(diǎn)，如解圖2所示，建立空間直角坐標(biāo)系，設(shè)B(xB，yB，0)，D(xD，yD，0)，AC中點(diǎn)O(1，0，0)，21得2n(m21)=2m(n21)，則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，結(jié)合PC⊥平面ABD，則可得PC⊥BD，因此BD⊥平面PAC，故BD⊥AC，故B，D關(guān)于平面PAC對(duì)稱，設(shè)△ABD的外心為S，顯然S應(yīng)在x軸上，故有(x02)2=(x0t)2+4t，整理得x······························12分同時(shí)PA在平面PAC中的垂直平分線恰為CE，因此球心T即為過S且垂直于平面ABD的直線與CE的交點(diǎn)，令v=t2，則v>2且v≠0，代入x0及R2表達(dá)式，得Rv2+6v且給定該球的半徑時(shí)，三棱錐P-BCD的體積有3個(gè)可能的值，等價(jià)于t有3個(gè)不同的解，即v有3個(gè)不同的解，①當(dāng)R時(shí)，關(guān)于w的方程Rw2+6w+20，4·、i2，方程v在區(qū)間(?2，0)有1解，v有唯一解2v2．故共有2組解，不滿足題意；···········································································16分 分別有一解.此時(shí)關(guān)于v的方程w=v在區(qū)間(?2，0)有一解，在(0，+∞)有2解，共3解，符合題意，代入x0的表達(dá)式為：x則x02024x0+4，后同解法一的討論．方法二3）