2024屆綿陽三診 文數(shù)答案

綿陽市高中2021級(jí)第三次診斷性考試文科數(shù)學(xué)參考答案及評(píng)分意見一、選擇題：本大題共12小題，每小題5分，共60CDACC二、填空題：本大題共4小題，每小題5分，共20．2．6215．16．3三、解答題：本大題共6小題，共701．1）①當(dāng)n16a14aa．1分1112②當(dāng)n2時(shí)，由6S14a6S14a分nnn1n1兩式相減，得6a4a4a·······················································3分nnn1∴21aannanan12(n，4分∴數(shù)列a}n1a為首項(xiàng)，2為公比的等比數(shù)列，························5分121∴a(1．·········································································6分nn2（）由（112(]142n1nbSn2n11(63分4n可知數(shù)列{}343為首項(xiàng)，4為公比的等比數(shù)列，分∴Tn434)nn614······································································10分n44n1分692382n3n12分．1）調(diào)試前，電池的平均放電時(shí)間為：2.5×0.02×5+7.50.065+12.5×0.085+17.50.04×小時(shí)，···········4分a調(diào)試后的合格率為：0.1×5+0.06×5=0.8a·······················5分a；6分22）由列聯(lián)表可計(jì)算2，10分K40607295%的把握認(rèn)為參數(shù)調(diào)試能夠改變產(chǎn)品合格率．··························12分?jǐn)?shù)學(xué)（文科）評(píng)分標(biāo)準(zhǔn)第1頁共6頁．1）∵E是的中點(diǎn)，AB=，AEPB，1分又平面∩PBCPB，且平面⊥平面PBC，AEPBC，2分過D作DF交于F，PCDPBC，且平面PCD∩PBC=PC，⊥平面PBC，········································································4分AEDF分又PCD，PCD，AEPCD；········································································6分2）∵∥=V=，==13S·d8分又∵平面PBC⊥平面C作CH交于H，⊥平面9分2在直角△中：dCH·sin22=210分2∴221VSsin分C332sin∠BAP時(shí)，體積的最大值為8312分?jǐn)?shù)學(xué)（文科）評(píng)分標(biāo)準(zhǔn)第2頁共6頁11．1）解：當(dāng)a122f(x)(xx)lnxxx1分24fxxx分()(此時(shí)切線斜率為：ke1；3分1所以曲線f(x)在(e，f(e))處的切線方程：2ye(exe)···············4分43xy2(eee0；分42）證明方法一：因?yàn)閒(x)(xaxa)，·································6分由f(x)0xaf(x)00xa．∴f(x)在a)單調(diào)遞減，在(a)單調(diào)遞增．∴5f(x)f(a)a，分2min4555f(x)a)e1，即證：a2a1aa)e，4441a只需證：ea1a，8分a2設(shè)1xg(x)x，即證：g(x)1在x(1)恒成立分1x2則[(x2)xxgx，()x1x3令h(x)(x2)xx1，10分2∴h(x)x2，x∴h(x)在(1)上單調(diào)遞增，則h(x)h0∴h(x)在(1)上單調(diào)遞增，則h(x)h0分∴g(x)0在(1)恒成立，則h(x)在(1)上單調(diào)遞增，∴g(x)g0，原不等式得證12分2：因?yàn)閒(x)(xaxlna)，6分由f(x)0xaf(x)00xa．∴f(x)在a)單調(diào)遞減，在(a)單調(diào)遞增．?dāng)?shù)學(xué)（文科）評(píng)分標(biāo)準(zhǔn)第3頁共6頁5∴2f(x)f(a)a分min4555f(x)a)ea1，即證：2)e1aa，a444即證：a2a)ea1(a，即證：a1a(a，ea1a只需證：x1x，ex1x令g(x)xex11xgx)，x即證：g(x)gx)，·····························································8分1xg(x)且x1ex11xg(x)0，ex1∴g(x)在x(1)單調(diào)遞減，9分又x(1)，1x)，x1lnx，只需證：x1lnx0·································10分令h(x)x1x，∴'1h(x)10h(x)在x(1)單調(diào)遞增，分x∴h(x)h0x1lnx0，所以原不等式得證．12分b32．1）離心率e1a22ba1，①································1分2當(dāng)=1，a1a122yb|b=3aa22，②····························3分聯(lián)立①②得:a，b14分故橢圓C方程為：x24y21；分2）設(shè)過，AB三點(diǎn)的圓的圓心為Q(0，)，(xyB(xy)，1122又F(，則|2=|QF2(x0)2(yn)2(03)2(n0)2，·················6分11數(shù)學(xué)（文科）評(píng)分標(biāo)準(zhǔn)第4頁共6頁又(xy)在橢圓11x24y21上故x214y，112帶入上式化簡(jiǎn)得到：y210，③7分11同理，根據(jù)可以得到：y10，④8分2=2222由③④可得：yy是方程3y210的兩個(gè)根，則1,21yy9分123x2y21設(shè)直線AB：xty1，聯(lián)立方程：4，xty1整理得：t2y230，⑤10分故31yy122t43，解得：t25，∴t5分l的斜率為：5．···························································12分5．1）方法一：令x0cos3sin0，3tan，········································································1分3∴2或2，kZ662分13當(dāng)2y23()4·································3分62213當(dāng)2y230，····································4分6221與y軸的交點(diǎn)坐標(biāo)為（，4，05分方法二：消參：由C1的參數(shù)方程得：x2)3sin)3)134，·············1分2(y222即曲線1的普通方程為：x2(y2)242分令x0y0或4，4分1與y軸的交點(diǎn)坐標(biāo)為（，4，05分?jǐn)?shù)學(xué)（文科）評(píng)分標(biāo)準(zhǔn)第5頁共6頁2）方法一：將曲線：x2(y2)24化為極坐標(biāo)方程，4sin，分CC2的極坐標(biāo)方程)4sin)22，334sin123sin(sin3cos)1，則1····················7分221整理得：)，所以或······························8分62666即或，9分62∴∠．································································10分263方法二：將2的極坐標(biāo)方程)2，3化為直角坐標(biāo)方程：3xy40，分2是過點(diǎn)（0,4）且傾斜角為3的直線，7分不妨設(shè)（0,4，因?yàn)闉橹睆剑浴?，9分2∴∠．································································10分263．1a33ab)bab3，①1分abf(x)xaxb（xa)(xb)ba2，3分且ab0，所以ab2，4分ab1；···················································