2024届绵阳三诊 文数答案

绵阳市高中2021级第三次诊断性考试文科数学参考答案及评分意见一、选择题：本大题共12小题，每小题5分，共60CDACC二、填空题：本大题共4小题，每小题5分，共20．2．6215．16．3三、解答题：本大题共6小题，共701．1）①当n16a14aa．1分1112②当n2时，由6S14a6S14a分nnn1n1两式相减，得6a4a4a·······················································3分nnn1∴21aannanan12(n，4分∴数列a}n1a为首项，2为公比的等比数列，························5分121∴a(1．·········································································6分nn2（）由（112(]142n1nbSn2n11(63分4n可知数列{}343为首项，4为公比的等比数列，分∴Tn434)nn614······································································10分n44n1分692382n3n12分．1）调试前，电池的平均放电时间为：2.5×0.02×5+7.50.065+12.5×0.085+17.50.04×小时，···········4分a调试后的合格率为：0.1×5+0.06×5=0.8a·······················5分a；6分22）由列联表可计算2，10分K40607295%的把握认为参数调试能够改变产品合格率．··························12分数学（文科）评分标准第1页共6页．1）∵E是的中点，AB=，AEPB，1分又平面∩PBCPB，且平面⊥平面PBC，AEPBC，2分过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时，体积的最大值为8312分数学（文科）评分标准第2页共6页11．1）解：当a122f(x)(xx)lnxxx1分24fxxx分()(此时切线斜率为：ke1；3分1所以曲线f(x)在(e，f(e))处的切线方程：2ye(exe)···············4分43xy2(eee0；分42）证明方法一：因为f(x)(xaxa)，·································6分由f(x)0xaf(x)00xa．∴f(x)在a)单调递减，在(a)单调递增．∴5f(x)f(a)a，分2min4555f(x)a)e1，即证：a2a1aa)e，4441a只需证：ea1a，8分a2设1xg(x)x，即证：g(x)1在x(1)恒成立分1x2则[(x2)xxgx，()x1x3令h(x)(x2)xx1，10分2∴h(x)x2，x∴h(x)在(1)上单调递增，则h(x)h0∴h(x)在(1)上单调递增，则h(x)h0分∴g(x)0在(1)恒成立，则h(x)在(1)上单调递增，∴g(x)g0，原不等式得证12分2：因为f(x)(xaxlna)，6分由f(x)0xaf(x)00xa．∴f(x)在a)单调递减，在(a)单调递增．数学（文科）评分标准第3页共6页5∴2f(x)f(a)a分min4555f(x)a)ea1，即证：2)e1aa，a444即证：a2a)ea1(a，即证：a1a(a，ea1a只需证：x1x，ex1x令g(x)xex11xgx)，x即证：g(x)gx)，·····························································8分1xg(x)且x1ex11xg(x)0，ex1∴g(x)在x(1)单调递减，9分又x(1)，1x)，x1lnx，只需证：x1lnx0·································10分令h(x)x1x，∴'1h(x)10h(x)在x(1)单调递增，分x∴h(x)h0x1lnx0，所以原不等式得证．12分b32．1）离心率e1a22ba1，①································1分2当=1，a1a122yb|b=3aa22，②····························3分联立①②得:a，b14分故椭圆C方程为：x24y21；分2）设过，AB三点的圆的圆心为Q(0，)，(xyB(xy)，1122又F(，则|2=|QF2(x0)2(yn)2(03)2(n0)2，·················6分11数学（文科）评分标准第4页共6页又(xy)在椭圆11x24y21上故x214y，112带入上式化简得到：y210，③7分11同理，根据可以得到：y10，④8分2=2222由③④可得：yy是方程3y210的两个根，则1,21yy9分123x2y21设直线AB：xty1，联立方程：4，xty1整理得：t2y230，⑤10分故31yy122t43，解得：t25，∴t5分l的斜率为：5．···························································12分5．1）方法一：令x0cos3sin0，3tan，········································································1分3∴2或2，kZ662分13当2y23()4·································3分62213当2y230，····································4分6221与y轴的交点坐标为（，4，05分方法二：消参：由C1的参数方程得：x2)3sin)3)134，·············1分2(y222即曲线1的普通方程为：x2(y2)242分令x0y0或4，4分1与y轴的交点坐标为（，4，05分数学（文科）评分标准第5页共6页2）方法一：将曲线：x2(y2)24化为极坐标方程，4sin，分CC2的极坐标方程)4sin)22，334sin123sin(sin3cos)1，则1····················7分221整理得：)，所以或······························8分62666即或，9分62∴∠．································································10分263方法二：将2的极坐标方程)2，3化为直角坐标方程：3xy40，分2是过点（0,4）且倾斜角为3的直线，7分不妨设（0,4，因为为直径，所以∠6，9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；···················································