2024届绵阳市高三第三次诊断性考试文科数学答案

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60分．CDACCACCBDCA二、填空题：本大题共4小题，每小题5分，共20分．613．214．15．4516．32三、解答题：本大题共6小题，共70分．117．解：（1）①当n=1时，61aa4=+，则a=．·································1分1112②当n2时，由61Sa4nn=+，得61Sa4nn−−11=+，·····························2分两式相减，得6an=−4an4an−1，······················································3分a，即n，分∴aann=−2−1=−2(n2)··················································4an−11∴数列{}a是以a=为首项，−2为公比的等比数列，·······················5分n121∴a=(−2)n−1．········································································6分n2121n+[1−−(2)]n14（2）由（1）得bS==2=+，·····························8分nn21+1−−(2)634n4可知数列{}是以为首项，4为公比的等比数列，···························9分334(1−4n)n∴T=+3····································································10分n61−4n44n+1−=+·······································································11分69223n++−3n8=．（也可不计算到此步）·····································12分1818．解：（1）调试前，电池的平均放电时间为：2.5×0.02×5+7.5×0.06×5+12.5×0.08×5+17.5×0.04×5=11小时，··········4分a调试后的合格率为：0.1×5+0.06×5=0.8，则=0.8，······················5分a+12∴a=48；····················································································6分100(2412−4816)2（2）由列联表可计算K2=4.7623.841，················10分40607228∴有95%的把握认为参数调试能够改变产品合格率．·························12分19．解：（1）∵E是BP的中点，AB=AP，∴AE⊥PB，················································································1分又平面PAB∩平面PBC=PB，且平面PAB⊥平面PBC，∴AE⊥平面PBC，·······································································2分过D作DF⊥PC交PC于F，∵平面PCD⊥平面PBC，且平面PCD∩平面PBC=PC，∴DF⊥平面PBC，·······································································4分∴AE∥DF，················································································5分又DF平面PCD，AE平面PCD，∴AE∥平面PCD；·······································································6分（2）∵AD∥BC∴VC-PBD=VD-PBC=VA-PBC=VC-PAB，1∴VC-PBD=VC-PAB=S△ABP·d，···························································8分3又∵平面PBC⊥平面PAB，过C作CH⊥PB交PB于H，∴CH⊥平面PAB，·······································································9分2在直角△CHB中：d=CH=BC·sin45=22=2，····················10分2221∴V=S=ABAPsinBAP，································11分C−PBD3ABP328∴当sin∠BAP=1时，体积的最大值为．······································12分31120．解：（1）解：当a=1时，fxxxxxx()()ln=+−−22，·······················1分24fx(x)x(1=+)ln，·········································································2分此时切线斜率为：k=+e1；···························································3分1所以曲线fx()在(e，f(e))处的切线方程：yx−=+e−(2e1)(e)··············4分43即：(e1)ee0+−−−=xy2；···························································5分4（2）证明方法一：因为fxxaxa()()(lnln)=+−，································6分由fx()0得到xa；由fx()0得到0xa．∴fx()在(0，a)单调递减，在(a，+)单调递增．5∴f()()x=fa=−a2，·····························································7分min4555要证f(x)−(1+lna)ea−1，即证：−aa21−(1+ln)ea−，4441+lna只需证：ea−11(a1)，·······················································8分a21+lnx设gx()=ex−1，即证：gx()1在x(1，+)恒成立.··············9分x2[(x−2)lnx+x−1]则gx()=ex−1，x3令h(x)=(x−2)lnx+x−1，·······················································10分2∴h(x)=lnx−+2，x∴hx()在(1，+)上单调递增，则h(x)=h(1)0∴hx()在上单调递增，则h(x)=h(1)0···························11分∴gx()0在恒成立，则在上单调递增，∴g(x)=g(1)0，原不等式得证.················································12分方法2：因为f(x)=(x+a)(lnx−lna)，···········································6分由fx()0得到xa；由fx()0得到0xa．∴fx()在(0，a)单调递减，在(a，+)单调递增．5∴f(x)=f(a)=−a2，····························································7分min4555要证fx(a)(1ln−)+ea−1，即证：−−+aa21(1ln)ea−，444aa1l+n即证：aaa21+(1ln)e(1)a−，即证：(1)a，ea−1axx1l+n只需证：，ex−1xx1l+nx令gx()=，则gx(1l+=n)，ex−1x即证：gx()g(x1+ln)，····························································8分1−x1−x又∵gx()=，且x1，则gx()=0，ex−1ex−1∴gx()在x(1，+)单调递减，···················································9分又x(1，+)，1+lnx(1，+)，∴即证xx+1ln，只需证：xx−1−ln0，································10分令h(x)=x−1−lnx，1∴hx'()=1−0，则hx()在x(1，+)单调递增，····················11分x∴h(x)=h(1)0，即xx−1−ln0，所以原不等式得证．··············12分b23b121．解：（1）离心率e1=−=，则=，①·······························1分a22a2a2−1a2−1当x=1，yb=，则|AB|=2b=3，②···························3分a2a2联立①②得:ab==21，，···························································4分x2故椭圆C方程为：+=y21；······················································5分4（2）设过F，A，B三点的圆的圆心为Q(0，n)，A()()x1，y1，Bx2，y2，又F(−3，0)，222222则|QA|=|QF|，即(x11−0)+(y−n)=(0+3)+(n−0)，················6分x2x2又A()xy，在椭圆+=y21上，故1+=y21，114412带入上式化简得到：3y211n1y0+−=，③·········································7分222同理，根据QB=QF可以得到：3y222n1y0+−=，④···················8分1由③④可得：yy,是方程32y1n2y0+−=的两个根，则yy=−，·····9分12123x2+=y21设直线AB：xt=+y1，联立方程：4，xt=+y1整理得：(t4y22)t2+y3+0−=，⑤···············································10分−31故yy==−，解得：t2=5，12t2+43∴t=5，··············································································11分5∴直线l的斜率为：．··························································12分522．解：（1）方法一：令x=0，即cos+3sin=0，3解得tan=−，················