2024届绵阳市高三第三次诊断性考试理科数学答案

绵阳市高中2021级第三次诊断性考试理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．CBACBDDCBCDA二、填空题：本大题共4小题，每小题5分，共20分．613．1014．15．4516．22三、解答题：本大题共6小题，共70分．10024124816−()217．解：（1）由列联表可计算K2=4.7623.841，········4分40607228∴有95%的把握认为参数调试能够改变产品合格率．···························5分（2）根据题意，设备更新后的合格概率为0.8，淘汰品概率为0.2．········6分可以认为从生产线中抽出的6件产品是否合格是相互独立的，···············8分设X表示这6件产品中淘汰品的件数，则XB(6，0.2)，······················9分060151可得：p=P(X1)=C660.80.2+C0.80.2································10分=0.85(0.8+1.2)=0.65536．·····················································12分18．解：（1）设{}an的公差为d，则1，1+d，2+2d成等比数列，·················1分∴(1+dd)2=1(2+2)，解得：d=1或d=−1，····································3分而，不满足，，成等比数列，d=−1a1a2a3+1∴d=1，······················································································4分∴数列{}an的通项公式ann=．·····················································5分n（2）令Dn=a1bn+a2bn−−1++an1b2+anb1=31−，·······························6分n+1∴Dn+1=+a1bn+1+a2bn+a3bn−1+anb21+an+1b=3−1，·······················7分n两式相减有：DDnn+1−=a1bn+−1+(bn+bn1++b1)=23，···················8分nn∴数列{}bn的前n+1项和为23，即Tn+1=23，·································9分又D1==a1b12，所以b1=2，·························································10分n−1∴b1+b2++bnn−1+b=23，····················································11分n−1∴Tn=23．···········································································12分19．解：（1）过C作CH⊥BB1交于H，············································1分∵C在平面ABB11A内的射影落在棱BB1上，∴CH⊥平面，又AB平面，···································2分∴CHAB⊥，·············································································3分又ABB⊥C1，且B1CCHC=，····················································4分∴AB⊥平面BCC11B；··································································5分3212（2）∵VSCHABCA−BCABBA=，则CH==1，························6分1111123过C作CQA⊥A1交AA1于Q，连结HQ，∵与的距离为则，AA1CC12CQ=2又∵CH⊥平面ABBA，则CH⊥HQ，··········································7分11在△中：222，则，RtCHQHQ=CQ−CH=2−1=1HQ=1又AA⊥CH且AA⊥CQ，11∴AA⊥CHQ∴AA⊥HQ1平面1又由（1）知：平面，∴BB，1∴AB⊥AA，则四边形ABHQ为矩形，1∴AB==HQ1，又四边形的面积为，则，分ABB1A13BB1=3··········································8分别以HB，HQ，HC为x轴、y轴、z轴建立如图所示空，，间直角坐标系设BH=x(x0)∴Bx(，0，0)，Ax(，1，0)，C(−3，0，1)，1∵AC=33∴AC2=(x+3)2+12+12=27，11解得x=2，···········································9分∴B(2，0，0)，A(−1，1，0)，C(0，0，1)，1∴AB=−(3，1，0)，BC=−(2，0，1)，1设平面ABC的法向量为n=()x，y，z，11ABn=30x−y=∴11，令则，，，分x=1，n1=(132)························10BCn1=−20x+z=易知平面ABBA的法向量n=(0，0，1)，······································11分112214∴cosnn12，==，141714∴平面ABC与平面ABBA所成锐二面角的余弦值为．···············12分1117b23b120．解：（1）离心率e1=−=，则=，①·······························1分a22a2a2−1a2−1当x=1，yb=，则|A|=B2b=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(x)y(1122B，)xy，，，又F(−30)，，222222则|Q|=A||QF，即(0)()(03)(0)xynn11−+−=++−，················6分x2x2又A(,)xy在椭圆+=y21上，故1+=y21，114412带入上式化简得到：3y11+2ny−1=0，③·········································7分222同理，根据QB=QF可以得到：3y22+2ny−1=0，④···················8分1由③④可得：yy,是方程3y2+2ny−1=0的两个根，则yy=−，·····9分12123x2+=y21设直线AB：x=+ty1，联立方程：4，x=+ty1整理得：(t22+4)y+2ty−3=0，⑤···············································10分−31故yy==−，解得：t2=5，12t2+43∴t=5，··············································································11分∴直线AB的方程为：xy5−1=0．···········································12分1121．解：（1）当a=1时，f(x)=(x2+x)lnx−x2−x，24∴f(x)=(x+1)lnx，则切线斜率k=e+1，······································2分1∴曲线f(x)在(e，f(e))处的切线方程：y−e2=(e+1)(x−e)，··········4分43即：(e+1)x−y−e2−e=0，·······················································5分4（2）证明方法一：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f(x)−(2+lna)ea−1，即证：−aa21−(2+ln)ea−，08482a2只需证：−lna−20(1a2)（*）·········································8分ea−12x2设g(x)=−lnx−2(1x2)，ex−14x−2x214x2−2x3−ex−1则gx()=−=，··········································9分eexx−−11xx42xx23−设h(x)=−1(1x2)，ex−12x32−10x+8x2x(x−1)(x−4)则hx()==，eexx−−11易知：hx()在(1，2)上单调递减，而h(1)=10，h(2)=−10，故必存在唯一x0(1，2)，使得hx(0)=0，······································10分∴当x(1，x0)时，hx()0，即gx()0；当x(x0，2)时，hx()0，即gx()0，∴gx()在(1，x0)上单调递增，在(x0，2)上单调递减．·······················11分8而g(1)=0，g(2)=−ln2−20，e∴gx()0在(1，2)上恒成立，即（*）式成立，原命题得证．·············12分方法二：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fxa()(2ln)e−+a−1，即证：−−+aa21(2ln)ea−，084822lnaa+只需证：−0(12)a，···············································8分ea−1a2xx2+ln设g(x)=−(1x2)，即证gx()0在x(1，2)恒成立.ex−1x2−+2xx1ln则g(x)=+(1x2)，ex−12x2(xx−+2)12ln令h()()x=gx，则h(x)=−(1x2)，························9分ex−13x又∵12x，2(xx−+2)12ln∴00，−，ex−13x2(xx−+2)12ln∴hx()=−0在(1，2)上恒成立.