六五文档>基础教育>试卷>2024届绵阳市高三第三次诊断性考试理科数学答案
2024届绵阳市高三第三次诊断性考试理科数学答案
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绵阳市高中2021级第三次诊断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.
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