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高三数学答案（第1页，共7页）2023年高考诊断性测试数学参考答案及评分标准一、选择题ACBBADBC二、选择题9.AC10.ACD11.ABD12.ABD三、填空题11.6012.437513.210xy14.5，1485四

、解答题17.解：（1）设数列na的公比为q，因为1323,,5aaa成等差数列，所以2111352aaqaq，······································································1分即2352qq，解得3q

或12q，因为na各项均为正数，所以0q，所以3q.·····································2分由4355Sa，得41213155331aa，解得11a.····

·······················4分所以11133nnnaa.·········································································5分（2）由（1）知，13nnbn

.··································································6分则01211323333nnTn所以12331323333nnTn，···········

································7分两式相减可得01123333nnnTn，·······································8分13313nnn，整

理可得211344nnnT.·······························································10分18.解：（1）因为2coscbAb，由正弦定理

得sin2sincossinCBAB.···2分又πABC，所以高三数学答案（第2页，共7页）sin()2sincossincoscossinsin()sinABBAABABABB.····4分因为ABC为锐角三角形，所以ππ(0,),(0,)22AB，ππ(,)

22AB，又sinyx在ππ(,)22上单调递增，所以ABB，即2AB.···············6分（2）由（1）可知，2AB，所以在ABD中，ABCBAD，由正弦定理得：2sinsin(π2)sin2ADABBBB，所以1cosAD

BDB，所以1sinsintan2cosABDBSABADBBB.····································9分又因为ABC为锐角三角形，所以π02B，π022B，π0π32B，解得ππ64B，··················

·····························································11分所以3tan(,1)3B，即ABD面积的取值范围为3(,1)3.·······················12

分19.解：（1）Logistic非线性回归模型1eabtuy拟合效果更好.·····················1分从散点图看，散点更均匀地分布在该模型拟合曲线附近；从残差图看，该模型下的残差更均匀地集中在以残差为0

的直线为对称轴的水平带状区域内.····················3分（2）将1eabtuy两边取对数得ln(1)uabty,···································5分则2012021()(

)138.32ˆ0.208665()iiiiiwwttbtt,ˆ0.208b,·····················7分ˆˆ()1.6080.20810.50.576awbt

.····································9分高三数学答案（第3页，共7页）zyxEOABCDV所以y关于t的经验回归方程为0.5760.20812.51ety.··································10分当22t时，体长

0.5760.20822412.512.512.281e1eymm.·························12分20.解：（1）证明：取BC中点E，连接,,BDDEVE，因为ABCD为菱

形，且60BAD，所以BCD为等边三角形，故DEBC.····1分又在等边三角形VBC中，VEBC，·······2分DEVEE，所以BC面DEV.··········4分VD面DEV，所以BCVD；·············5分（

2）由VEBC，DEBC，可得DEV就是二面角ABCV的平面角，所以60DEV，··········································································6分在DEV中，3VEDE，所以DEV为边长为

3的等边三角形，由（1）可知，面DEV底面ABCD，取DE中点O，以O为坐标原点，以,,DAOEOV所在的方向为,,xyz轴的正向，建立空间直角坐标系Oxyz，····7分在VOE中，33,22OEOV，可得3(2,

,0)2A，3(1,,0)2B，3(1,,0)2C，3(0,0,)2V，故(2,0,0)CB，33(1,,)22CV，33(2,,)22AV.·········8分设(,,)xyzn为平面VBC的一个法向量，则有2033022xxyz，令3y，则

1z，得(0,3,1)n，·················10分设直线VA与平面VBC所成角为，则有||337sin|cos,|1427||||AVAVAVnnn，故直线VA与平面VBC所成角的正弦值为3714.····

··································12分高三数学答案（第4页，共7页）21.解：（1）设(,)Pxy，由题意||22|22|PFx，因为22||(2)PFxy，所以22(2)22|22|xyx，···················2分即2

22(2)|22|2xyx，两边平方并整理得22142xy.故点P的轨迹C的方程为22142xy.··················································4分（2）设直线l方程为11(2)2ykxk≤≤，联立221421xyyk

x，消y并整理得，22(21)420kxkx，设1122(,),(,)AxyBxy，则122421kxxk，122221xxk，·············5分又121222()221yykxxk，可得线段AB中点坐

标为2221(,)2121kkk，所以线段AB中垂线的方程为22112()2121kyxkkk，令0y，可得2(,0)21kNk，·····················································

·····6分对于直线1ykx，令0y，可得1(,0)Mk，所以22211||||21(21)kkMNkkkk.··················································7分高三数学答案（第5页，共7页）又222

22122224821||1||1()82212121kkABkxxkkkkk，·············································································

·························9分所以2222||282628(1)14||11ABkkkMNkk，··························10分令251

[,5]4tk，则22668(1)148141yktkt，因为6814ytt在5[,5]4上单调递增，所以22[5,170]55y，故||44[5,170]||55ABMN.·······

·····················12分22.解：（1）21()cos1(1)fxaxx，················································1分因为0为()fx的一个极值点，所以(0)20fa，

所以2a.·············2分（2）①当10x时，()2110fx，所以()fx单减，所以对(1,0]x，有()(0)1fxf，此时函数()fx无零点；··················

························3分当02x时，32()2sin(1)fxxx，()fx在(0,)2上单调递减，又(0)20f，32()202(1)2f，由零点存在定理，存在0(0,)2x，使得()0fx，且当0

(0,)xx时，()0fx，即()fx单调递增，当0(,)2xx时，()0fx，即()fx单调递减.又因为(0)0f，所以0(0,]xx，()0fx，()fx在0(0,)x单增；因为高三数学答案（第6页，共7页）0()0

fx，21()102(1)2f,所以存在10(,)2xx，当01(,)xxx时，()0fx，()fx单增，当1(,)2xx时，()0fx，()fx单减.所以，当1(0,)xx时，()fx单增，()(0)1fxf；当1(,)2xx时，()fx

单减，1()()202212fxf，此时()fx在(0,)2上无零点；···················5分当(,)2x时，21()2cos10(1)fxxx，所以()fx在(,)2单减，又()02

f，1()001f，由零点存在定理，函数()fx在(,)2上存在唯一零点；·····················································································

···········6分当x时，1()2sin2101fxxxx，此时函数无零点；综上，()fx在区间(1,)上存在唯一零点.···········································7分②因

为21()2104(1)4f，由（1）中()fx在(0,)2上的单调性分析，知14x，所以()fx在(0,)4单增，所以对(0,)4x，有()(0)1fxf，即12sin11xxx，所以11sin(1)21xxx

.··································8分令21(2)xkk≥，则2222211111111sin()2111kkkkkkkk···9分所以22111111111sin()()()2334121nk

knnn···················10分高三数学答案（第7页，共7页）因为sinxx，1(0,]4x，所以2211111sin(1)1kkkkkk，··········11分所以221111

111sin(1)()()12231nkknnn1，所以1121n221sin1nkk.······························································12分