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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11.6012.437513.210x
y14.5,1485四、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01211323
333nnTn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2coscbA
b,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:2sinsi
n(π2)sin2ADABBBB,所以1cosADBDB,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()1.6080.20810.50.5
76awbt.····································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0.5760.20822412.512.512.281e1eymm.·························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BCV
D;·············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,33,22O
EOV,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(,
,)xyzn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||||AVAVAVnnn,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222(2)|22|2xyx,两边平方并整理得22142x
y.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221421xyykx
,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2222212222
4821||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()0fx,21()102(1)2f,所以存在10(,)2xx,当01(,)xxx时,()0fx,()fx单增,当1(,)2
xx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()02f,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14
x,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()()()2334121nkknnn···················10分高三数学答案(第7页,共7页)因为s
inxx,1(0,]4x,所以2211111sin(1)1kkkkkk,··········11分所以221111111sin(1)()()12231nkknnn1,所以1121n221sin
1nkk.······························································12分