【文档说明】山东省德州、滨州市2022-2023学年高三下学期一模数学试题答案.pdf，共(8)页，1.062 MB，由小赞的店铺上传

转载请保留链接：https://www.doc5u.com/view-61e77b31f7d5e9c543505dea3e4fded1.html

以下为本文档部分文字说明：

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412131

55331aa，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）ABBAABABABBsin()2sincossincoscossinsin()sin.····4分因为ABC为锐角三角形，所以AB22(0,),(0

,)ππ，AB22(,)ππ，又yxsin在22(,)ππ上单调递增，所以ABB，即AB2.···············6分（2）由（1）可知，AB2，所以在ABD中，ABCBAD，由正弦定理得：BBBADAB2)sin2πs

insin(2，所以BADBDcos1，所以BSABADBBBABD2cossintan1sin.····································9分又因为ABC为锐角三角形，所以B20π，B20

2π，B23π0π，解得B64ππ，···············································································11分所以B3tan(,1)3，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yuabt1e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yuabt1e两边取对数得yabtuln(1),···································5分则

ttbwwttiiiii()6650.208ˆ138.32()()1220120,b0.208ˆ,·····················7分awbt()1.6080.20810.50.576ˆˆ.················

····················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0.5760.208224

12.512.512.281e1ey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DE

V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222(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221421xyykx，消y并整理得，2

2(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22222122224821||1||1()8221212

1kkABkxxkkkkk，·············································································

·························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()c

os1(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，2

1()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()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14x，所以()fx在(0,)4单增，所以对(0,)4x，有()(0)1fxf，即12sin11xxx，所以11si

n(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xxsin，x4(0,]1，所以kkkkkk(1)1sin1111122，··········11分所以

knnnkn2231sin(1)()()11111111221，所以n2111kknsin1122.···························································

···12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com