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高三数学试题答案第1页（共8页）吉林市普通中学2022—2023学年度高中毕业年级第二次调研考试数学试题参考答案一、单项选择题：本大题共8小题，每小题5分，共40分.12345678DCADDDBC二、多项选择题：本

大题共4小题，共20分.全部选对的得5分，部分选对的得2分，有选错的得0分.三、填空题：本大题共4个小题,每小题5分，共20分.13.,4114.715.2697；1a16.)2,0[1

1题解析：解析：23288)(xxxxxf.令0)(xf，则2x；令0)(xf，则2x且0x；)(xf的增区间为：），（2，减区间为：），），（，（200.对于A选项：6)2(

f且)(xg有三个零点，6a，即A选项正确；对于B选项：当0x时，016)82()82()()(22xxxxxxfxf，)()(xfxf.)()(31xfxf，)()()(331xfxfxf.)(xf

在），（0上单调递减，31xx，即031xx，即B选项错误；对于C选项：令2)0()(2)(2)(，，xxfxfxg.0)4(12)(4)(2)(2)(2222xx

xxfxfxg，)(xg在2)0(，上递减，即0)0()(gxg.202x，0)()(4)(2222xfxfxg，)()(422xfxf.)()(32xfxf，)()(432xf

xf.又)(xf在），（2上单调递增，324xx，432xx即，C选项正确；对于D选项：)()(21xfxf，2221218282xxxx，即082212121））（（xxxxx

x.21xx，212116xxxx.）（）（）（）（21221212212124164xxxxxxxxxx令021txxt，，则331022212224232216416tttttxx）（）（）（，故D选项正确

.[另解：由)()()(321xfxfxf，可得0163131xxxx，故B错；323216xxxx，由基本不等式可得C对.]15题解析：（1）由斐波那契数nF除以4的余数按原顺序构成的数列}{na，是以6为最小正周期的数列，所以2697133782023

321aaaa9101112ADBDACDABC高三数学试题答案第2页（共8页）（2）由斐波那契数nF的递推关系可知：2n时12nnnFFF，且121FF，aF2024所以1)()()(22024202320243423202221a

FFFFFFFFFFF.（也可直接由性质12321nnaaaaa得出结论）15题说明：本题源自人教A版选择性必修第二册P10—P11页阅读与思考部分内容.斐波那契数列}{na满足：11a，12a

，21n-nnaaa2)(*n,Nn，该数列通项公式为：])251()251[(51nnna（用无理数表示有理数的典型代表）常考性质：（1）12232221nnnaaaaaa，即112nnnkkaaa.（2）n-naaaaa212531

，即nnkkaa2112.（3）1122642nnaaaaa，即11212nnkkaa.（4）12321nnaaaaa，即121nnkkaa.（5）ka4

为3的倍数.四、解答题17．【解析】（Ⅰ）设在男生、女生中分别抽取m名和n名，则15001009001500900nm，解得：40,60nm.·····························

································································2分记抽取的总样本的平均数为，根据按比例分配分层随机抽样总样本平均数与各层样本平均数的关系，可得：)(142.15100402.1310060

cm所以，抽取的总样本的平均数为cm14.·······································································5分（Ⅱ）男生样本的平均数为2

.13x，样本方差为36.1321s；女生样本的平均数为2.15y，样本方差为561722.s；由（Ⅰ）知，总样本的平均数为14.记总样本的样本方差为2s，则16]})142.15(56.17[40])142.13

(36.13[60{1001222s························9分所以，估计高三年级全体学生的坐位体前屈成绩的方差为16.·········································10分（注：本次考试给出了分层随

机抽样的方差公式，但是此公式需要记忆，请老师在教学中强调并指导学生完成公式的推导.）18．【解析】（Ⅰ）（法一）利用余弦定理因为6coscosBcCb，由余弦定理，得622222222aa

cbcacabcbab··································································4分（法二）利用正弦定理因为6coscosBcCb，由正弦定理，得

6)(2BcosCsinCcosBsinR6)(2CBsinR62AsinR即6a················································································

············4分（Ⅱ）选择①：2646sinsinCcsinBbsinAa所以Bsinb26，Csinc26.················································

····················5分所以）（4sinsin218sinsin218bcsinA21SABCBBCB高三数学试题答案第3页（共8页）BcosBBBBBsBB2992sin9sin18co

ssin18sin22co22sin2182）（94229）（Bsin···················································································9分因为ABC是锐角

三角形所以2020CB又BC43,所以243020BB,所以24B.···············10分所以43424B，所以142sin22）（B.·····················

···················11分所以2942299）（Bsin.所以929S18ABC··············································

····································12分（注：利用余弦定理，解出929bc，仅得出ABCS的最大值，给9分）选择②因为10cb，则bc10因为ABC是锐角三角形，所以

02co02co02co222222222abcbasCacbcasBbcacbsA，即010b360b103603610222222222222222

）（）（）（bcbabbcabbacb所以534516b········································································7分因为bbabcbasCco31652222，所以

bbbCcosCsin23)8)(2(41所以bbbbabsinCSABC3)8)(2(4321161042bb，）（534516b···················································10分由二

次函数16102xxy的性质可得，当5x时，函数取最大值9；当516x时，25144y.所以12548ABCS··································································

··························12分法二：ABC中，610BCABAC，所以点A的轨迹为以B，C为焦点，10为长轴长的椭圆的一部分.以BC中点为坐标原点，BC所在直线为x轴建立平面直角坐标系.可得A点所在的椭圆为:116252

2yx，如图································································6分不妨设A在x轴上方，坐标为），（00yxAbcacbsAco2222bcbccb23

622）（1322bc264bcbc02571253214322）（cb，高三数学试题答案第4页（共8页）当且仅当cb时，即A点在短轴端点2A时，取得最小值，此时角A最大是锐角,此时40y，12210yaSABC···········

··················································································8分当AB与x轴垂直时，即A点在图中1A时，角B为直角，此时5160y，548210yaSABC当AC与x轴垂直时，即A点

在图中3A时，角C为直角，此时5160y，548210yaSABC······························································

······························································10分综上，A位于除去端点的弧31AA上时，满足ABC是锐角三角形.021yaSABC，）（45160y····················

························································11分所以12548ABCS同理可得A在x轴下方时，12548ABCS，综上，12548ABCS········

·················································································12分19．【解析】法一（几何法）（Ⅰ）因为四边形ABCD为菱形，所以BDAC,因为FCFA

,且O为AC中点，所以OFAC,·····················································2分OOFBD,BDEFOF,BD平面所以AC平面BDEF······

·····················································································5分（Ⅱ）因为四边形BDEF为矩形，连

接OE,12BF,BD,且O为BD中点,所以22EF,EOFO,222EFFOEO,所以FOEO.由（1）可知OACEO,ACFO,所以FO平面AEC又EA平面AEC,所以EAFO.过O作AEOG，连

接GF,又OOFOG,所以EA平面OFGGF平面OFG,所以GFEA,则OGF为二面角CAEF的平面角·················································

···················8分在EOARt中，OAOEEAOG,56EAOAOEOG在FOGRt中，5425622OFOGFG·················································10分则

46FGOGOGFcos所以二面角CAEF的余弦值为46.··································································12分法二(向量法）（Ⅰ）问同上（Ⅱ

）由（Ⅰ）可知：AC平面BDEF，ED平面BDEF,所以EDAC.因为四边形BDEF为矩形，所以BDED,且OACBD,所以ED平面ABCD因为60DAB,取AB中点M,连DM,则CDDM.以D为坐标原点，DM,DC,DE

所在直线分别为x，y，z轴，建立空间直角坐标系·······································································································

······················6分高三数学试题答案第5页（共8页））（000,,D,）（013,,A，）（100,,E,）（020,,C,）（113,,F）（113,,AE,）（120,,AF,）（033,,AC设平面AEF的一个法向量为）（1111z,y,xn

则0011AFnAEn得0203zyzyx取3y，则36x,z得）（6331,,n···································

·······························································8分设平面AEF的一个法向量）（2222z,y,xn则0022ACnAEn得03303yxzyx取3x，则21z,y得）

（2132,,n···································································································10分设

二面角CAEF为，由图可知二面角为锐角则464133693122121nnnncos所以二面角CAEF的余弦值为46··············································

·······················12分法三由（Ⅰ）可知：AC平面BDEF，ED平面BDEF,所以EDAC.因为四边形BDEF为矩形，所以BDED,且OACBD,所以ED平面ABCD取EF中点

N，以O为坐标原点，OA，OB，ON所在直线分别为x，y,z轴建立空间直角坐标系，如图所示·······························································································

···6分）（000,,O,）（003,,A，）（110,,E,）（003,,C,)110(,,F）（113,,AE,）（113,,AF,）（0032,,AC设平面AEF的一个法向量）（1111z,y,xn则0011AFnAEn得

0303zyxzyx取3x，则03y,z得）（3031,,n·········································································

························8分设平面AEC法向量为）（2222z,y,xn则0022ACnAEn得03203xzyx取1z，则01x,y得）（110

2,,n·································································································10分设二面角C

AEF为，由图可知二面角为锐角高三数学试题答案第6页（共8页）则4629332121nnnncos所以二面角CAEF的余弦值为46·····································································12

分20.【解析】（Ⅰ）31a311S又数列nSn为以2为公差的等差数列nSn12n即nnSn22································································

···················3分2n时，14)1()1(22221nnnnnSSannn1n时，31a符合上式数列}{na的通项公式为14nan.·····································

····································6分（Ⅱ）)341141(2)1()34)(14()14()1()2()1(1nnnnnaaabnnn

nnnn···························8分9244]38131[21)]381181()151111()11171()7131([212nnnnnTn数列}{nb的前n2项和92442nnTn

·····························································12分（注：写成6161612nTn，本次考试不扣分，但教学中请强调结果最简化）21.【解析】解：（Ⅰ）由已知可得：2212)(22xyx，整理化简可得：

3322yx，即1322yx所以动点M的轨迹方程为：1322yx·····················································4分（Ⅱ）方案一：由OQOP可设直线OP的方程为kxy，直线OQ的方程为x

ky1·······················5分由3322yxkxy，可得22223333kkykx2，所以2222233(1k)kyxOP同理可得122233

(1k)kOQ,又由02|OP|且02|OQ|可得3312k······················7分法一：所以32)3(1133112222kkkOPOQ2·····························

····································8分所以)11)((23222222OPOQOQOPOQOP所以6)12(223)(223222222OPOQOQOPOQOP········

································10分当且仅当3OQOP时等号成立所以22OQOP的最小值为6···········································································

12分法二：所以1))(3(3)6(113)3(13)3(12222222222kkkkkkkOQOP······································8分令4)341(2tkt，高三数学试题答案第7页（共8页）则1)211

16(61616364))(3(46222222tttttttOQOP·························10分所以当211t时，即1k时，22OQOP有最小值为6.···········

······························12分法三：310332)3103(23103)12(61))(3(3)6(1242242424222222kk

kkkkkkkkkkOQOP103332222kk······································································

······8分设)331(1)(xxxxf222111)(xxxxf令310)(xxf令1310)(xxf)(xf在)131(,递减，在)31(,递增············································

·······························10分)31()3()()1(ffxff即310)(2xf即10)1(3622kk所以622OQOP即22OQOP的最小值为6···········································

········12分法四：因为0130322k,k所以6)2133()6(11))(3(3)6(113)3(13)3(1222222222222222kkkkkkkkkkOQOP····10分当且仅当13322kk时，即1k时，

22OQOP有最小值为6.··························12分方案二：当直线m的斜率不存在时，设直线m的方程为0xx则)(),(0000y,xQy,xP2323023)33(20202020202020y,

xxxxyxOQOP6)(2202022yxOQOP··················································································5分当直线m的斜率存在时，设直线m的方程为tkx

y设其与动点M的轨迹相交于)(11y,xP，)(22y,xQ所以tkxyyx3322联立得：0)-0(332)(32222kt-ktx-xk·····························6分其中

0-32k，03)12(3))(4(34Δ222222kttktk①由韦达定理可得：22132kktxx，2213kt-xx23········································

···········7分由OQOP可知：02121yyxxOQOP所以03332)()(1))((222221212121kkttxxktxxktkxtkxxx

OQOP2所以33222kt②，·························································································8分由①②可得222222212222)(33)12()(1)1kktkxxk（PQ

OQOP所以）96166(1969)106（242242422kkkkkkkOQOP······································10分当0

k时，66166(12222）k9kOQOP当0k时，622OQOP高三数学试题答案第8页（共8页）所以22OQOP的最小值为6···········································································

····12分22.【解析】解：（Ⅰ）函数)(xf的定义域是)0(，，0)11(112)(22xxxxf，)(xf在)0(，上单调递减.·································································

··················4分（Ⅱ）12)(xxlnxg，定义域是)0(，，则xxxxg212)(，令0)(x'g，则20x；令0)(x'g，则2x.)(xg在2)0(，上单调递增，在)(2，上单调递减.················

·····································5分0(1)g，.lng,lng0324(4)0232(3)(3,4),0x使0)(0xg，即1200xxln.

当01xx时，0)(xg；当10x或0xx时，0)(xg···································7分0))((2nmxxxg当01xx时，02nmxx；当10

x或0xx时，02nmxx.1和0x是方程02nmxx的两个不等实数根.042nm，由韦达定理mx01，.nx01,xn,xm001.nm1·············································

·························8分法一nm111211121112112nnenlnnnenlnnnenlnnnm又由1200xxln1200xex又0xn21nen原式1)(112

nfnnnln（其中(3,4)0xn）.········································10分由（Ⅰ）知)(xf在区间(3,4)上单调递减且..lnf,.lnf050411241(4)35035321(3)01])([nf.

即011122nnenlnm.···············································································12分法二由1nmnm1

11211121112112nnenlnnnenlnnnenlnnnm又由1200xxln1200xex又0xn21nen，12nnln原式nnnn1111

···················································································10分)31,41(1(3,4)0nxn01n即011122nnenlnm.··

·············································································12分