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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[11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xxxxfxfxg，)(xg在2)0(，上递减，即0)0()(gxg.202x，0)()(4)(2222xfxfxg，)()

(422xfxf.)()(32xfxf，)()(432xfxf.又)(xf在），（2上单调递增，324xx，432xx即，C选项正确；对于D选项：)()(21xfxf，2221218

282xxxx，即082212121））（（xxxxxx.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n

F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2697133782023321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aFFFFFFFFFFF.（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nnkkaa2

112.（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1500100

9001500900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：)(142.15100402.1310060cm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])1

42.13(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62222

2222aacbcacabcbab··································································4分（法二）利用正弦定理因为6cos

cos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）BcosBBBBBsBB2992sin9sin18cos

sin18sin22co22sin2182）（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1

42sin22）（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02co02co02co22

2222222abcbasCacbcasBbcacbsA，即010b360b103603610222222222222222）（）（）（bcbabbcabbacb所以5345

16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:1162522yx，如图································································6分不妨设A在x轴上方，坐标为），（00yxAbcacbsAco2222bcbccb23622

）（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ya

SABC·····························································································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51

60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B

DEF,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）（1102,,n····························································································

·····10分设二面角CAEF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n

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

2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xky1·······················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OPOQOQOP

OQOP········································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)21116(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kkk

kkkkkkkkkOQOP103332222kk············································································8分设)331(1)(xxxxf222111)(x

xxxf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kkk

kkkkkkkOQOP····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tkxy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))((22222121212

1kkttxxktxxktkxtkxxxOQOP2所以33222kt②，·················································

········································8分由①②可得222222212222)(33)12()(1)1kktkxxk（PQOQOP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111211121112112nnenlnnnenlnnnenlnnn

m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