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高三数学试题答案第1页（共6页）吉林市普通中学2022—2023学年度高三毕业年级第三次调研测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．12345678ADBCBDAC8．提示：lnxlnex22xlnxxex2

2易证ttetf)(在),0(上单调递增，)()2(xlnfxfxlnx2，即xex2易知xexy2的最小值为e21，e21.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．11．提示：根据题意，切线方程为44xy，的水平截面为圆环，外半径为41t，内半径为t，故截面面积22)41()41(tttS)10(t，利用祖暅原理，可以构

造一个上底面半径为43，下底面半径为1，高为1的圆台。与圆台的体积相等，直接用圆台体积公式求V，也可以借助两个圆锥体积之差求V，如图。12．提示：xxgxf2)12()(，2)12(2)(xgxf)1(xf与)(xg均为偶函数)()()2

()(xgxgxfxf，0)()(0)2()(xgxgxfxf，①0)0(0)1(gf，1)1(02)1(2)1(ggf，即A错误；2)12(2)(

xgxf②2)23(2)2(xgxf将①带入得：2)32(2)(xgxf，即2)32(2)(xgxf③由②③得：2)12()32(xgxg2023)2023(1)1(

gg，，即B正确；42)3(22)3(2)2(gf，即C正确；42)21(22)12(2)1()(xgxgxfxf9101112ADABCBDBCD41

trt高三数学试题答案第2页（共6页）198992)10050()10051()10049()10098()1002()10099()1001()100(991fffffffifi即

D正确.三、填空题：本大题共4小题，每小题5分，共20分．其中第16题的第一个空填对得2分，第二个空填对得3分．13．12014．1515．]4415[]143[,,16．4；152（注：可以用不等关系表示）

四、解答题17．【解析】（Ⅰ）211a，42a．···························································································2分令10221

0242n，解得：12n，不符合题意，舍去；令102423n，解得：342n，符合题意．因此，1024是数列}{na的第342项．································································5分（Ⅱ）12

22432112nnnaaaaaaS322)86(62421nn)86104()2221(32nn2)864)(1(41)41(21

nnn6115364)23)(1()14(612nnnnnn．·········································10分（注：未列举和分组，而直接用和公式的扣2分）法二：32122nna，

又41212nnaa，所以数列}{12na是以21为首项，4为公比的等比数列.262nan，又6222nnaa，所以数列}{2na是以4为首项，6为公差的等差数列.12nS为数列}{12na的前n项和与数列}{2n

a的前1n项和的总和.2)864)(1(41)41(2112nnSnn6115364)23)(1()14(612nnnnnn·····················

·······················10分（注：未用定义证明数列为等比、等差数列，而直接用和公式的扣2分）18．【解析】（Ⅰ）由题可知，653222AOC在AOC中，由余弦定理得AOCcosOCOAOCOAAC

2222322311211）（226AC···························································································4分（注：32

AC此次考试不扣分，但请加以强调结果需化简）法二：在ABC中，421AOBACB，12521AOCABC，2AB由正弦定理得ABCsinACACBsinAB高三数学试题答案第3页（共6页）22612542ACsinACsin······

···························································4分（Ⅱ）设AOB，则34AOC)34(11211121

sinsinSSAOCAOB)6(23)232321)]3([21sincossinsinsin（······································································

·····································8分设阴影部分面积为S，优弧BC所对的扇形BOC面积为扇形S，则32)322(1212扇形S32)6(23)(sinSSSSAOCAOB扇形········

······························10分O点在ABC内部36566当26时，即32时，2332minS即阴影部分面积的最小值是2332·······································

·························12分法二：4321BOCsinOCOBSBOC劣弧BC所对的扇形的面积3132212BOCS扇形433BOCBOC

SSS扇形弓形设aBC,bAC,cAB在BOC中，由余弦定理：32222BOCcosOCOBOCOBBC3BC即3a·······························································

······················8分在ABC中，由余弦定理：BACcosbccba2222bcbcbcbccb23223bc，当且仅当3cb时，等号成立···················

·································10分阴影部分面积为ABCSSS弓形BACsinbc21)433(bc43433223324334332即阴影部分面积的最小值是2332

································································12分19．【解析】（Ⅰ）证明：取AD中点H，连接EH，HO.O是BD中点AB

HO21||//EF平面ABCD，EF平面ABFE，平面ABCD平面ABABFEAB//EFEFAB2FECBADOH高三数学试题答案第4页（共6页）ABEF21||EFHO||四边形EFOH是平行四边形EH

//FOEH平面ADE，FO平面ADE//FO平面ADE························································································4分（注：也可以用面面平行证

明）（Ⅱ）ADEHEDAEADOH，HEHOHAD平面EFOHAD平面ABCD平面ABCD平面EFOH在平面EFOH中，过O作OHOG平面ABCD平面HOEFOHOG平面ABCD取AB中点N，取BC中点Q，连接ON，OQ

.以O为原点，ON，OQ，OG所在直线为x轴，y轴，z轴建立空间直角坐标系·····6分如图，则)022(,,A，)022(,,B，)022(,,C，)022(,,D，)210(,,E，)2

10(,,F)0022(,,AD，)212(,,AE，)212(,,BF设)22()212(,,,,BFBM）（10)2222(,,M)222(2

,,CM·······································································8分设)(z,y,xn是平面ADE的一个法向量)120(

02202200,,nzyxxAEnADn········································10分211422)222(322222

CM,ncos化简得：210122或1（舍）当M是BF的中点时，使得CM与平面ADE所成角正弦值为21142····················12分（说明：（1）建系方法不限，阅卷参考：若按下图①图②建系，平面AD

E法向量均为：)12(0,,n；（2）（Ⅰ）也可用空间向量法证明）图①图②20．【解析】（Ⅰ）零假设为0H：该球队胜利与甲球员参赛无关.·················································

······1分503.10288302518324010)830102(5022≈·················································

····3分因为879.72·····························································································4分所以依据005.0的独立性检验，我们推断0H不成立，所以认为

该球队胜利与甲球员参赛有关，此推断犯错误的概率不大于005.0.····························································6分（Ⅱ）（ⅰ）证明：)|()|()|()|()()()()()()()()()()()()(

)()()()()()()()()()()()()()()()()|()|()|()|(BAPBAPBAPBAPBPBAPBPBAPBPBAPBPABPBPBAPBPBAPBPBAPBPABPBAPBAPBAPABPAPBAPAPBAP

APBAPAPABPABPABPABPABPR················································································································

····8分FECBADOxyzHFECBADOxyzExFECBADOyzHQGNM高三数学试题答案第5页（共6页）（ⅱ）51)(B|AP·····································

····························································9分43)(B|AP··································

·····························································10分.12143554151)|()|()|()|(BAPBAPBAPBAPR·····································

·······················12分21．【解析】（Ⅰ）设点),(yxP，则)1,(xM.由题知||||PMPF,即：|1|)1(22yyx.整理得：yx42.则曲线C的方程为yx42.··········

·····································································4分法二：由题知，点P到点)1,0(F的距离等于其到直线1y的距离相等，则点P的轨迹为以)1,0(F为焦点，以1y为准线的抛物线.则曲线C的方

程为yx42···············································································4分（Ⅱ）由题，AB为圆4)2(22yx的直径，则OBOA.易知直线OA存在斜率，设为k，且0k，则直线OB的斜

率为k1.又有，OA所在直线为kxy，联立kxyyx42，解得：01x或kx42.联立kxyyx4)2(22，解得：03x或2414kkx································

··········6分所以，222224221)2(||41|)2(4|1||1||kkkkkkkxxkAS.同理可得，222221)12(4)1(1]2)1[(|1|4||kkkkkkBT···················

·························8分四边形ABST的面积||||)252(8)1(||)12)(2(8||||21224222kkkkkkkkkBTASs||1||]1

)||1|(|2[8||1||)522(8222kkkkkkkk······················································10分令),2[,||1||tkkt.)12(8)12(82tttts.因为s在

),2[上单调递增，所以当2t时，s有最小值36.即，当1k时，四边形ABST面积的最小值为36···········································12分法二：略解由yxmyx42得24myS.由

4)2(22yxmyx得142myA.·································································6分22221)12(41mmm|yy|m|AS|AS

.同理可得:1||)2(4111)12(4||22222mmmmmmBT.···············································8分高三数学

试题答案第6页（共6页）|BT||AS|SABST21四边形||1||)252(8)1(||)2)(12(822222mmmmmmmm········································10分令

21|m||m|t)12(8)18(2)(2tttttS在)2[,t上单调递增.36)421(8)(mintS，即四边形ABST面积的最小值为36.······················12分22．【解析】（Ⅰ）)4(2)()

()(xsineecosxsinxxmsinxexmxxx，·······························2分令0)(xm，则430x；令0)(xm，则x43.)(

xm∴在)43,0(上单调递增，在),43(上单调递减.·········································4分)(xm∴在),0(上的最大值为：4322)43(em·······················

························5分（Ⅱ）证明：)(xgy的图像与直线)0(kkxy的三个公共点如图所示，且在)23(，内相切，其切点为：).23(),(，，sinA··········································

·····················7分当)23(，x时，.|)(cosycosxyxsinxgyx，，sincos，即tan.·····························

······································9分)1(2)1()1(2)1()(2)()(2222)121(2)21(22)2(32222222222232

tantantansincossincostansincoscossincossincossincossinsincoscossinsinsincoscossinsinsincoscoscossincoscossinc

oscossin得证············································································································12分法二：)1(2)1()1(2)1()(2)()(2

22)2()2(32222222222tantantansincossincostansincostancoscossincoscossi

ncoscossin得证············································································································12分获得更多资源请

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