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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9101112ADABCBDBCD41tr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102210242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）1222432112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}{2

na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，1

2521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)(sinSSSSAOCA

OB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BOCBOCSSS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ABHO21||//EF平面ABCD，EF平面ABFE，平面ABC

D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，)210(,,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,n

cos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ADE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：)|()|()|()|()()()()()()()()()()()()()()()()()()()()()()()()()()()()()|()|(

)|()|(BAPBAPBAPBAPBPBAPBPBAPBPBAPBPABPBPBAPBPBAPBPBAPBPABPBAPBAPBAPABPAPBAPAPBAPAPBAPAPABPABPABPABPABPR

·······································································································

·············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k

xyyx42，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

tantantansincossincostansincoscossincossincossincossinsin

coscossinsinsincoscossinsinsincoscoscossincoscossincoscossin得证········································································

····································12分法二：)1(2)1()1(2)1()(2)()(222)2()2(32222222222tantant

ansincossincostansincostancoscossincoscossincoscossin得证···································································

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