江苏省盐城市2022届高三年级第三次模拟考试-数学试题【含答案】

7344:uId:7344824282:fId:824282扫描全能王创建7344:uId:7344824282:fId:824282扫描全能王创建扫描全能王创建扫描全能王7344:uId:7344824282:fId:824282扫描全能王创建扫描全能王创建7344:uId:7344824281:fId:824281扫描全能王创建盐城市2022届高三年级第三次模拟考试数学参考答案一、单选 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 题号12345678答案ACCBBABD二、多选题题号9101112答案ABDBCBCDBC三、填空题13．250014．515．4316．0,四、解答题17．（1）解：选①23分S4a1a2a3a4a1(1qqq)120······································12，两式相除得到，a1a3a1(1q)301q4a(1qn)所以q3，（用S1需要考虑q1）.·····································2分n1q选②3分a4a1q81·····················································································1q327aaa(1q2)30，两式相除得到，1311q210得到10q327q2279q327q2q3279q2(q3)(q3)(q23q9)(q3)(10q23q9)0所以q3.··························································································2分选③22，分anan1an12an1(an3an1)(an4an1)0··································1得到，an3an10所以q3.·························································································2分由①（②或③）得，2得到，q3a1a3a1(1q)30a13所以n，分an3(nN*)·········································································4（）证明：由（）可得n21an323n23n11分bnnn1nn1······························6(an1)(an11)(31)(31)3131111111所以Sbbbn12n313213213313n13n1111············································································8分43n111·······················································································10分47344:uId:73441a2c2b218．解：（1）根据余弦定理可得：cosB······································2分2ac由条件知a2c2b2ac1所以cosB，···············································································4分2因为0B2所以ABC（不交代范围扣1分）··················································5分3（2） 方法 快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载 一：设BAC=，2因为ABC，DBC=，所以ABD，3622在ABD中，BD=2，BAC=，所以AB=tan1122则S=ABBD=2=···················································7分ABD22tantan在BCD中，BDC，BCD23BCBD根据正弦定理知：sinBDCsinBCD2sin22cos4cos得BC313cossinsincossin322所以12cos2分SBCD=BCBDsin=··························9263cossin3tan则223SABCSABDSDBC2tan3tan3tantan223分=2·········································································1133tan24383当且仅当tan=时，S的最小值为········································12分2ABC3方法二：2由（1）可知：ABC3所以123分Sacsinac····························································7ABC234因为BD2且DBC，6824834:fId:82483421a得SBDasinDBC26211a所以SSSBDcBDasinc························9分ABCABDDBC22623a则acc················································································11分4232a43根据基本不等式得：ac，当且仅当c时取等号32383SABC的最小值为········································································12分3方法三：根据条件以BD所在直线为x轴，以BA所在直线为y轴，建立如图所示的直角坐标系，3aa则A(0，c)，D(2，0)，C(,)·····························································7分22ayc2根据A,D,C三点共线得：····················A·····························9分23a22D3aBx化简得：acc·······································································11分42C32a43所以ac当且仅当c时取等号32383所以SABC的最小值为··································································12分319．解：（1）令“从12名工作人员中选出4人作为组长，至少有2个组长来自A部门”为事件E，····························································································1分C4C3C1C2C266666则P(E)4··································································2分C128·····················································································3分118答：从12名工作人员中选出4人作为组长，至少有2个组长来自A部门的概率为11·····································································································4分1（2）由题意知，A部门有6人，每人到甲村的概率为，41安排到甲村的A部门人数X服B(6,)，X=0，1，2，3，4，5，6························2分413729P(X0）C0()0()6，6444096131458P(X1）C1()1()5，6444096131215P(X2）C2()2()4，644409613540P(X3）C3()3()3，644409613135P(X4）C4()4()2，6444096uerr:uId:uerr31318P(X5）C5()5()1，6444096131P(X6）C6()6()0·······························································9分6444096X012345672914581215540135181P409640964096409640964096409672914581215540135181E(X)0123456409640964096409640964096409661443··············································································11分4096213或者E(X)np6423答：数学期望值为.注：（不答扣1分）················································12分2120．解：（1）由双曲线的渐近线方程为yx2x2不妨设双曲线方程为C:y2························································2分4因为(22,1)在双曲线C上代入方程解出1x2所以双曲线方程为：y21·····························································4分4（2）当直线l垂直与x轴时，直线l方程x=2此时M点坐标为(2,0)，则M到y的距离为2·············································6分当直线l斜率存在时设方程为y=kx+tykxt联立方程组2消去得：x2yy14(14k2)x28ktx4t240因为直线和右支双曲线相切所以：64k2t216(t21)(14k2)0化简得：t24k21············································································8分1yxt联立方程组2的A点横坐标为xA1ykxtk2t同理可得得B点横坐标为x····················································10分B1k2方法一：tt6kt8kxxAB112kk14kt2264k216(t21)(xx)216ABt2t2iwrier:fId:iwrier4xAxB4xxxAB2M2综上所述：点M到y轴的距离的最小值为2············································12分方法二：tt4t2xxAB112kk4k122又t24k21所以：xAxB4xx所以xABxx2M2AB因为xAxB，故等号取不到综上所述：点M到y轴的距离的最小值为2···········································12分方法三：x2由（1）可知双曲线C:y214设双曲线上任意一点的坐标为其中CPx0,y0x02xx则过点P的切线方程为0yy1····················································6分401yx12与切线方程yx联立方程组2xx0yy14044解得：xA，同理可得xB···········································8分x02y0x02y0则点A，B的中点M的横坐标xAxB1444x0xM·····································10分22x02y0x02y0x04y0因为点P在双曲线C上，x2所以0y21即x24y2440004x0则xM=x02x04y0综上所述：点M到y轴的距离的最小值为2···········································12分21．（1）证明：平面PAD平面PAB，平面PAD平面PAB=PA，PAPB，PB平面PABPB平面PAD又PB平面PBC平面PAD平面PBC.······································································4分（2）解:方法一：在等腰梯形ABCD中，过点C，D分别作CE，DF垂直于AB,垂足为E，F,延长AD，BC，交于点Q.连接PQ，在平面PAD中过点D作DGPA，垂足为G，连接FG.平面PAD平面PAB，DGPA，DG平面PABUERR:uId:UERR5又DFABFGABDFG为二面角PABD的平面角平面PAD平面PBC，交线为PQPD在平面PBC上的射影为PQDPQ为PD与平面PBC所成的角······················································6分在等腰梯形ABCD中不妨设AB=4，由已知条件得AF1,DF3，ADBD,3在RtDFG中，cosDFG3FG1,DG2，····························································Q···············8分在RtADG中，由AD2,DG2得DAG4DC由（1）知PB平面PADPBAD，又FADBDABAD平面PBDEADPDAPDG4在APQ中APDDPQ，P4PD与平面PBC所成角为45·····························································12分方法二：1四边形ABCD为等腰梯形，设AB=CD=AB=2a，2DCADBD，由（1）知PB平面PADFAD平面PADABPBADAD平面DPBEDP平面DPBADDPP过点D作DEAP，交AP于E，作DFAB，交AB于点F，连接EF，平面PAD平面PAB，DE平面PADDE平面PABDFE为二面角P-AB-D的平面角，···················································6分3则cosDFE3在RtDFE中，DF3a，则EFa，DE2a，AEFAPB90，EAFPABRtAFE∽RtAPB，AFa得到APBP22a···········································································8分过P在平面PAD内作PZAP，易证PZ平面APB，分别以PA，PB，PZ为x，y，z轴，建立如图所示的空间直角坐标系，P(0,0,0)，A(22a，0,0)，B(0,22a,0)，D(2a，0,2a)IWRIER:fId:IWRIER61根据DCAB(2a,2a,0)2求出C(0,2a,2a)，PD(2a,0,2a)···············································10分PB(0,22a,0),PC(0,2a,2a)设平面PBC的法向量为n(x,y,z)x10x2ay2az0得到令x1，得到y0，即n(1,0,0)0x2ay0z0z0PDn2a2设直线PD与平面PBC所成角为，则sin|cosPD,n|||||，|PD||n|2a12所以直线PD与平面PBC所成角的大小为45··········································12分22．解：（1）因为函数f(x)在(0,)单调递增所以f(x)=exxa0在(0,)恒成立·················································2分又因为f(x)ex10所以f(x)f(0)a1所以a+1≥0即a≥-1···········································································4分（2）方法一：因为f(x)=exxa0，f(x)=ex10，所以f(x)在(0,)单调递增f(x)f(x)又因为21f(x0)x2x1ex2ex11所以x0分ex0a(x2x1)a················································6x2x12xxx1x2xxf(x)f(12)ex0xae212a0202xxxxxx212112xxxxxx1222221212eee22e(1)(ee(x2x1))x2x1x2x1xx令21t0得h(t)etet2t其中h(0)0····································10分2h(t)etet20恒成立，h(t)h(0)0xx所以f(x)f(12)002又因为f(x)在(0,)单调递增2x0x1x2·················································································12分方法二：x0f(x0)ex0af(x)f(x)ex2ex1xx2121a·······················································6分x2x1x2x121(x1)2令g(x)x2lnx,(x1)，则g(x)0故xx22022-05-15T20:39:44.63772971x2x1x1x2xxx21e2e2g(x)g(1)0，即x2，令xe2，则1lnxx2x1x2x1x1x2xxxxex2ex121e2e221所以e2e2·················································8分x2x1x2x1xxf(x)f(x)ex2ex1xx12xx由21x021212f(x0)ex0ex2x1x2x122xx令h(x)exx(x0)，则h(x)ex10即函数单调递增，故x1202即分2x0x1x2··················································································128