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高三数学参考答案（第1页，共6页）2021年高考诊断性测试数学参考答案一、单选题BDCCBADD二、多选题9.ABC10.BD11.BC12.ACD三、填空题13.2214.1.7515.[133,)16.56四、解答题17.解：若选①:（1）由已知2212bTT，3324bT

T，所以322bqb，通项2212222nnnnbbq.··········································2分故111ab.·······················································

····································3分不妨设{}na的公差为d.则121414dd，··············································4分解得2

d，所以21nan.·····································································6分（2）由[lg]nnca，则123450ccccc，67501ccc，51

521002ccc，·············································································9分所以123100cccc

145250145.···········································10分若选②:（1）由已知2212bTT，3324bTT，所以322bqb，

通项2212222nnnnbbq.··········································2分故111ab.·················································

··········································3分不妨设{}na的公差为d，则4341282d，···········································4

分解得4d，所以43nan．·····································································6分（2）由[lg]nnca，则1230ccc，4525

1ccc，26271002ccc，············································································9分所以123100cccc122275172

.···········································10分若选③:（1）由已知2212bTT，3324bTT，所以322bqb，通项2212222nnn

nbbq.··········································2分故111ab.····················································

·······································3分不妨设{}na的公差为d，则2(17)(14)(112)ddd，·································4分因为0d，解得2d，所

以21nan．····················································6分高三数学参考答案（第2页，共6页）（2）由[lg]nnca，则123450ccccc，67501ccc

，51521002ccc，·············································································9分所以12

3100cccc145250145.···········································10分18.解：（1）()fxsin3cosxx2sin()3x，··········

··································1分()fx图象向右平移6个单位长度得到2sin()6yx的图象，·························2分横坐标缩短为原来的12（纵坐标不变）得到2sin(2)6

yx图象，所以()2sin(2)6gxx.···········································································3分令222262kxk

，······························································4分解得36kxk.所以()gx的单调递增区间为[,]()36kkk

Z；··································5分（2）由（1）知，()26cg，··································································

····6分因为21sin()cos()cos()3664BBB，所以1cos()62B.又因为(0,)B，所以7(,)666B.当1cos()62B时，63

B，6B.··················································7分此时由余弦定理可知，2422cos126aa.解得311a.

·····················································································8分所以13112(311)sin262ABCS.····························

·············9分高三数学参考答案（第3页，共6页）当1cos()62B时，263B，2B.············································10分此时由勾股定理可得，12422a.······

·············································11分所以1222222ABCS.················································

·················12分19.解：（1）证明：因为面PAD面ABCD，面PADI面ABCDAD，AB面ABCD，ABAD，所以AB面PAD，··················

·············································1分又PD面PAD，所以PDAB，·····························································2分又PDBM，ABBMBI

，所以PD面ABM，····································3分AM面ABM，所以PDAM；····························································

·4分（2）取AD中点O，以O为坐标原点，分别以,,OAABOPuuruuuruuur方向为,,xyz轴正方向，建立如图所示的空间直角坐标系，··························································

··················5分设OPa，则有(1,2,0),(1,2,0),(1,0,0),(0,0,)BCDPa，可得(2,0,0)CBuur，(0,2,0)CDuuur，(1,2,)CPauur，设111(,,)xyzm为平面PBC的一个法向量

，则有00CBCPuurguurgmm，即11112020xxyaz，令1ya，则(0,,2)am.················7分设222(,,)xyzn为平面PCD的一个法向量，则有00CDCPuuurgu

urgnn，即22222020yxyaz，令2xa，则(,0,1)an.··············9分因为||10||||5gmnmn，可得22210541aa，·························

·············10分解得1a.·····················································································

··········11分所以112AP.·············································································12分20.解：（1）根据题意可得，的所有可能

取值为24,25,26,27,28,29,30.111(24)1010100P,133(25)2101050P,123317(26)2+1051010100P,11327(27)

2210510525P,高三数学参考答案（第4页，共6页）31227(28)21055525P,214(29)25525P,111(30)5525P,·····················

·······················································5分的分布列如下：24252627282930P11003501710072572542512513177741()242526

2728293027.41005010025252525E.·····························································································7

分（2）当每两天生产配送27百份时，利润为1317(2420360)(2520260)(2620160)1005010013172720(1)514.410050100百元.·································

··················9分当每两天生产配送28百份时，利润为1317(2420460)(2520360)(2620260)10050100712(2720160)2820492.82525百元

.·········································11分由于514.4492.8，所以选择每两天生产配送27百份.····································12分21.解：（1）由12AFF

为直角三角形，故bc，又121242FFAScb，可得4bc，······················································2分解得2bc，所以28a，所以椭圆C的方程为22184xy；····················

············································4分（2）当切线l的斜率不存在时，其方程为263x.高三数学参考答案（第5页，共6页）将263x代入22184xy，得263y，不妨设2626

(,)33M，2626(,)33N，又26(,0)3P,所以83PMPNuuuruuurg，同理当263x时，也有83PMPNuuuruuurg.·············································

········5分当切线l的斜率存在时，设方程为ykxm，11(,)Mxy，22(,)Nxy，因为l与圆228:3Oxy相切，所以22631mk，即22388mk，·························7分将ykxm代入22184xy，

得222(21)4280kxkmxm，所以122421kmxxk，21222821mxxk，·················································

······8分又()()PMPNPOONPOOMuuuruuuruuuruuuruuuruuurgg2POOPONOPOMONOMuuuruuuruuuruuuruuuruuuruuurggg，222POPOPOONOMuuuruuuruuuruuuruu

urg2ONOMPOuuuruuuruuurg·······································································9分又12121212()()OMONxxyyxxkxmkxm

uuuruuurg221212(1)()kxxkmxxm2222222(1)(28)42121kmkmmkk22238821mkk，····································

·····························10分将22388mk代入上式，得0OMONuuuruuurg，···············································11分综上，8

3PMPNuuuruuurg.··············································································12分22.解：（1）()si

nfxxx，··········································································1分因为(sin)1cos0xxx，所以()fx在(,)单增，又(0)0f，所以当(,0)

x时，()0fx，()fx单调递减；当(0,)x时，()0fx，()fx单调递增；·····································································

································2分故当0x时，()fx取极小值(0)1f，无极大值.··········································

3分高三数学参考答案（第6页，共6页）（2）2()(cos1)(e)2xxgxxa，·······················································

········4分由（1）知，()(0)fxf，即2cos102xx.············································5分当0a时，e0xa,()0gx，()gx在(,)单增；···········

················6分当0a时，令e0xa，得lnxa.于是，当(,ln)xa，e0xa，()0gx，()gx单减，当(ln,)xa，e0xa，()0gx，()gx单增.综上，当0a时，()gx在(,)单增；当0a时，()

gx在(,ln)a单减，在(ln,)a单增.······················································································

····7分（3）令()()e1xhxfxbx，则()e(1)sin1xhxbxx,[0,)x.()ecos1xhxxb，()hx的导函数()esinxhxx.因为

[0,)x，所以()1sin0hxx，()hx在[0,)单调递减.···········8分当1b时，对0x，()(0)10hxhb，所以()hx在[0,)上单调递减，所以对0x，()(0)0hxh.·

·································································9分当1b时，因为()hx在[0,)单调递减，(0)10hb，当x时，()hx.故0(0,)x，

使0()0hx，··············································10分且0(0,)xx时，()0hx，()hx单调递增，所以0()(0)0hxh，与0x，()0hx矛盾.·

·····································································································11

分所以实数b的取值范围是[1,).····························································12分