【文档说明】山东省烟台市2021届高三下学期3月高考诊断性测试数学试题答案.pdf,共(6)页,239.253 KB,由管理员店铺上传
转载请保留链接:https://www.doc5u.com/view-5a39b3fbe98d4ed83f5cced2a11131e7.html
以下为本文档部分文字说明:
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,3324bTT,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2d,所以21nan.···············
······················································6分(2)由[lg]nnca,则123450ccccc,67501ccc,5152
1002ccc,·············································································9分所以123100cccc14525
0145.···········································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,45251ccc,2
6271002ccc,············································································9分所以123100cccc12227517
2.···········································10分若选③:(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,515210
02ccc,·············································································9分所以123100cccc145250145.···············
····························10分18.解:(1)()fxsin3cosxx2sin()3x,············································1分()fx图象向右平移6个单位长度得到2
sin()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)6yx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[,]()36kkkZ;··································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,63B,6B.··················································7分此时由余弦定理可知,2422
cos126aa.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,ABBMBI,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
)CBuur,(0,2,0)CDuuur,(1,2,)CPauur,设111(,,)xyzm为平面PBC的一个法向量,则有00CBCPuurguurgmm,即11112020xxyaz,令1ya,则(0,,
2)am.················7分设222(,,)xyzn为平面PCD的一个法向量,则有00CDCPuuurguurgnn,即22222020yxyaz,令2xa,则(,0,1)an.·
·············9分因为||10||||5gmnmn,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+10510
10100P,11327(27)2210510525P,高三数学参考答案(第4页,共6页)31227(28)21055525P,214(29)25525P,11
1(30)5525P,············································································5分的分布列如下:24252627282930P11003501710
072572542512513177741()2425262728293027.41005010025252525E.··························
···································································7分(2)当每两天生产配送27百份时,利润为1317(2420360)(2520260)(2620160)100501001317272
0(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)(2620
260)10050100712(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83PMPNuuuruuurg,同理当263x时,也有83PMPNuuuruuurg.·········
············································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()()PMPNPOONPOOMuuuruuuruuuruuuruuuruuurgg2POOPONOPOMONOMuuuruuuruuuruuuruuuruuuruuurggg,222POPOPOONOMuuuruuuruuuruuuruuur
g2ONOMPOuuuruuuruuurg·······································································9分又1212121
2()()OMONxxyyxxkxmkxmuuuruuurg221212(1)()kxxkmxxm2222222(1)(28)42121kmkmmkk22238821mkk,
·································································10分将22388mk代入上式,得0OMONuuuruuurg,······
·········································11分综上,83PMPNuuuruuurg.················································
······························12分22.解:(1)()sinfxxx,··········································································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