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

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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
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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,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,51521002ccc,···························

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

1ab.···························································································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,26271002ccc,·················

···························································9分所以123100cccc122275172.···········································1

0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,51521002ccc,·······························································

··············9分所以123100cccc145250145.···········································10分18.解:(1)()fxs

in3cosxx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)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[,]()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,63B,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,)C

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

a,······································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:24252627282930P11003

501710072572542512513177741()2425262728293027.41005010025252525E.··················································

···········································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(2

1)4280kxkmxm,所以122421kmxxk,21222821mxxk,·······················································8分又()()PMPNPOONPOOMuuuruuu

ruuuruuuruuuruuurgg2POOPONOPOMONOMuuuruuuruuuruuuruuuruuuruuurggg,222POPOPOONOMuuuruuuruuuruuuruuurg2ONOMPO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,83PMPNuuuruuurg.················································

······························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

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