【文档说明】北京市顺义区2023-2024学年高一下学期期末考试数学试卷 PDF版含答案.pdf,共(12)页,2.314 MB,由管理员店铺上传
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#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGRVADuARCwZNIBIA=}#}{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEAEIAAAAANABAA=}#}{#{QQABTYawwwCwkJZACB57AQX
+CEoQkJKRJUgGRVADuARCwZNIBIA=}#}{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEAEIAAAAANABAA=}#}{#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGRVADuARCw
ZNIBIA=}#}第1页/共6页顺义区2023—2024学年度第二学期期末质量检测高一数学试卷答案一、选择题1-5CDCBA6-10DBCAA二、填空题11、4355i+12、1413、2;3−14、0;315、①②③三、解答题16.参考答案与评分标准:解(Ⅰ)因为e1,e2是两
个单位向量,其夹角为120°,则|e1|=1,|e2|=1,e1·e2=-12.-----------------------------------------2分又a2=(2e1-e2)2=4e21-4e1·e2+e22=7,-----------------------------4分所
以|a|=7,-----------------------------5分同理b2=(3e1+2e2)2=9e21+12e1·e2+4e22=7,-----------------------------7分所以|b|=7.------------------------
-----8分(Ⅱ)由题得,a·b=(2e1-e2)·(3e1+2e2)=6e21+e1·e2-2e22=72.------------10分设a与b的夹角为θ,则cosθ=a·b|a|·|b|=727×7=12.-------------------------
----12分因为θ∈[0,π],所以θ=π3,则向量a与b的夹角为π3.---------------------13分17.参考答案与评分标准:(Ⅰ)由()()22sincos2cosfxAxxxAR=+,知(0)2f=即条
件①不满足.------1分且有()()12sin112cos2sin2+++=++=xAxxAxf,=−A1tan,22,-----------------------------------------------------
----------4分所以()fx的最大值为112++A,---------------------------------------------------------------------5分由条件②:()fx的最大值为12+,得12112+=++A,解得:1=A.--
-------------------------6分{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEAEIAAAAANABAA=}#}{#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGR
VADuARCwZNIBIA=}#}第2页/共6页当1=A时,()142sin2++=xxf,128+=f满足条件③,--------------7分当1−=A时,()2sin(2)14fxx
=−−+,18=f不满足条件③,---------------8分所以()fx满足条件②和条件③,且()142sin2++=xxf.------------------9分因此,函数(
)fx的最小正周期为==22T.--------------------------------10分(Ⅱ)由Zkkxk+++−,224222,得到Zkkxk++−,883---------------------11分∵
,0x,∴函数()fx在区间,0上的单调递增区间为8,0,,85-----13分18.参考答案及评分标准:解:(Ⅰ)在正方体1111ABCDABCD−中,因为11//ABCD,且11ABCD=,............
......2分所以四边形11ABCD为平行四边形.所以11//BCAD..................4分又1BC平面1ADE,EADAD11平面所以1//BC平面1ADE..................................5分(Ⅱ)在正方体1111ABC
DABCD−中,四边形CBB1C1为正方形所以CB1⊥BC1.......................6分AB⊥平面CBB1C1CB1平面CBB1C1{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEAEIAAAAANABAA=}#}
{#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGRVADuARCwZNIBIA=}#}第3页/共6页所以AB⊥CB1.......................8分ABBC1=B.....
..................9分CB1⊥平面ABC1D1;...........................................10分(Ⅲ)...........................
................13分19.参考答案及评分标准:解(Ⅰ)由()xxxxfsin2324sin24sin+−+=得:()xxxxxxxxxfsin232sin2cos21sin
232sin222cos222sin222cos2222+−=+−+=()+=+=6sinsin23cos21xxxxf所以----------
---------------------------------------3分因为()()CfBf=,所以+=+6sin6sinCB-------------4分在ABC中,因为(),0,CB,所以+67,66
B,+67,66C,---5分又∵cb,∴CB,所以=+++66CB,解得:32=+CB.---------7分因为πABC++=,所以3=A.-----------
-----------------------------------------------------8分(Ⅱ)由(Ⅰ)知3=A,又因为5=a,7=+cb,在ABC中,由余弦定理得3cos2cos222222bccbAbccba−+=−+=,--
---10分所以()bcbccbbccb349325222−=−+=−+=,解得:8=bc,-------------13分所以ABC的面积为3223821sin21===AbcSABC.----------------
---15分20.参考答案及评分标准解:(Ⅰ)因为四边形ABCD是正方形,所以//ABCD.................................1分又,ABCDEFCDCDEF平面平面,所以//ABCDEF平面........
..........................2分又平面ABFE平面CDEFEF=,ABABFE平面,....................3分{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEA
EIAAAAANABAA=}#}{#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGRVADuARCwZNIBIA=}#}第4页/共6页所以//ABEF..............................
......................................4分(Ⅱ)EDEA=.取AD的中点N,连接MN,NE因为N是AD中点,M是HB中点,所以//MNAB.又底面ABCD为正方形,所以ADNM⊥...
...............................5分因为EDEA=,所以ADNE⊥又NMNEN=,所以ADNME⊥平面..................................6分又因为MENME平面,所以ADME⊥.又EMBH⊥
且BHAD与是相交线,所以MEABCD⊥平面.................................7分ME平面AME所以平面AME⊥平面ABCD;.................................................
..................8分(Ⅲ)过M点作TG//BC,因为中点,为中点,为BHMDCHAB,4=所以.3,1======AGDTEFTCHTBG又4=AD,由(Ⅱ)可知,MEABCD⊥平面,四棱锥E-ADTG体积𝑉𝐸−𝐴𝐷𝑇𝐺
=13𝑆ℎ...............................9分=13×4×3×2√3=8√3..........10分因为EF//GB,EF//TC且EF=GB=TC,所以四边形EFTC为平行四边形,...................................11分
四边形EFGB也是平行四边形.所以ET//FC.ET平面BCFFC平面BCF所以ET//平面BCF同理EG//平面BCFET,EG平面ENG{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEAEIAAAAAN
ABAA=}#}{#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGRVADuARCwZNIBIA=}#}第5页/共6页ETEG=E所以平面ETG//平面BCF所以五面体GBCTEF为三棱柱..........
..................................12分在三棱柱BCF-TGE中,BG⊥TGMEABCD⊥平面GB平面ABCDBG⊥MEMETG=MBG⊥平面ETG...........................................
.13分𝑉棱柱𝐸𝑇𝐺−𝐵𝐶𝐹=12×4×2√3×1=4√3........................................14分所以五面体ABCDEF的体积为12√3....................
.....15分21.参考答案及评分标准(1)根据向量集的定义,即可写出1Y.在1Y中,检验任意1aY,存在2aY,使得120aa=,即可得出答案;(2)在Y中取()1,2ax=,可得()21,ab=−或()2,1ab=−,根据数量积的坐标公式结合条件即得;(3)取()11,pxxY
=,设(),qstY=,根据条件可得st、中一个必为1−,另一个数是1,从而1X,然后利用反证法,即得.【详解】(1)由已知可得,()11,1,2,1|,,1,,2Yaastst==−−
()()()()()()()()()1,1,1,1,1,2,1,1,1,1,1,2,2,1,2,1,2,2=−−−−−−.------------------------------2分因为()()1,11,10−−−
=,()()1,11,10−−−=,()()1,22,10−=,()()1,11,10−=,()()1,22,10−=,()()2,21,10−=,即对任意11aY,存在21aY,使得120aa=,所以,1X具有性质P.-------------------
-----------------------------------------------------------4分(2)因为1,1,2,x−具有性质P,取()1,2ax=,由120aa=,则Y中的()21,ab=
−或()2,1ab=−.当()21,ab=−时,由()(),21,0xb−=可得,2xb=.-----------------------------5分因为1,1,2,bx−,所以1b=或2b=,所以2x=或4x=.又2x,则4x=;-------------------------
-------------------------------------------------7分当()2,1ab=−时,有()(),2,10xb−=可得,2xb=.因为2x,所以不存在,舍去.综上所述,4x=.---------------
----------------------------------------------------9分{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCAAQgOhEAEIAAAA
ANABAA=}#}{#{QQABTYawwwCwkJZACB57AQX+CEoQkJKRJUgGRVADuARCwZNIBIA=}#}第6页/共6页(3)因为数集121,,,,nXxxx=−,其中120nxx
x,取()11,pxxY=,设(),qstY=,由0pq=得()10xst+=,则0st+=,则s和t中有一个数是1−,则s和t中有一个数是1,即1X,----------------------------------------------------11
分假设1(1)kxkn=,则101nxx,再取()1,nexxY=,(),fstY=,则10nsxtx+=,所以s和t异号,且其中一个值为1−,若1s=−,则11nxtxtx=,矛盾;若1t=−,则1nnxsxsx=,矛盾;则假设1(1)kxkn=不成立,可得
当1nx时,11x=.--------------------------------------------------15分{#{QQABQYYUggCgAIJAAAhCAQWaCEIQkBCA
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