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1上林县中学2020年秋季学期高一年级期末考试直升班数学试题考生注意：1.本试卷分选择题和非选择题两部分。满分150分，考试时间120分钟。2.考生作答时，请将答案答在答题卡上，选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；非选择题请用直

径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答，超出答题区域书写的答案无效，在试题卷、草稿纸上作答无效。3.本卷命题范围：必修1，必修2，必修4第一章~第二章。一、选择题：本大題共12小题，每小题5分，共60分.在每小题给出的四个选

项中，只有一项是符合题目要求的.1.已知1tan4=，则sincossin+=（）A.5B.54C.6D.652.若一个圆锥的轴截面是面积为1的等腰直角三角形，则该圆的侧面积为（）A.2πB.22πC.2πD.4π3.已知直线l过点(1,1)且平行于直线480xy+−=，则直

线l的方程是（）A.430xy−+=B.450xy−−=C.450xy++=D.450xy+−=4.若集合M满足1M，*3|1MxxN剟，则符合条件的集合M的个数为（）A.2B.3C.4D.55

.已知，是不同的平面，m，n是不同的直线，则下列命题不正确的是（）A.若m⊥，//mn，n，则⊥B.若//mn，m=，则//n，//nC.若//mn，m⊥，则n⊥D.若m⊥，m⊥，则//6.已知cos

46a=，sin134b=，()cos43c=−，则（）2A.bcaB.bacC.cabD.cba7.已知某商品的进货成本为20（元/件），经过长时间调研，发现售价x（元）与月销售

量y（件）满足函数关系24000150yxx=+.为了获得最大利润，商家每月应进货的件数为（）A.80件B.100件C.120件D.160件8.在平面直角坐标系xOy中，圆C与圆O；221xy+=外切，且与直线250xy−+=相切，则C的面积的最小值为（）A.4π5B.35π−C.35

π2−D.(625)π−9.已知函数23,0()(1)22,0axaxfxxaxax+=−−−+„在R上单调增，则实数a的取值范围为（）A.20,5B.2,15C.(0,1]D.2,2

510.如图，多面体1111ABCDABCD−为正方体，则下面结论正确的是（）A.11//ABBCB.平面11CBD⊥平面1111ABCDC.平面11//CBD平面1ABDD.异面直线AD与1CB所成的角为30°11.已知函数()3fxxa=−−，若函数(())ffx无零

点，则实数a的取值范围为（）A.(,6)−−B.(,6]−−C.(,0)−D.(,0]−12.已知函数π()sin12fxx=−（0）在区间ππ,44−上单调递增，则实数的取值范围为（）3A.51,3B.15,23C.5,2

3D.50,3二、填空题：本大题共4小题，每小题5分，共20分.13.幂函数()(1)mfxmx−=−的增区间为__________.14.已知函数()fx的定义域为R，且对任意1x，2xR（12xx），均有()()12120fxfxxx

−−成立，则不等式()2log(1)(3)fxf−的解集为__________.15.关于函数()2sin|sin|fxxx=+有下述四个结论：①函数()fx的最小正周期为π②函数()fx的最小值为1−③点(π,0)是函数()fx图象的一个对称中心④直

线π2x=−是函数()fx图象的一条对称轴其中所有正确的结论的序号是__________.16.如阁，在四棱锥PABCD−中，平面PAD⊥平面ABCD，PAD△为等边三角形，四边形ABCD为矩形，24ABAD==，则四棱锥PABCD−的外接球的表面积为__________.三、解答题：本大题共6

小题，共70分.解答应写出必要的文字说明、证明过程及演算步骤.17.（本小题满分10分）已知(1,3)a=−，(2,4)b=−，makb=−，(1)2nkab=−−.当k为何值时：（1）//mn；（2）mn⊥.18.（本小题满分12分）已知圆C的圆心坐标为(,0)a，且圆C与y轴相切.（1）已知

1a=，(4,4)M，点N是圆C上的任意一点求MN的最小值；（2）已知0a，直线l的斜率为43，且与y轴交于点20,3−.若直线l与圆C相离，求a的取值范围.419.（本小题满分12分）已知函数23()2xxfx

ba−=++（0a）为定义在R上的奇函数.（1）求实数a，b的值：（2）解关于x的不等式4()3()2fxfx++„.20.（本小题满分12分）已知函数()sin()fxAx=+（0A，0，π

π22−）的部分图象如图所示.（1）求函数()fx的解析式；（2）求函数()fx的单调递减区间.21.（本小题满分12分）如图，四棱锥PABCD−的底面ABCD是正方形，PAD△为等边三角形，M，N分别是AB，AD的中点.且平面PAD⊥平面ABCD.（1）证明：CM⊥平面PNB

；（2）设点E是棱PA上一点，若//PC平面DEM，求PEEA的值.522.（本小题满分12分》已知函数()2222()log16loglog64xfxx=+.（1）求函数()fx的值域；（2）关于

x的方程()20fxax−+=恰有三个解，求实数a的取值集合；（3）若()()12fxfxm==，且2120xx，求实数m的取值范围.2020年秋季学期高一年级期末考试·直升班数学参考答案、提示及评分细

则1.Asincos115sintan+=+=.2.A设底面圆的半径为r，高为h，母线长为l，由题可知，22rhl==，则21(2)12r=，1r=，2l=，所以圆锥的侧面积为π2πrl=.3.D设所求直线方程为40xyc++=，依题意410c++=，

解得5c=−，所以所求直线的方程为450xy+−=.4.C由题意可知，{1}M=或{1,2}或{1,3}或{1,2,3}.5.B若//mn，m=，则n，//n或//n，n或//n，//

n.6.Dcos46a=，sin46cos44b==，cos43c=，有cba.7.D由题可知，利润224000150800001000()(20)150fxxxxxx=−+=−++，令1tx=，则2()800001000150gt

tt=−++，其图象对称轴为1160t=，即160x=.8.C由题可知，(0,0)到直线250xy−+=的距离为225512=+，又因为圆C与圆O：221xy+=外切，所以圆C的径的最小值为51−，圆C的面积的最小值为2π(51)35π42−−=.69.A由题意可知0102223aaa

a−−+„…，解得20,5a.10.C若11//ABBC，因为11//ABCD，所以11//BCCD，矛盾，故A错误；因为1BB⊥平面1111ABCD，所以平面11BBDD⊥平面1111A

BCD，则平面11CBD⊥平面1111ABCD也是错的，故B错误；因为11//ABCD，11//ADCB，所以平面11//CBD平面1ABD，故C正确；因为1111ABCDABCD−为正方体，所以145BCB=.又//ADBC，所以AD与1CB所成的角为45°，故D错误，11.A设(

)fxt=，则()||30ftta=−−=的解为1,23ta=，由题意可知，1,2()fxt=，无解，即min3()3afx=−，解得6a−.12.D由题意有2πππ2π44T=−−=

…，可得02„，令πππ2π2π2122kxk−−+剟（kZ），有15π17π2π2π1212kxk−+剟（kZ），令0k=，可得函数()fx的一个增区间为5π7π,1212−.又由7π7ππ

12244…，必有5ππ124−−„，可得503„.13.(,0)−由11m−=，有2m=，2()fxx−=，增区间为(,0)−.14.(7,1)−由题意知，()fx单调递减，有2log(1)3x−，得71x−.15.①②由3sin,sin0()sin,si

n0xxfxxx=…，可得函数()fx的图象为：7可知函数()fx的最小正周期为2π，最小值为1−，点(π,0)不是函数()fx的一个对称中心，直线π2x=−是函数()fx图象的一条对称轴.16.64π3如图，取AD的

中点E，BC的中点F，连EF，PE，在PE上取点G，使得2PGGE=，取EF的中点H，分别过点G，H作平面PAD、平面ABCD的垂线，两垂线相交于点O，显然点O为四棱锥PABCD−外接球的球心，2AD=，4AB=，可得3P

E=，33GEOH==，22125AH=+=，22343(5)33OA=+=，故四棱锥PABCD−外接球的表面积为24364π4π33=.17.解；(1,3)(2,4)(2

1,43)mkkk=−−−=−−+··································································2分(1)(1,3)2(2,4)(3,35)nkkk=−−−−=

−−+······································································4分（1）若//mn，有(21)(35)(3)(43)kkkk−

−+=−−+，整理为220kk−−=解得1k=−或2；··········································································································7分

（2）若mn⊥，有(21)(3)(43)(35)0kkkk−−−−+++=，整理为271890kk++=解得：9327k−=.·····································································

······························10分18.解：（1）由题可知，圆C的方程为22(1)1xy−+=，························································2分又2

2(41)(40)5MC=−+−=，··················································································4分所以MN的最小值为514−=.········

·················································································6分（2）直线l的方程为4233yx=−，即4320

xy−−=.··························································7分因为直线l与C相离，8所以圆心(,0)Ca到直线l的距离dr，即22|42|||43a

a−+，·················································9分又0a，则245aa−−，解得2a−.······························

···········································11分所以a的取值范是(2,0)−.····························································

·································12分19.解：（1）由题意可知，2(0)01fba−=+=+，整理得21ba=+.···········································2分又由(1)

(1)ff−=−，即13232122bbaa−−+=−−++，整理得152442baa=+++，························4分即15224421aaa+=+++，解得1a=，所以211ba==+，当1ab==时，经检验，()()fxfx−=−恒成立，所

以1ab==；·········································6分（2）由（1）可知，()22123()12121xxxxfx−−=+=++，·················································

··········7分不等式4()3()2fxfx++„时化为()()22142132142xxxx−+++„·········································································

······················9分有()()()2242213221xxxxx−+++„有123x…，得21log3x…故不等式4()3()2fxfx++„的解集为21log,3+.········

·················································12分20.解：（1）由函数可知，1A=，函数()fx的周期为2π3πππ416162T==−=，解得4=·····2分可得()sin(4)fxx=+，代入点π,016

的坐标有，πsin04+=.·····································4分又由ππ22−，有ππ3π444−+，有π04+=，得π4=−9故有函数()fx的解析式为π()sin44f

xx=−.··································································6分（2）令ππ3π2π42π242kxk+−+剟（kZ），··············

···················································9分解得π3ππ7π216216kkx++剟（kZ）故函数()fx的单调递区间为π3ππ7π,216216kk++（kZ）.·

··············································12分21.（1）证明：在正方形ABCDD中，M，N分别是AB，AD的中点，∴BMAN=，BCAB=，90MBCNAB==.

∴MBCNAB≌△△，···································································································1分∴BCMNBA=.又90BCMBMC+

=，∴90NBABMC+=，∴CMBN⊥.··························································································

·····················2分∵PAD△为等边一角形，N是AD的中点，∴PNAD⊥.···································································································

·············3分又平面PAD⊥平面ABCD，PN平面PAD，平面PAD平面ABCDAD=，∴PN⊥平面ABCD.·············································

························································4分又CM平面ABCD，∴CMPN⊥.··········································

·····································································5分∵BN，PN平面PNB，BNPNN=，∴CM⊥平面PNB.················

·······················································································6分（2）解：连接AC交DM于点Q，连接EQ.∵//PC平面D

EM，PC平面PAC，平面PAC平面DEMEQ=，∴//PCEQ.··················································································································8

分∴::PEEACQQA=.····································································································9分在正方形ABCD中，//AMCD

且2CDAM=，10∴::2CQQACDAM==.···························································································10分∴2PEEA=.·····

············································································································12分22.解：（1）易知()fx的定

义域为(0,)x+，设2logxt=R，则()()222()2log4log6(24)(6)2(2)3232fxxxttt=+−=+−=−−−…，所以()fx的值域为[32,)−+；·········································

··············································4分（2）设2logxt=R，由（1）可知，()()(24)(6)fxgttt==+−，令()0gt=，解得12t=−，26t=，·····························

·····················································5分所以2log2x=−或2log6x=，解得：14x=或64x=，·······································

···············6分因为()20fxax−+=恰有三个解，所以214xax−+=或264xax−+=恰有三个解，即2640xax−+=恰有一解，所以24640a=−=，解得16a=，所以a的取值集合为{16,16}−；···

····················································································8分（3）设211logxt=，222logx

t=，因为212xx，所以2221loglog1xx+，即211tt+.·········9分则()(24)(6)gtttm=+−=的两根为1t，2t，整理得228240ttm−−−=，所以124tt+=，12122mtt=−−

，·················································································10分所以()2211212648(24)042641mttttttm=++−=+−=+，11解得63,2m

−+.···································································································12分