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2020~2021学年第二学期期中调研测试高二数学2021.04一、单项选择题：本大题共小题，每小题分，共计40分，每小题给出的四个选项中，只有一个是正确的，请把正确的选项填涂在答题卡的相应位置上。1.函数3()2lnfxxxx=++的单调递区间为（）A.(0,3)B.(0,1

)C.(1,3)D.(3,)+2.用数字0，1，2，3可以组成无重复数字的四位偶数有（）A.12个B.10个C.20个D.16个3.函数1()sin2fxxx=−的大致图象可能是（）A.B.C.D.4.在

12nxx−的展开式中，只有第5项的二项式系数最大，则展开式中x的系数为（）A.7−B.358−C.358D.75.已知函数2()fxx=的图象在1x=处的切线与函数e()xgxa=的图象相切，则实数a=（）A.eB.ee2C.e2D.ee6.将编号为1，

2，3，4，5，6，7的小球放入编号为1，2，3，4，5，6，7的七个盒子中，每盒放一球，若有且只有三个盒子的编号与放入的小球的编号相同，则不同的放法种数为（）A.315B.640C.840D.50407.

已知函数21()exfx−=，1()ln2gxx=+，若()()fmgn=，则mn−的最大值是（）A.ln212+−B.1e2−C.ln(2e)2D.1e2−−8.已知21001210()gxaaxaxax=++++，290129()hxbbxbxbx=++++，若1

9(1)(12)xx+−=10(1)()()xgxhx−+，则9a=（）A.192B.19102C.18102−D.1832−二、多项选择题：本题共2小题，每小题5分，共10分.在每小题给出的四个选项中，有多项符合题目

要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.给定函数()()1xfxex=+.下列说法正确的有（）A.函数()fx在区间(,2)−−上单调递减，在区间(2,)−上单调递增B.函数()fx的图象与x轴有两个交点C.当210ae−时，方程(

)fxa=有两个不同的的解D.若方程()fxa=只有一个解，则0a10.下列结论正确的是（）A.6本不同的书分给甲、乙、丙三人，每人两本，有222642CCC种不同的分法；B.6本不同的书分给甲、乙、丙三人，其中一人1本，一人2本，一人3本，有123653CCC种不同的分法；C.

6本相同的书分给甲、乙、丙三人，每人至少一本，有10种不同的分法；D.6本不同的书分给甲、乙、丙三人，每人至少一本，有540种不同的分法.11.设6260126(21)(1)(1)(1)xaaxaxax+=+++++++，下列结论正确的是（）A.6

012563aaaaa−+−+=B.23100aa+=C.1a，2a，3a，…，6a中最大的是2aD.当999x=时，6(21)x+除以2000的余数是112.已知函数()(1)lnfxxxx=−−，下述结论正确的是（）A.()fx存在唯一极值点0x，且0(1,2)x

B.存在实数a，使得()2faC.方程()1fx=−有且仅有两个实数根，且两根互为倒数D.当1k时，函数()fx与()gxkx=的图象有两个交点三、填空题：本题共4小题，每小题5分，共20分.13.二项式612xx+

的展开式中，常数项为___________.14.若函数32()1fxxaxx=−++在(2,)+上单调递增，则实数a的取值范围是___________.15.如图，用五种不同的颜色涂在图中不同的区域内，要求每个区域只能涂一种颜

色，且相邻（有公共边）区域涂的颜色不同，则不同的涂色方案一共有__________种.（用数字作答）.16.已知函数1()2lnfxaxxx=−−，若()fx在[1,]e上单调减函数，则实数a的最大值为_________.若0a，在[1,]e上至少

存在一点0x，使得()0020efxx−成立，则实数a的最小值为___________.四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.17.（本题满分10分）已知函数2()xfxe

xa=−+，xR的图象在点0x=处的切线为ybx=.（1）求函数()fx的解析式；（2）设2()()gxfxxx=+−，求证：()0gx；18.（本题满分12分）已知从3312nxx−的展开式的所有项中任取两项的组合数是21。（1）求展开式中所有二

项式系数之和2）若323112naxxx+−的展开式中的常数项为72，求a的值。19.按照下列要求，分别求有多少种不同的方法？（列式并用数字作答）（1）5个不同的小球放入4个不同的盒子，每个盒

子至少放一个小球（2）6个不同的小球放入4个不同的盒子，每个盒子至少一个小球；（3）6个相同的小球放入4个不同的盒子，每个盒子至少一个小球；（4）6个不同的小球放入4个不同的盒子，恰有1个空盒.20.（本题满分12分）已知函数()(1)lnafxxaxx=−

−+（aR）.（1）当01a时，求函数()fx的单调区间；（2）是否存在实数a，使()fxx恒成立，若存在，求出实数a的取值范围；若不存在，说明理由.21.（本题满分12分）在杨辉三角形中，从第2行开始，除1以外，其它每一个数值是它上面的两个数

值之和，该三角形数阵开头几行如图所示.第0行1第1行11第2行121第3行1331第4行14641第5行15101051第6行1615201561（1）在杨辉三角形中是否存在某一行，使该行中三个相邻的数之比是3:4:5？若存在，试求出是第几行；若不存在，请说明理由；（2）已知n，

r为正整数，且3nr+，求证：任何四个相邻的组合数Crn，1Crn+，2Crn+，3Crn+不能构成等差数列.22.（本题满分12分）已知函数()(ln1)fxxx=−，()gxaxb=+（a，bR）.（1）若1a=时，直线()yg

x=是曲线()fx的一条切线，求b的值；（2）若bea=−，且()()fxgx在[,)xe+上恒成立，求a的取值范围；（3）令()()()xfxgx=−，且()x在区间2,ee上有零点，求24ab+的最小值.2020~2021

学年第二学期期中调研测试高二数学参考答案2021.04一、单项选择题：本大题共小题，每小题分，共计40分，每小题给出的四个选项中，只有一个是正确的，请把正确的选项填涂在答题卡的相应位置上。1.B；2.B；3.A；4.D；5.B；6.A；7.A；8.D二、多项选择题：本题共2小题

，每小题5分，共10分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.AC；10.ACD；11.ABD；12.ACD三、填空题：本题共4小题，每小题5分，共20分.13.402；14.

13,4−；15.180；16.（1）221ee+（2）241ee−四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.17.（本题满分10分）【解析】（1）2()xfxexa=−+，()2xfxex

=−，························································1分由已知得(0)10(0)1fafb=+===，·······················

·································································2分解得11ab=−=，故2()1xfxex=−−·····································

·········································4分（2）2()()1xgxfxxxex=+−=−−，()10xgxe=−=得0x=.································

····6分当(,0)x−时，()0gx，()gx单调递减；当(0,)x+时，()0gx，()gx单调递增.·································································8分∴min()(0)0

gxg==，从而2()fxxx−+，即()0gx.··············································10分18.解析：3312nxx−的展开式

共有1n+项，由题意得21C21n+=，解得6n=，·················2分所以展开式中所有二项系数之和为6264=.·················································

····················4分（2）由（1）知，6332233111122naxaxxxxx+−=+−3312nxx−的展开式的通项为626331663

11C()C22rrrrrrrTxxx−−+=−=−，···················6分令6203r−=，或2，解得3r=或0r=，·························

·············································8分因为展开式中的常数项为3306617CC22a−+=，··················································

·········10分解得1a=−.············································································································12分19.【解析】（

1）2454CA240=；·····················································································3分（2）2211346421642222CCCCCA1560AA+

=；或342222644642CACCCA1560+=；··································6分（3）24C410+=；或25C10=；············

·······································································9分（4）222321236426315433CCCCCCCA2160A++=

.或()143323222463633642CCACCACCC2160++=.········12分20.【解析】（1）函数()fx的定义域为(0,)+，221()(1)()1aaxaxfxxxx+−−=+−=，·········1分21.【解析】（1

）存在.··································································································1分杨辉三角形的第n行由二项式系数Ckn，0k=，1，2，…，n组成

.若第n行中有三个相邻的数之比为3∶4∶5，则1C3C14knknknk−==−+，1C14C5knknknk++==−，·····························································

·····3分即373nk−=−，495nk−=，解得27k=，62n=.即第62行有三个相邻的数2662C，2762C，2862C的比为3∶4∶5.················································5分（2）证明若有n

，r（3nr+），使得Crn，1Crn+，2Crn+，3Crn+成等差数列，则122CCCrrrnnn++=+，2132CCCrrrnnn+++=+，·················································

··················6分即2!!!(1)!(1)!!()!(2)!(2)!nnnrnrrnrrnr=++−−−+−−，2!!!(2)!(2)!(1)!(1)!(3)!(3)!nnnrnrrnrrn

r=++−−+−−+−−··············································7分所以211(1)(1)()(1)(1)(2)rnrnrnrrr=++−−−−−++，211(2)(2)(2)(1)(2)(3)rnrnrnr

rr=++−−−−−−++整理得2(45)4(2)20nrnrr−++++=，2(49)4(1)(3)20nrnrr−+++++=.两式相减得23nr=+，·····································································

·························9分所以23Crr+，123Crr++，223Crr++，323Crr++成等差数.··································································10分由二项式系数的性

质可知31223232323CCCCrrrrrrrr+++++++==.···················································11分这与等差数列的性质矛盾，从而要证明的结论成立.···

·······················································12分22.【解析】（1）当1a=时，()gxxb=+，()lnfxx=，设切点()()00,

Axfx，则()fx在点A处的切线为()()0000ln1lnyxxxxx−−=−，·············································2分化简得()00lnyxxx=−，因为()gxxb=+是()fx的一条切

线，0ln1x=，0xb−=，解得0xe=，be=−；···································································3分（2）当bea=

−时，令()()()(ln1)()hxfxgxxxaxe=−=−−−，则()lnhxxa=−.············4分若1a，则当xe时，()0hx恒成立，()hx在[,)e+上单调递增，()()0

hxhe=，即()()fxgx符合题意；···································································5分若1a时，由()0hx=，得axee=，当1axe时，()0hx，()hx在,aee

上单调递减，()()0hxhe=，与已知()hx在[,)xe+上恒成立矛盾，舍去.········································6分综上，1a且0a.······

····························································································7分（3）()(ln1)xxx

axb=−−−，()lnxxa=−.若1a，则()0x在区间()2,ee上恒成立，()x在区间2,ee上单调递增，因为()x在区间2,ee上有零点，所以()222()00eaebeeaeb=−−

=−−，解得22aebeae−−.······················································································

········8分所以222244(2)414abaaeaeee+−=−−−，当1a=时，等号成立，此时bae=−.············································

·································9分若12a时，当aexe时，()0x，()x在()1,ae上单调递减，当2aexe时，()0x，()x在()1,ae上单调递增.因为()x在区间2,ee上有零点，所以(

)(1)0aaaaeeaaebeb=−−−=−−，所以abe−，所以2244aabae+−，·······································································10分令2()4ahaa

e=−，12a，则()()24220aahaaeae=−=−，所以()ha在（2）上单调递减.所以222()2444haee−=−.若2a，则()0x在区间()2,ee上恒成立，()x在区间2,ee上单调递减.因为叫()x在区间2,ee上有零点，所以()

222()00eaebeeaeb=−−=−−，解得22eaebae−−.··································································································11

分所以()222222242444424444abaaeeaeeeee−+−+=−+−−，当22ae=时，等号成立，此时242bee=−；······································

··························12分综上，24ab+的最小值是2444ee−.当01a时，由()0fx，得0xa，或1x，由()0fx，得1ax，·····

····················································································2分故函数()fx的单调递增区间为(0,)a和(1,)+，单调递减区间为(,1)a，···········

···················4分当1a=时，22(1)()0xfxx−=恒成立，故函数()fx的单调递增区间为(0,)+.···················5分（2）()fxx恒

成立等价于(1)ln0aaxx++恒成立，令()(1)lngxaaxx=++，当10a+=时，即当1a=−时，()1gx=−，故()0gx在(0,)+内不能恒成立，················6分当10a+时，即当1a−时，则(1)1ga=−，故()0gx在

(0,)+内不能恒成立，········7分当10a+时，即当1a−时，()(1)(1ln)gxax=++，由()0gx=解得1xe=，···············8分当10xe时，()0gx；当1xe时，()0gx.

······················································10分所以min11()0agxgaee+==−，解得11ae−.····················································

··11分综上，当11ae−时，()0gx在(0,)+内恒成立，即()fxx恒成立，所以实数a的取值范围是1,1e+−.······································

····································12分