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试卷第1页,总6页2021年葫芦岛市普通高中学业质量监测考试高二数学参考答案及评分标准一、单项选择题12345678BACBCCBD二、多项选择题9101112ABBCDBCBD三、填空题13.1,2ab(不唯一)14.415.0.482516.13,3e四、解答题17.(
本小题满分10分)(1)选择①由已知得S3=3a1+3d=4,又a1=1,故d=13---------------------1所以23nan,--------------------------------------------
-------------------------332,22nannnbb-------------------------------------------------------------------5
选择②由已知得d=a7-a17-1=13,又a1=1,----------------------------------------1所以23nan,-------------------------------------------------
---------------------332,22nannnbb-------------------------------------------------------------------5选择③由已知得S3=3a1+3d=4,又
d=13,-----------------------------------1所以23nan,-----------------------------------------------------------------------2试卷第2页,总6页32,22nannnbb
------------------------------------------------------------------------5(2)223nncn----------------------------------
-------------------------------6T10=21-1+22-43+23-53+……+210-10+23=(21+22…+210)-(1+43+…+10+23)-------------------------------------------
--------------8=2(1-210)1-2-10(1+123)2=2021--------------------------------------------------------------10注:如果列举10项直接求
和也给分18.(本小题满分12分)(1)设“选出执行空间站任务3名航天员性别不同”为事件M,则761)(373433CCCMP-------------------------------------
---------------------------------4(2)X=0,1,2,3--------------------------------------------------------------------------5X~H(N
=7,n=3,M=3),故33437kkCCPXkC(3,2,1,0k),所以:3043374035CCPXC,-------------------------------------------------------
------621433718135CCPXC,--------------------------------------------------------------712433712235CCPXC,----------
----------------------------------------------------80343371335CCPXC,----------------------------
----------------------------------9所以X的分布列X0123试卷第3页,总6页P35435183512351----------------------1079733)(NnMXE--------------------------------
-------------------------------------1219.(本小题满分12分)(1)由3221()3fxxaxbxa可得2()36fxxaxb又1x为极值点,所以(1)360,63fabba又极值为0,即2(1)1311faba
,则2320aa--------------2可得:29ab或13ab--------------------------------------------------------------4当2,9ab时,32()69
5fxxxx,22()31293433(3)(1)fxxxxxxxx(,1)1(1,3)3(3,)()fx00()fx↗极大值(1)1f↘极小值(3)5f↗当1,3ab时,32()332fxxx
x222()3633213(1)0fxxxxxx(不恒为0)()fx在R上单调递增,无极值.综上2,9ab.-----------------------------------------------------
------------------6(2)由(1)知,F(x)=x3-6x2+9x-5,且F(2)=-3.F′(x)=3x2-12x+9所以k=F′(2)=-3-----------------------------------------
-----------------------------------8l的方程:y+3=-3(x-2),即3x+y-3=0-------------------------------------------
-------10又(m,n)在l上,故3m+n=3(1m+13n)=13(1m+13n)(3m+3)=13(103+nm+mn)≥169,当且仅当m=n=34时,取“=”试卷第4页,总6页故,1m+13n的最小值为169-
---------------------------------------------------------1220.(本小题满分12分)(1)令2tx,则模型二可化为y关于t的线性回归问题,则1491625115t
,1.32.85.78.913.86.55y,-----------------2则由参考数据可得51522150.520.55iiiiitytyctt,6.50.52110.8dyct
,--------------------------------------------------4则模型二的方程为20.50.8yx;----------------------------------------------------6(2)由模型二的回归方
程可得,(2)10.510.81.3y,(2)20.540.82.8y,(2)30.590.85.3y,(2)40.5160.88.8y,(2)50.5250.813.3y,---------------------------------
-------------------------------85(2)2222221ˆ)000.40.10.50.423.7iiiyy(,---------------------------10故
模型二的拟合效果更好.-----------------------------------------------------------------1221.(本小题满分12分)(1)233,32311
nnnnnnnaaNnaa11),(所以2331nnnnaa-1,----------------------------------------------------
----------------------2又因为131a∴数列nna3是以1为首项,2为公差的等差数列.----------------------------------4(2)由(1)知123nann,所以nnna312)(所以nnnS31235
333132)(…………①14323123533313nnnS)(………②由①-②得1432312)3333(232nnnnS)(试卷
第5页,总6页整理得1S3(1)3nnn----------------------------------------------------------8(3)由(2)知nnna312)(
,∴不等式nann3842,对任意Nn恒成立,等价于nnnn3)12()12(32)(对任意Nn恒成立,整理得:nn332,-------------------------------------------------------------
------10令nnnb332法一:则2n时,112325124333nnnnnnnnbb∴3n时,10nnbb>;4n时,10nnbb<.∴2n或3n时,nb取最大值为
91,∴91.∴λ的取值范围是:,91.-----------------------------------------------------------12法二:2n时,11212132369
3nnnnnbnnbn∴当3n时,11nnbb<,∴2n或3n时,nb取最大值为91,∴91.∴λ的取值范围是:,91.-------------------------------------------------------
----1221.(本小题满分12分)(1)F(x)=(x-1)ex-ax+a=(x-1)(ex-a)①a≤0时,在(-,1)上,F(x)<0,F(x)单调递减;在(1,+)上,F(x)>0,F(x)单调递增;------
--------------------------------------------------------------------------------1②0<a<e时,在(-,lna)上,F(x)>0,F(x)
单调递增;在(lna,1)F(x)<0,F(x)单调递减;在(1,+)F(x)>0,F(x)单调递增;-----------------------------------------------2③a=e时,在(-,+)上,F(x)>0,F(x)单调递增
;-----------------------------3④a>e时,在(-,1)上,F(x)>0,F(x)单调递增;在(1,lna)上,F(x)<0,F(x)单调递减;在(lna,+)上,F(x)>0,F
(x)单调递增;---------------------------------------4(2)f(x)x-2+32≥G(x)ex-12x2-2(a+1)x-2(a+1)2+32≥0令F(x)=ex-12
x2-2(a+1)x-2(a+1)2+32,则F(x)=ex-x-2(a+1)试卷第6页,总6页令G(x)=F(x)=ex-x-2(a+1)则G(x)=ex-1≥0,由x≥0可知G(x)≥0,从而F(x)单调递增∴F(x)≥F(0)=-2a-1,且F(x)≠e2-2a-4-
---------------------------------6①当-2a-1≥0即a≤-12时,F(x)≥0恒成立,∴Fmin(x)=F(0)=52-2(a+1)2由题意:应用52-2(a+1)
2≥0解得:-1-52≤a≤-1+52∴-1-52≤a≤-12合题意;---------------------------------------------------------------8②当-2a-1<
0即a>-12时,F(0)=-2a-1<0,由ex≥ex>2x可知e2(a+1)>4(a+1))F(2(a+1))=e2(a+1)-2(a+1)-2(a+1)>4(a+1)-4(a+1)>0(若考生没有取点,而是用“趋近”的方式说明,扣1分)所以存在唯一x0(0,2(a+1))使
得:F(x0)=0,即ex0-x0-2(a+1)=0当x(0,x0)时,F(x)<0,当x(x0,+)时,F(x0)>0所以F(x)在(0,x0)上单调递减,在(x0,+)上单调递增;故Fmin(x)
=F(x0)=ex0-12x02-2(a+1)x0-2(a+1)2+32=ex0-12x02-(ex0-x0)x0-(ex0-x0)22+32=-12e2x0+ex0+32---------------------------------------------
--------------------10由题意,应有:-12e2x0+ex0+32≥0-12(ex0-1)2+2≥00<ex0≤30<x0≤ln3令(x)=ex-x,则(x)=ex-1≥0(∵x≥0)∴(x)在[0,+)上单调递增∵0<x0≤ln3
∴(0)<(x0)≤(ln3)1<(x0)≤3-ln3∴2(a+1)=ex0-x0=(x0)(1,3-ln3]∴-12<a≤1-ln32综上:a的取值范围是[-1-52,1-ln32]--------------------------------
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