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济宁市实验中学2022级高三上学期开学考数学试题参考答案一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一个选项是符合题目要求的.12345678BACBBDDA二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.
全部选对的得6分,部分选对的得部分分,有选错的得0分.9.BCD10.AD11.ACD三、填空题:本题共3小题,每小题5分,共15分.12.(,4−13.1−14.1,2e+四、解答题:本题共5小题,共
77分.解答应写出文字说明、证明过程或演算步骤.15.(1)()()()322,41,13,ab=−−−−=−−,...........................................................................................
..............3分故()2233322ab−=+−=rr;.................................................................................................
.....................6分(2)设a与b夹角为,()()1,21,11210cos10141110abab−−−−+====++,.............................
...............................................................12分故a与b夹角的余弦值为1010.................................................................
....................................................13分16.(1)因为()fx是二次函数,且关于x的不等式()0fx的解集为|31,Rxxx−
,所以()(3)(1),0fxaxxa=+−,...............................................................................
....................................2分所以当=1x−时,min()(1)44fxfa=−=−=−,所以1a=,...........................
..............................................4分故函数()fx的解析式为2()(3)(1)23fxxxxx=+−=+−........................................
.....................................6分(2)因为函数()gx与()fx的图象关于y轴对称,所以2()()23gxfxxx=−=−−,..............
.....................................................................................................7分当0x时,()gx的图象恒在直线4ykx=−的上方,所以()4gxkx−,在(
)0,+上恒成立,即2234xxkx−−−,所以12kxx+−,..............................................................................................9分令1()2(0)hxxx
x=+−,则min()khx,..............................................................................................10分因
为11()2220hxxxxx=+−−=(当且仅当1xx=,即1x=时,等号成立),............................14分所以实数k的取值范围是(),0−...............................................
..................................................................15分17.(1)由于2()3sincoscos444xxxfx=+3111sincossin()22222262xxx=++=++,....
..........2分令)(2236222Zkkxk+++,.................................................................................................4
分整理得)(438432Zkkxk++,............................................................................................
.....6分所以函数的单调递减区间为28[4,4]().33kkkZ++...................................................................................7分
(2)由于3()2f=,所以13sin()2622++=,..............................................................................................9分则sin()126+=,即2
()262kkZ+=+,....................................................................................
..........11分解得24()3kkZ=+,......................................................................
.......................................................13分则222cos()cos(4)cos(4)1().333kkkZ−=−−=−=..................................
.....................................15分18.(1)由*121(N)nnaan+=−得+−=−*112(1),(N)nnaan,.............................
......................................3分又112a−=,所以1na−是首项为2,公比为2的等比数列......................................................
..............7分(2)由(1)知,11222nnna−−==,所以2log(1)nnban=−=........................................................
........9分所以11111(1)1nnbbnnnn+==−++,.................................................................
.................................................11分123nnSbbbb=++++11111111223111nnnnn=−+−++−=−=+++.................
................................................15分当Nn时,111nSn=−+单调递增,故112nS...............................................................
............................17分19.(1)当1a=时,()e1xfxx=−−,定义域为R,求导可得()e1xfx=−,........................
.........1分令()0fx=,得0x=,当0x时,()0fx,函数()fx在区间(),0−上单调递减,.......................................................2分当0x时,()0fx,函数()fx在区
间()0,+上单调递增,.......................................................3分所以()yfx=在0x=处取到极小值为0,无极大值...........
..................................................................4分(2)方程()1e0xfxax+=−=,当0x=时,显然方程不成立,所以0x,则exax=,..............................
...................................................................................................5分方程有两个不等实根,即ya=与()exgxx=有2个交点
,.....................................................................6分()()21exxgxx−=,当0x或01x时,()0gx,()g
x在区间(),0−和()0,1上单调递减,并且(),0x−时,()0gx,当()0,1x时,()0gx,当1x时,()0gx,()gx在区间()1,+上单调递增,0x时,当1x=时,()gx取得最小值,()1eg=,........................
.........................................................8分作出函数()ygx=的图象,如图所示:因此ya=与()exgxx=有2个交点时,e
a,故a的取值范围为()e,+.............................................................................................................................
.....10分(3)证明:0a,由()e0xfxa=−=,得lnxa=,所以函数()yfx=在(),lna−上单调递减,在()ln,a+上单调递增.由题意12xx,且()()12fxfx=,则()1,lnxa−,()2ln,
xa+.要证122lnxxa+,只需证122lnxax−,而122lnlnxaxa−,且函数()fx在(),lna−上单调递减,故只需证()()122lnfxfax−,又()()12fxfx=,所以只需证()()222lnfxfax−,...
..........................................................................12分即证()()222ln0fxfax−−,令()()()2lnhx
fxfax=−−,即()()2ln2e1e2ln1ee22lnxaxxxhxaxaaxaaxaa−−=−−−−−−=−−+,()2ee2xxhxaa−=+−,由均值不等式可得()22ee22ee20xxxxhxaaaa−−=+−−=,当且仅当2eexx
a−=,即lnxa=时,等号成立.所以函数()hx在R上单调递增......................................................................................
.....................15分由2lnxa,可得()()2ln0hxha=,即()()222ln0fxfax−−,所以()()122lnfxfax−,又函数()fx在(),lna−上单调递减,