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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)()()()3
22,41,13,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|3
1,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()()23
gxfxxx=−=−−,................................................................................................................
...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)hxxxx=+−,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()3s
incoscos444xxxfx=+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()()2
1exxgxx−=,当0x或01x时,()0gx,()gx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()ex
gxx=有2个交点时,ea,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()()222ln0fxf
ax−−,令()()()2lnhxfxfax=−−,即()()2ln2e1e2ln1ee22lnxaxxxhxaxaaxaaxaa−−=−−−−−−=−−+,()2ee2xxhxaa−=+−,由均值不等式可得()22ee22
ee20xxxxhxaaaa−−=+−−=,当且仅当2eexxa−=,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()()122ln
fxfax−,又函数()fx在(),lna−上单调递减,