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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()()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()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123nnS
bbbb=++++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()()21exxgx
x−=,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()exgxx=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()()12
fxfx=,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()22ee22ee20xxxxhxaaaa−−=+−−=,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()()122lnfxfax−,又函数()fx在(),lna−上单调递减,