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唐山一中2020—2021学年度第一学期期中考试高一年级数学试卷答案选择:1-5ABDBD6-8AAB9-12ADBCACABD填空:13.31,2,14.315.216.()()--13+U,,17.(1)1{|1}ABxx=−I;(){|1}RABx
x=−ð(2)01a【详解】(1)由02131−=x得1x,所以{|1}Axx=,R{|1}Axx=>ð由{|12}=−Bxx所以1{|1}ABxx=−I..................................................
.............................................3分所以(){|1}RABxx=−ð............................................................
...................................5分(2)因为CB,且0a所以2a≤2,1a所以a的取值范围为:01a.................................................................
........................10分18.(1)4k(3)k2【详解】(1)若()fx在(2,)x+单调递增,则22k,所以4k........................................4分(2)因为()0fx在(0,)x+上
恒成立,所以210−+xkx在(0,)x+恒成立,即1+kxx在(0,)x+恒成立.........................................................
................................8分令1()gxxx=+,则11()22=+=gxxxxx,当且仅当1x=时等号成立所以k2..............................................................
..............................................................12分19.(1)a=1(2)x=-1【详解】(1)-12,12aa==..................
...........................................................................................4分(2)()()()1,2xfxgxfx==
即,112042xx−−=21,0220,0,212,12xxttttttx=−−====−>,>...................................
.......................................................................8分.................................
............................................................................................................12分20.(1)2
m=,()4fxx−=;(2)111(,)(,3)322−.【详解】(1)由题意,函数()24−=mmfxx(实数mZ)的图像关于y轴对称,且()()23ff,所以在区间(0,)+为单调递减函数,所以240mm−,解得04m,...................
......................................................................3分又由mZ,且函数()24−=mmfxx(实数mZ)的图像关
于y轴对称,所以24mm−为偶数,所以2m=,所以()4fxx−=.............................................................................
.................................................6分(2)因为函数()4fxx−=图象关于y轴对称,且在区间(0,)+为单调递减函数,所以不等式()()212+−fafa,等价于
122aa−+且120,20aa−+,.............8分解得1132a−或132a,所以实数a的取值范围是111(,)(,3)322−.....................
...................................................12分21.选法见解析;2a=,0b=;(1)证明见解析;(2)103t.【详解】(1)①由()()220fx
fx−++=得()fx对称中心为()2,0即得2a=,0b=;②(i)当1a时,()xfxab=+在1,2上单调递增,则有224abab+=+=得220aa−−=,得2a=,0b=;(ii)当01a时,()xfxab=+在1,2上单调
递减,则242abab+=+=得220aa−+=,无解,所以2a=,0b=;③由()24fxxax=−+得()()2125fxxaxa+=+−+−,因为()1fx+在1,1bb−+上是偶函数,
则202a−=,且()()110bb−++=,所以2a=,0b=;............................................................................................................2分由
①或②或③得()222xgxx=+,()1,1x−,任取()()()()()()()()()()121221121212222212121221121212,(1,1),-112122222222-110100xxxxxxxxxxgxgxxxxxxxxxxxgxgxgxgx
−−−−=−=++++−−−且<<<,则∵<<<,则>,<,∴<,即<则()gx在()1,1−上单调递增.................................................
................................................7分(2)因为()()222xgxgxx−−==−+,则()gx为奇函数.由()()120gtgt−+即()()21gtgt−又因为()gx
在()1,1−上单调递增,则121,111,21,tttt−−−−解得103t..................12分22.(1)20(2)32(3)6x=或45x=【详解】(1)由题知:当108x=米时,点F在线段C
D上,22||1086=−=DF所以12ABCDABCDABEECFADFSSSSSSS=−=−−−所以1(10)641642420==−−−=Sf(平方米)............................
............................2分(2)由题知,当8x(米)时,点F在线段AD上此时:132=ADESS(平方米)...........................................
............................................3分当8x(米)时,点F在线段CD上,[8,82)x,令2||64[0,8)==−tDFx所以12ABCDABCDABEECFADFSSSSSSS=−=−−−所以221
()64162(864)464==−−−−−−Sfxxx232264322=−−=−xt...........................................................................
.................5分因为[0,8)t,所以132232=−St,等号当且仅当0t=时,即8x=时取得所以()fx最大值为32..........................................................................
.............................7分(3)因为1264+=SS,所以:()12121292592564+=+=+SSWSSSS.................9分212112121925192534[342]164
64=+++=SSSSSSSS(万元)等号当且仅当21121292564,+==SSSSSS时取得,即124=S时取得当8x(米)时,点F在线段AD上,1424==Sx,6x=当8x(米)时,点F在线段CD上,213226424=−−=Sx,45x=综上的
W取最小值时6x=或45x=.......................................................................12分