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2024-2025学年高一数学上学期期中模拟卷(江苏专用)参考答案第一部分(选择题共58分)一、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,只有一项是符合题目要求的。12345678BCABDAAB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.91011BCABDABC第二部分(非选择题共92分)三、填空题:本题共3小题,每小题5分,共15分。12.((),80,−−+13.114.()0,+
四、解答题:本题共5小题,共77分。解答应写出文字说明、证明过程或演算步棸。15.(13分)【解析】(1)21ln233lg25lg2lg50(lg2)0.125e−−++++2322ln31lg5lg2(lg10lg5)(lg2)e8−
=+++++......................................................................................2分22lg5lg
2(1lg5)(lg2)43=+++++2lg5lg2(lg5lg21)7=++++...............................................................................
..................................4分2lg52lg27=++9=;..............................................................
...........................................................................................6分(2)∵2363412xy==,∴6lg122lg3x=,6lg123lg4y=,......
..........................................................................................................9分∴32xy+6lg126lg1223lg3lg4lg12lg1232lg3l
g4xyxy++==lg3lg4lg12lg3lg4lg12lg12lg3lg4+=.......................................................................
............................................................11分lg3lg41lg12+==....................................................................
.................................................................13分16.(15分)【解析】(1)由题意得:()()10100010.2%101000xx−+,.......................
..................................2分即25000xx−,又0x,所以0500x.即最多调整500名员工从事第三产业....................................
................................................................4分(2)从事第三产业的员工创造的年总利润为310500−xax万元,从事原来产业的员工的年总利润为()11010001500xx−+万
元,....................................................6分则()()31010100010.2%500xaxxx−−+........................
....................................................................8分所以223110002500500xaxxxx−+−−所以221000500xaxx++,即210001500xax++恒成
立,...............................................................................................................10分因为2100021000
24500500xxxx+=,当且仅当21000500xx=,即500x=时等号成立.........................................................................
...........13分所以5a,又0a,所以05a,即a的取值范围为(0,5...........................................................................................
..............................15分17.(15分)【解析】(1)函数()fx为R上的奇函数,........................................................
..................................1分因为()()2fxxxxfx−=−+=−,所以函数()fx为R上的奇函数.................................................................
............................................4分(2)222,0()22,0xxxfxxxxxxx−=−=−−,................................
..............................................................5分图象如图所示,...........................................
.................................................8分(3)()3Ma=,若()()3Magaa==−=,则3a=−,此时(3)3g−=,2(3)(3)2(3)33f−=−−−−=−,满足题意;...
...................................................10分若()()23Mafaaaa==−=,显然0a时22()2(1)11Maaaa=−−=−++,()3Ma=无解;12分因此0a,2()23Maaa=−=,解得3a=,此时(
3)33g=−满足题意...................................14分所以3a=−或3a=....................................................................................
.............................................15分18.(17分)【解析】(1)因为12a=,所以不等式101axx−+可化为11201xx−+,也即(2)(1)0xx−+,.........2分解得:12x−,故(
12)P=−,.............................................................................................................4分(2)由不等式101axx−+可化为()()110ax
x−+,............................................................................5分因为关于x的不等式101axx−+的解集为()1,1,2P=−−−+,....................
.............................6分所以1−和12−是方程(1)(1)0axx−+=的两根,所以2a=−......................................................
..........................................................................................8分(3)因为不等式11x−可化为:111x−−,解得:02x,所以(0,2)Q=,又因为“x
Q”是“xP”的充分非必要条件,所以Q是P的真子集,........................................................................................9分当0a=时,(1)P=−+,,满足题意;....
..........................................................................................11分当0a时,1(1)Pa=−,,要使Q是P的真子集,则有12a,所以102a;.........
...................13分当10a−时,1()(1,)Pa=−−+,,满足Q是P的真子集,..........................................
.........14分当1a=−时,(1)(1,)P=−−−+,,满足Q是P的真子集,.......................................................15分当1a−时,1(1)(,)Pa
=−−+,,满足Q是P的真子集,........................................................16分综上所述:实数a的取值范围为1(,]2−....................................
.........................................................17分19.(17分)【解析】(1)解:由函数()21xfxaxb+=+为奇函数,且()12f−=−,可得()12f=,则2222abab=−−+=+,解得
1,0ab==,可得()1fxxx=+,...........................................2分经检验,有解析式可知,定义域|0xx,关于原点对称,可得()()()110fxfxxxxx+−=++−+=−,所以()fx是
奇函数,满足题意....................................4分函数()fx在()0,1上单调递减,在()1,+上单调递增,.....................................................................5分
证明如下:任取()12,0,1xx,且12xx,则()()()121212121212111xxfxfxxxxxxxxx−−=+−+=−,因为()12,0,1xx,且12xx,所以120xx−,1201xx,所以1210xx−,
所以()()120fxfx−,即()()12fxfx,所以函数()fx在()0,1上单调递减,同理可证明函数()fx在()1,+上单调递增........................8分(2)解:由题意,函数()22112hxxtxxx=+−+,令
1zxx=+,可得222yztz=−−,...........10分由(1)可知函数1zxx=+在1,12上单调递减,在1,2上单调递增,所以52,2z,因为函数222yztz=−−的对称轴方程为0zt=,所以函数222yztz=−−在52,2上单调
递增,.................................................................................12分当2z=时,222yztz=
−−取得最小值,min42yt=−+;当52z=时,222yztz=−−取得最大值,max1754yt=−+.所以()min42hxt=−+,()max1754hxt=−+,.........................................................
.............................14分又因为对任意的121,,22xx都有()()12154hxhx−恒成立,所以()()maxmin154hxhx−,即()171554244tt−+−−+,解得32t−
,........................................16分又因为0t,所以302t−,所以实数t的取值范围是3[,0)2−................................................17分