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高一质量检测数学评分细则考查时间:120分钟考查内容:第一、二、三章一.选择题(本题共12小题,1-10题为单选,在每小题给出的四个选项中只有一个选项符合题目要求.11-12为多选,至少有两个选项符合题目要求.每小题5分,共计60分)123456
789101112ACDDBDAABBCDBD二.填空题(本题共4小题,每小题5分,共20分.)13.8314.815.816.2,13二.填空题(本题共6大题,共70分.)17.(本题10分)解:(1)当1m时,22xxB.................
......................................................2分∴32xxBA................................................................
.................................5分(2)∵BA,∴B,则有:311212mmmm,....................................................................
...................................................8分解之得:m≤2∴实数m的取值范围是2mm;........................................................
.........................10分18.(本题12分)解:(1)设()fxkxb,................................................
............................................2分则(1)(1)fxkxbkxkb,又因为(1)64fxx,所以64kkb,6k,2b,.................................
..4分所以()62fxx.............................................................................................................6分(2)设1xu,1x
u则.....................................................................................2分22()(1)(1)32fuuuuu,................
.....................................................4分所以2()32fxxx................................................................................
..................6分19.(本题12分)【解析】(1)因为1x,10x,..............................................................................1分11111211311
1xxxxxx,..................................................3分当且仅当111xx时,即当2x时等号成立,...............................
................................5分11xx的最小值为3;...............................................................................
..........................6分(2)由100xx知010x......................................................................................
.....1分当0x或10时,100xx;当010x时,100x,由基本不等式可得101052xxxx....................3分当且仅当10xx,即当5x时等号成立.......................
.........................................................5分综上,10xx的最大值为5..................................................
..................................................6分20.(本题12分)解:(1)由20,()2xfxxx时,当0,0xx时2()2fxxx....................
...................2分又函数()fx为偶函数,2()2fxxx故函数的解析式为222(0)(){2(0)xxxfxxxx................................
..........................................................................................4分...............6分(2)由函数的图像可
知,函数()fx的单调递增区间为...............................................8分单调递减区间为,...................................................................
............................................10分函数()fx的值域为1,.........................................................................
.................................................12分21.(本题12分)【解析】(1)4444134mmf,解得1m;........
.................................................................2分(2)因为4fxxx,定义域为0xx,关于原点对称,.........................................
................3分又44fxxxfxxx,因此,函数yfx为奇函数;................................5分(3)设
120xx,则12121212214444fxfxxxxxxxxx1212121212441xxxxxxxxxx,因为120xx,所以120xx,所以12fxfx,..
...............................................................10分因此,函数yfx在0,上为单调增函数.........................................................
...........................12分22.(本题12分)试题解析:(1)为奇函数,且有定义,则,..............................................2分则,,得,...............................
..............4分所以解析式...............................................................................
.............6分(2)在恒成立,即在恒成立,...............8分其中,............................................................
..............................10分分母在取得最小值,..........................................................
...............................11分得到,即....................................................................................................12分