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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)123456789101112ACDDBDAABBCD
BD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,1xu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