【文档说明】吉林省长春市农安县2020-2021学年高一上学期期末考试数学答案.pdf,共(4)页,137.869 KB,由管理员店铺上传
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农安县基础年级期末质量检测高一数学试题数学评分细则考查时间:120分钟考查内容:第一册一.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6
0分)123456789101112BABBACDBBCABDCD二.填空题(本题共4小题,每小题5分,共20分.)13.414.315.416.(1,4)二.填空题(本题共6大题,共70分.)17.(本题10分)解:(1)151515153log2loglog20log
24153log(2204)2...........................................................................................3分15log151;.................
..........................................................................5分(2)1020.5231(2)2(2)(0.01)5411221411()()49100
...........................................................................................3分12114310................................
...........................................................4分1615......................................................................
......................5分18.(本题12分)解:解:(1)由题意可得:5OP................................................................................2
分由角的终边上的点的性质可得,53sin;.................................................................3分54cos..............................
..........................................................4分(2)由(1)可知53sin,54cos,再结合诱导公式得:)cos(2-sin(cos(2-2cos()()π)π)πf
cos2sincos2sin...................................................................6分243243...........................
........................................8分115.....................................10分所以115)(f....................................
...............................................12分19.(本题12分)解:(1)当1m时,不等式()0fx为220xx,...............................2分即(21)0xx,解
得0x或12x,...............................4分因此所求解集为1(,0),2...............................6分(2)不等式()10fx即2(1)0mxmxm,.............
..................7分由题意知10m且3,32是方程2(1)0mxmxm的两根,...............................9分因此33213321mmmm,,................
...............11分解得97m................................12分20.(本题12分)解:(1)由函数的图象可得A=2,........................................
.....1分34T=34•2=1112-6,求得ω=2...............................................3分再根据五点法作图可得2×6+φ=2,∴φ=6
,..............................................5分故f(x)=2sin(2x+6).............................................
..6分令2kπ-2≤2x+6≤2kπ+2,k∈z,............................................8分求得kπ-3≤x≤kπ+6,....................................
..........9分故函数的增区间为[kπ-3,kπ+6],k∈z..............................................10分若x∈[-2,0],则2x+6∈[-56,6],...............................
...............11分∴sin(2x+6)∈[-1,12],故f(x)∈[-2,1]...............................................12分21.(
本题12分)解:(1)2223sincossincos3sin2cos22sin26fxxxxxxxx,..........4分所以,函数fx的最小正周期为22T,............
...........5分最大值为2;........................6分(2)解不等式3222262kxkkZ,....................8分可得263kxkkZ
,....................10分因此,函数fx的单调递减区间为2,63kkkZ.....................12分22(本题12分)解:(1)()11xafxe为奇函数,则1(0)102f
a即2a,............................................2分所以2()11xfxe,经检验符合题意,............................................3
分设12xx,则12211212222()()()011(1)(1)xxxxxxeefxfxeeee,............................................5分所以12()()fxfx,所以()fx在R上单
调递增;............................................6分(2)因为2(1)(2)0fxfx.所以22(1)(2)(2)fxfxfx,.............................
...............8分所以212xx,...........................................10分解可得,11311322x,故不等式的解集113113{|}22xx....................
........................12分