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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60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)1
51515153log2loglog20log24153log(2204)2...........................................................................................3分
15log151;...........................................................................................5分(2)1020.5231(2)2(2)(0.01)5411221411()()4
9100...........................................................................................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)πfcos2sincos2sin.........................................
..........................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263k
xkkZ,....................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)102fa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