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期末数学参考答案选择题1-8ADAACACC9-12AD、ACD、AB、ABC填空题13.),53(14.714.02,16.2解答题310331211211211211tan1tan1tan1tan1sincos
sincossincossincos)1(17原式解:(2)原式2721)21(2tantan2cossinsincos22222
=情给分)分,选做两题,过程酌(以上每小题)原式(5513454131414212411tan4tan2tancossincos4cossin2sin3222222分)(,、单减区间为
:,的单增区间为:分)((满足:单调递减时当分)(满足:单调递增时当分))由(12.....().........32663)(10....(..........).........326)2236222)(7.(..........).........63)(226222)()2(4.
(....................1)62sin(2)(62221sin21sin2)0(1.18ZkkkkkxfZkkxkZkkxkxxfZkkxkZkkxkxxfxxff
分)分)时,当时,当分))令(分),的定义域为且的定义域满足:)解(12.....(....................0)(481)(10......
(..........09481298....().........91(991)20(326.(..........20)(20420420)(39)3()2()2()(119maxminmaxmin22
xhxhytyttttytxtxhxxxxhxfxfxhxxx217224882{{0)8(241.20ababaabaabaxbax的两个根为与)由题意可知解(分)即恒成立,在即为的解集为)知由(分),的值域为单调递减在对称轴分)12
.....(..........20024920)721(41721c0721R01)2(6.(].........425[)(4)0()(25)1()(1,0)(013(..........421)(min2222maxmin2
ccxxcxxcxxcbxaxxffxffxfxfxxxxf分)次过滤后达到要求所以至少经过分)又分))(时,即当分)那么有次后杂质的含量为解:根据题意设过滤12..(....
......1311.(..........83.124771.03010.023010.0213lg2lg22lg213lg2lg22lg214lg3lg2lg2143lg40lg40log8......(..........401log)43(log401)43(001.
04304.0%1.04(..........)43(04.0)411(04.0.21434343yxxxxyyyyyy22解(1):当x=y=0时,则f(0)+f(0)=f(0),∴
f(0)=0,f(x)+f(-x)=f(0)=0,即f(-x)=-f(x),∴f(x)在(-1,1)上是奇函数.........(3分).(2)任取-1<x1<x2<0,∵当x∈(-1,0)时,有f(x)>0.∴f(x1)-f(x2)=f(x1)+f(-x2)=f(21211xxxx
)-1<x1<x2<00)1)(1()(1212121xxxxxx0)1(01111012121212121121221xxxxfxxxxxxxxxxxx即f(x1)>f(x2),∴f(x)在(-1,0)上是减函数.......
........(8分)分)(),所以不等式的解集为:),解得同时满足由上单减。在(上的奇函数,所以为(单减,又在()知由(12............2321()21,23(2321()21,23(41111411)(
)41()1,1)()1,1)()0,1)(2)3(222xxxxxxfxfxfxfxf