江苏省苏州市2019-2020学年高一上学期期末数学试题(解析版)

2019一、单项选择题：本题共8小题，每小题5分，共40分，在每小题给出的四个选项中，只有一项是符合题目要求的.CAU=A=则（）UB.CAU=,A=.U.1f(x)）4xA.1f(x)(,4)4x0x4,4x.3,bc3c（a）3c<a<bb<a<c<<a<<aDcc.b100c(0.8)31a331,bcaD.,,033..cos（）23344-5555D.345,2245cos()sin,2.3,x01fxxff（）logx,x022113333B.1113ff(,2221ff1f(1).23,.在3tanB1tanAtanBC）3tanAAtanCAB).1tanAtanB3(tanAtan),AB3CAB),1AB3C..3DEABADAB2DCE（）515152525A,E,CAEAC由设,用,,DEABAD.,E,CAEAC,设DEDAAEACAD1(ADDC)AD(ADAB)AD21(1)ADAB23DEABAD,512251,.3551.yfxxkx)=(xk)000yfxaf(x)对k若对2a21x（）32325252(21)(21)2121D)52a50xx与aa令t20得2xx0000.ta50a2a(x)f对2,21xf(2x)ff(x)x0,00aaa,2141212xx00)52a50a0a2(xx0020ta50令t,,2x04(0aa5552tt05a(21)(2.a212aa5att120.二、多项选择题：本题共4小题，每小题5分，共20分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，部分选对的得3分，有选错的得0分.A=x|B2},若Ba）12BAax2a.由2B2,2A,2a2a1.2a2.（）1y2yexy=x||yxx311y((0,)当x(0,),y故xx;2x(0,)2R,当2,;yxxyx333yx(0,);yexRx(0,)yex.,当,.e(eaee，0a（）12112212,,.用a121212aee)11221212211525,0a(1,0)若若则12,;20121212021,012a(0,1)则12,;21551212,(1,0)(1,0)(1,0)..1,)=>，且在区间,ω,φ326为（）6ω=2φ=ω=φ=ω=φ=ω=φ=6261,,逐项判断是否过点32(,)126.1,6f()sin()sin136f(x)sin(2x),;323211,f(x)sin(2x)f()sin()sin,,233262322x2k,2k(kZ)xk,k(kZ)令,2222(,)126f(x)sin(2x);211f(x)sin(6x)f())sin,,,6366232116x2k,2k(kZ)xk,k(kZ)令,62293183116x2k,2k(kZ)xk,k(kZ)令,622183183f(x)sin(6x)(,)126;6511f(x)sin(6x)f())sin,,,6366232116x2k,2k(kZ)xk,k(kZ)令,62293183k1,x,当9185(,)126f(x)sin(6x).6.三、填空题：本题共4小题，每小题5分，共20分.C.设C(x,y)AC,CBC.C(x,y)(xy(8x,3y)则,22(8x)xxy1.32(3y)y.(x)x2fxn∈n=.2.f247f3810,f(2)f(3)0,(x)x22.fxn.5=α=.394,sin,cos,3.55:12sincos,934sin20,2)(,),,923sin0,cos0,1,55:,669.49:4.yfxfxxxxbx=2b=22.9121054f(2x)f(2x).(x)求xab,=2f(2x)f(2x)fx当x当x1,f(3)(1)(93)0abf得①2,f(4)f(0)(164ab)0得②a7,bab5:f(x)(x2x)(x27x12)x48x319x212x,()4fx,x3x2x44()0(2x8x0x或x,fx令,x224444f(x)(,,)(,)2222,449f(x)f()f()min22491210:4.四、解答题：本大题共6小题，共70分，解答时应写出文字说明、证明过程或演算步骤.A=x|−+<B=x|0<x<若a=求；若B=a.=x0xa(,2][4,)BAaa对当a=3A由B=A知a.Ax(x3)(x1)0x1x3若a=则,=x0x;BAB=,a2aa1A{xax2},即,a1a0a2.2a4a2aa2aA即a1,即a1,;A{x2ax},a12a0a4a4,,.a(,2][4,).1317已知锐角,,cos14cos.求α+β)求α.23cos()983cos()sin)(1)sin,,,.33sin11(),221437sin1cos1()22713343);7713133431)coscossinsin1471472,coscos则,3.A=，DE.求求ADBC.EBEC56EBECADBC.4AD,(1)以,AC,以,AC,.14ABACABACcos60424,21(),BCACAB21ADBC(ABAC)(ACAB)2121[AC2AB2](164)6;21EA21131()(AC)22441EA21113()(AC)22443113()444435354228..tHHtH+B>；5hh.Ht)40cost50(0t30)h40米15Ht)At)B设A与303655ht.Ht)At)B(A0,0,B【详解】(1),,AB90AB得.AB10,,3015Ht)40sin(t()500t30150Ht)1040sin(0)50sin1即,又t,,152t)40sin(t)5040cost50(0t30)H152151Ht)40cost5030,costt5,151525;30365,5若甲在摩天轮上坐了t(5tt5:h40cost50[40cos(t5)50]151513cost]21540[costcost5)]40[cost151521540cos(t)1531040当t即t,.h153.3113()444435354228..tHHtH+B>；5hh.Ht)40cost50(0t30)h40米15Ht)At)B设A与303655ht.Ht)At)B(A0,0,B【详解】(1),,AB90AB得.AB10,,3015Ht)40sin(t()500t30150Ht)1040sin(0)50sin1即,又t,,152t)40sin(t)504