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2023-2024学年第一学期联盟校第一次学情调研检测高三年级数学参考答案及评分标准1-8BACDBACC9.BCD10.ACD11.ABC12.ACD13.514.1915.35-16.,20,17.【详解】(1)sin110°sin20°cos2155°-sin2155°=
sin70°sin20°cos310°=cos20°sin20°cos50°=12sin40°sin40°=12...............................................................................6分(2)解:
Q、都为锐角,则0,225311sin1cos11414,2143cos177,...........8分1143153393sinsinsincossinc
os14771498...................................................................
.............10分18.【详解】(1)要使函数有意义,则0216014log0143xxx,......3分解得14114224xxxx...........................
........5分故142Axx.....................................6分(2)xA是xB的充分不必要条件,|131RBxmxmm
,,则集合A是集合B的真子集..............................8分则有413211131mmmm,解得312m,.....................
........11分所以实数m的取值范围是312mm............................12分19.【详解】(1)设阴影部分直角三角形的高为ycm,所以阴影部分的面积163360002Sxyxy,所以12000xy,.......2分又60x
,故200y,由图可知20220ADycm,350230ABxcm..................4分海报纸的周长为2220230900cm.......................
........5分故海报纸的周长为900cm...........................................6分(2)由(1)知12000xy,0x,0y,35020360501000326050100049000ABCDSxyxyxyxyxy
,.................................................................9分当且仅当65xy,即100xcm,120ycm时等号成立,此时,350ABcm,1
40ADcm....................................11分故选择矩形的长、宽分别为350cm,140cm的海报纸,可使用纸量最少.20.【详解】(1)∵f(x)≤|f(﹣)|,即当x
=时函数f(x)取到最值,又f(x)=asin2x+cos2x=,其中tanφ=(a≠0),∴[f(﹣)]2=a2+1,代入得[asin2(﹣)+cos2()]2=a2+1,即()2=a2+1,解得(a+)2=0,∴a
=﹣,..............3分f(x)=﹣sin2x+cos2x=﹣2sin(2x﹣),.........................5分(2)由(1)可得:f(x)=﹣2sin(2x﹣),由复数的几何意义知:A(﹣2,﹣4),B(﹣2,f(t)),.
....................7分{#{QQABaY6AogiIAABAAAgCEwHwCgMQkBAAAAoGRAAEMAABQAFABAA=}#}{#{QQABJYQAggiAABAAAAhCAwXgCgCQkAEAAIo
GgAAEMAABwRFABAA=}#}∴S△ABC==|f(t)+4|=﹣2sin(2x﹣)+4,当2x﹣=2kπ﹣,k∈Z,即x=kπ﹣,k∈Z时,S△OAB有最大值6;......9分当2x﹣=2kπ+,k∈Z,即x=kπ+,k∈Z时,S△OAB有最小值2;...
.....11分∴S△OAB∈[2,6]..............................................................12分21.【详解】(1)因为2cos3s
inacbCbC所以sin2sinsincos3sinsinABBCBC,则sin2sinsincos3sinsinBCBBCBC,即sincoscossin2sinsincos3sinsinBCBCBBCBC,所以2sin3sinsincossinBB
CBC,又0,B,则sin0B,所以3sincos2BB,即sin16B,.....................2分由0,B,得5,666B,.................
..........3分所以62B,所以23B;.................................................5分(2)因为2222cosbacacB
,所以229acac,..........................................6分因为D为AC的中点,所以12BDBABC,则222221922444acacac
BDBABCBABC,..........8分因为23sinsinsinabcABC,所以23sinaA,23sin23sin3cos3sin3cCAAA,则
1cos223sin3cos3sin33sin262AacAAAA33sin23cos236sin236AAA,.......................10分因为0,3A
,所以52,666A,所以1sin2,162A,则0,3ac,所以9239,444ac,所以33,22BD.......................................
.............12分22.【详解】(1)(i)当0a时,21ln2fxxx,22ln2f,1fxxx,132222f,......2分故曲线yfx在点22f,处的切线方程为
32ln222yx,即322ln220xy;......3分(ii)21ln2fxxx,0,x,211xfxxxx,令0'xf,解得
0,1x,令0fx,解得1,x,..................5分当1,eex时,max112fxf,又221111ln1e2ee2ef,2211eelnee
122f,其中222211111e1e1e20e2e222eff,故2min1ee12fxf,{#{QQABaY6AogiIAABAAAgCEwHwCgMQkBAAAAoGRAAEMAABQAFA
BAA=}#}{#{QQABJYQAggiAABAAAAhCAwXgCgCQkAEAAIoGgAAEMAABwRFABAA=}#}故fx的单调递增区间为0,1,单调递减区间为1,;fx在区间1,ee上的最大值为12,最小值为21e12
;...............7分(2)21ln22xgxaxxa,对1,x,21ln202axxax恒成立,变形为ln122xaxax对1,x恒成立,....
..................9分令,1,lnxhxxx,则21lnxhxx,当1,ex时,0hx,lnxhxx单调递增,当e,+x时,
0hx,lnxhxx单调递减,其中10h,lne1eeeh,当1x时,ln0xhxx恒成立,......10分故画出lnxhxx的图象如下:其中122yxaa恒过点2,1A,又210111h
,故lnxhxx在1,0处的切线方程为1yx,......11分又2,1A在1yx上,结合图象可得此时1yx在,1,lnxhxxx上方,另外由图象可知当122yxaa
的斜率为0时,满足要求,当122yxaa的斜率小于0时,不合要求,故要想满足ln122xaxax,需要10,12a,解得11,22a,a的取值范围是11,22
...................................................12分{#{QQABaY6AogiIAABAAAgCEwHwCgMQkBAAAAoGRAAEMAABQAFABAA=}#}{#
{QQABJYQAggiAABAAAAhCAwXgCgCQkAEAAIoGgAAEMAABwRFABAA=}#}