【文档说明】湖北省荆州中学等四校2022届高三模拟联考(四)数学试题参考答案.pdf,共(3)页,435.969 KB,由envi的店铺上传
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2022届高三模拟联考(四)数学参考答案选择题:题号123456789101112答案BDACBACDABCABBCDBCD填空题:13.(,2]14.1m15.23(3)aba16.252055317.解:因为222222
cosabcbcbcbcA,所以1cos2A,.................................................2分又0180A,所以60A,..........................................
............................................................3分又因60B,所以ABC为等边三角形,故ABBCAC,由5BD,可得55CDBCAB,故133
sin53242ADCSACCDACDABAB,解得2AB或3;...................................................................
..........................................................5分(2)解:由(1)得:当2AB时,3CD,则2222sinADACCDACCDACD
149232192,所以19AD,设ACD△外接圆的半径为R,由正弦定理可得2192sin3ADRACD,所以193R,所以ACD△外接圆的面积为2193R,.........................
..............................................................8分当2AB时,3CD,则2222sinADACCDACCDACD194232192
,所以19AD,同理ACD△外接圆的面积为193,综上所述,ACD△外接圆的面积为193....................................................
...................................10分18.(1)证明:因为2AC,132CC,14AC,所以22211ACCCAC,即1ACCC.又因为1BCBB,11//BBCC,所以1B
CCC,ACBCC,所以1CC⊥平面ABC.因为1CC平面11BBCC,所以平面ABC平面11BBCC........................................................5分(2)解:连接AM,因
为2ABAC,M是BC的中点,所以AMBC.由(1)知,平面ABC平面11BBCC,所以AM平面11BBCC.以M为原点建立如图所示的空间直角坐标系Mxyz,...........................
................................6分则平面11BBCC的一个法向量是(0,0,1)m,(0,0,3)A,(0,2)3,0N,1(1,20)3,C.设1(01)APtACt,(,,)Pxyz,(,,3)APxyz,1
3(1,2,3)AC,代入上式得xt,23yt,3(1)zt,所以(,233)3,Pttt.设平面MNP的一个法向量为111,,nxyz,(0,,0)32MN,(,23,33)MPttt,由00nMNnMP,得
111120233()031ytxtytz.令1zt,得(33,0,)ntt......................................................
...........................................................10分因为二面角PMNC的平面角的大小为30,所以32||||mnmn,即22323(1)
ttt,解得34t.....................................................................11分即点P的坐标为3323,,424P
所以点P为线段1AC上靠近1C点的四等分点.....................................................................................12分19.解(1)当11a时,不论2a取7还
是10都不能与12或8构成等比数列,不合题意,..............................2分当12a时,当且仅当234,8aa时符合题意,当116a时,不论2a取7还是4都不能与5或12构成等比数列,不合题意,...................
..........4分∴2q,∴12(2)nna....................................................................................................................
..6分(2)21(2)22(2)1(2)33nnnS,∴1122(2)33nnS,2122224(2)(2)3333nnnS,∵1124244(2)(2)3
333nnnnSS222(2)33n2nS,∴12,,nnnSSS或21,,nnnSSS成等差数列,∴数列nS中的任意连续三项按适当顺序排列后可以成等差数列...........
...............................................12分20.解:(1)(i)因为25010025,所以21000,10YN,因为220.9545P,所以10.9545
20.022752P,因为9801000210,所以98020.02275PYPY;.........................................
.....................................................4分(ii)由第一问知98020.02275PYPY,庞加莱计算25个面包质量的平均值为978.72g,978.72980,而0.022750.05
,为小概率事件,小概率事件基本不会发生,这就是庞加莱举报该面包师的理由................................................................................................
..............................................6分(2)设取出黑色面包个数为随机变量,则的可能取值为0,1,2,.........................
......................................7分则143254390365387105p;124235337122365387630p,121232592365387630p,故分布列为:
其中数学期望393375991630012105616023E.........................................................................
...........12分012p391053376305963021.解.(1)根据椭圆的定义可得:482a,解得:22a将2,2代入方程22281xyb,得21212b解得:24b椭圆C的方程为:22841xy.....................
.......................................................................................................4分(2)由题知,002,0xy,设1122,,,Mxy
Nxy,则直线PM的方程为00(2)2yyxx由0022(2),21,84yyxxxy得2002200242240xxyyyy200122200020442222yyy
xxyy2012200422yyxy220022000042443yyxyxx..........................8分0103yyx,同理可得0203yyx...........................
........................................................................................9分21002112OPNOFNPFyyySMFSyy△
△00121yyyy000000()133yyyyxx003317xx................11分211OPNOFNPFSMFS为定值7...............................................
.....................................................................................12分22.(1)解:2(1)()(0),(1),(1)xexaf
xxfebfaxx,所以yfx在1x处的切线方程为()(1)yebaxyaxeba即,比较系数可得2,1ab.........................
...................................................................................................5分(2)=1,=-1>0,0,0,0,
xxxexxexxxx令则,令得令得则0x是()x的极小值点同时也是最小值点,()(0)=01(0)xxexx所以,所以当且仅当取等号.................................
..................7分ln()()()ln1(ln)10xxxehxfxgxhxxxexxx令,即当且仅当ln0=xx“”取“”................................
..................................................................................9分所以()(),fxgx,则有()(),
fmgm而()(),()()fmgngmgn..................................................11分又11,()gxgxx单调递增,所以mn.....................................
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