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2021年哈三中高三学年第四次模拟考试理科数学答案一、选择题123456789101112BCBADCCADCBD二、填空题13.214.5415.2021216.21010三、解答题17.(1)//PO平面
CEFPO平面APB且平面APB平面EFCEFEFPO//13AEAP13AFAO2OF...........................................
..................6分.(2)平面PAB平面ABC且平面PAB平面ABABCABPO且PO平面PABPO平面ABC以O为坐标原点,OAOCOP、、分别为x轴、
y轴、z轴正方向建系0,0,2F0,3,0C0,0,3B3,0,0P2,3,0CF3,3,0BC0,3,3CP设平面CBP的一个法向量为,,nxyz则00nBCnCP
解得11,1n,设直线CF与平面CBP所成角为23539sincos,39133CFn.........................................12分PABCOEF18.(1)177x67
y.........................................4分.55.13617715.16715.1xya..................................
........6分.(2)721407iiyy.....................................8分.252.3610.870.9407R该线性回归方程的拟合效
果是良好的..........................12分.19.(1)nnnnaaaa33112nnbb31且312121aabnb是以3为首项3为公比的等比数列;nnb3.................................
.......6分(2)2)13(331331nnnS;13113121)13)(13(349111nnnnnnnnSSb41)13121(21)131131131131131131(2111322
nnnnT...........12分20.(1)12a,12xfx,1ln12gxxx1ln22xFxfxgxxx22211110222xFxxxx
Fx在0+,单调递增,且1=0F,综上=1,=0,=xFxfxgx01,0,xFxfxgx,1+,0,xFxfxgx,.....
.....................................4分(2)0mn,1mn,要证lnlnmnmnmn,即证lnlnmnmnmn,即证lnmnmnmn,设mtn,且1t,则即证21ln2lntttt
,即证1ln0122tttt由(1)知,1+,0xFx,成立,所以不等式成立,证毕...........................................8分(2)11ln2aaxxax在1+,上恒成立,则1
12ln1axxxx在1+,上恒成立,①1x时,aR使得上式成立,②1+x,时,120xx则1ln112xxaxx在1+,上恒成立,设1ln1112xxhxxx
x1+x,时,有1111ln1112222xxxxxxx,则11ln1112211222xxxxhxxxxx,所以12a.........................................12
分21.(1)1148PFQFPQa,=2a,123FPFSbc解得①3,1bc时,椭圆的标准方程为22143xy,②1,3bc时,椭圆的标准方程为2214xy...................................
.......4分(2)设,0,,0Aaba,1122,,,PxyQxy由题意知直线斜率不为0,设:lxmyc,22221xmycxyab,整理得22222222220bmaymcbybcba,212222222122222
mcbyybmabcayybma(*).........................................6分121212,yykkxaxa,由题知21=kk,则有
21211221212121===yxaymycamyycayyxaymycamyycay,将(*)代入整理得22222121222122222212112222+2=mcbmbcacamcbmyycaycayb
mabmamyycaymbcacaybma222222221122222222222211222222+2+2=mbcacamccacamb
mccacaycaybmabmambcambcacaycaybmabma2222212221222222112221===1camaccaymaccayacacebmaacacemaccamaccay
caybma1213,411e,13,25e.....................................12分22.(1):310lxy;2:4Cyx................
........................5分(2)直线'l的一个参数方程为12232xtyt(t为参数)代入到2:4Cyx中得232804tt设DE、对应的参数分别为1122(0)(0)tttt、则1283tt,121283PDPEtttt
........................................10分23.(1)322224224()232522(2)22(2)xxxxfxxx当
且仅当222422(2)xxx即4x时=“”成立........................................5分(2)由题意可知()fx值域为()gx值域的子集且()5,fx则0a()gx
在1,单调递增()(1)22gxga即225a解得72a........................................10分