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2021.11高三期中答案一、选择题题号123456789101112答案DCADABCDBCDADACAD二、填空题13.214.8−15.823m16.2+1n，33232nn+−三、17.（本小题满分10分）解：(1)由正弦定理得：sincos3

sinsinsinsinBABAAC+=+·······sinsin()sincoscossinCABABAB=+=+sincos3sinsinsinsinsinsincoscossinBABAACAABAB+=+=++3sinsinsinsinsinsincosB

AACAAB=+=+·············3sin1cosBB=+2sin()16B−=3B=·······································5分（2）由余弦定理2222cosbacacB=

+−得：2232cos3acac=+−2()393acacac=+−=−2ac=········································8分13sin22ABCSacB==·············

··········10分18.（本小题满分12分）解：（1）设0x0x−2()(1)fxx−=−2()(1)fxx=−−又()fx是定义在R上的奇函数，(0)0f=22(1),(,0)()00(1),(0,)xxfxxxx−−−==++

·······································6分（2）()(2)0xxfaefe−+−−()(2)xxfaefe−−−−恒成立()fx是定义在R上的奇函数()(2)xxfaefe−+·······················

······················8分又()fx在R上递增2xxaee−+··················································10分2()2xxaee+令,(

0)xett=22att+恒成立220tt+0a···············································12分19.（本小题满分12分）解(1)53()sin(2)2sin()cos()644fxxxx

=−−−+13cos2sin2(sincos)(sincos)22xxxxxx=++−+sin(2)6x=−····································4分单调增区间为,,63kkkZ−+········

·····················6分(2)70,12x,2,66tx=−−1sin(2),162tx=−−,|sin|yt=的图象如图要使在|sin|ytm=−在,6−有两个零点112m······

·························12分20.（本小题满分12分）解：（1）连接ECBD，交点为O，连接MO//ABCD，BECD=6−121CBED为平行四边形O为DB中点，又点M为棱PB的中点//MOPD······

····································2分PD面MCE，MO面MCE//PD面MCE···········································4分（2）四

边形CBED为平行四边形，四边形ABCD为等腰梯形EDCBPD==又45BAD=PDDE⊥························6分又面PDE⊥平面EBCD，面PDEEBCDDE=，PD面P

DEPD⊥面CBED··························8分所以，可以如图以D为原点，,DPDC为,zy轴，在平面CBED中过D做与面DCP垂直的有向直线为x轴，建立空间直角坐标系.(0,0,2)P，(1,1,0)E，(1,3,0)B，(0,2,0)C(1,3,2

)PB=−，(1,1,2)PE=−，(1,1,0)EDBC==−−设1111(,,)nxyz=为平面PDE的一个法向量则11111200xyzxy+−=+=解得1(1,1,0)n=−同理得：平面PBC的一个法向量2(2,2,2)n=−····

·················10分设平面PBE与平面PBC的夹角为则12122cos||2||||nnnn==平面PDE与平面PBC的夹角为45·····················12分21.（本小题满分12

分）解：（1）123(1)1(1)221nnnnnnanaaan+−−+−=+=+，为奇数，为偶数21+1+1nnba−=2212123(1)1(1)1122nnnnnbaa++−−+−+=+=+22-1+21211=1=22

2(1)2(1)nnnnnaaaab−−=+++=+=+（）1数列1nb+为以2为首项，公比为2的等比数列.·····························6分DPEBC（1）由（1）知12nnb+=21nnb=−2312

22322(123)nnSnn=++++−++++令231222322nnTn=++++234121222322nnTn+=++++2341222222nnnT

n+−=+++++−112(21)2(1)22nnnnn++=−−=−−1(1)22nnTn+=−+···················································10分(1)1232nnn+++++=·················

·····························11分1(1)(1)222nnnnSn++=−−+···········································12分22.（本小题满分12分）解:(1)21ln()2xfxxx+=−+2211ln

ln()44xxfxxxxx−−−=−+=−+·····························2分211ln(1)44xfxx−−=−+=−,(1)1f=−切线方程为43yx=−+···

··································5分（2）不等式()()gxfx在(0,)+恒成立即21lnxxeax+−恒成立令21ln()xxhxex+=−，(0,)x+22222ln2ln()2xxxxexhxexx+

=+=····································6分令2221()2ln,()4(1)0,(0,)xxxxexxxxexx=+=+++()x在区间(0,)+为增函数，且122242()20,(1)20eeeee−

−=−=20(,1)xe−，满足022000()2ln0xxxex=+=，则0(0,),()0,()xxhxhx为减函数0(,),()0,()xxhxhx+为增函数··································9分所以，020min001ln()(

)xxhxhxex+==−又因为022000()2ln0xxxex=+=02000ln20xxxex+=，001ln20000ln12lnxxxxeexx=−=··························10分又

因为xyxe=在(0,)+为增函数所以，0012lnxx=，0201xex=min0()()2hxhx==2a····················································1