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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：sincos3sinsinsinsinBABAAC+=+·······sinsin()sincoscossinCABABAB=+=+sincos3sinsinsinsinsinsincosco

ssinBABAACAABAB+=+=++3sinsinsinsinsinsincosBAACAAB=+=+·············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···············································1

2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1

2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PDEPD⊥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=222(1)2(1)nnnnnaaaab−−=+++=+=+（）1数列1

nb+为以2为首项，公比为2的等比数列.·····························6分DPEBC（1）由（1）知12nnb+=21nnb=−231222322(123)nnSnn=++++−++++令231222322nnTn=++++234121

222322nnTn+=++++2341222222nnnTn+−=+++++−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+=−+2211lnln()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+222

22ln2ln()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()2ln0xx

xex=+=02000ln20xxxex+=，001ln20000ln12lnxxxxeexx=−=··························10分又因为xyxe=在(0,)+为增函数所以，0012lnxx=，0201xex=

min0()()2hxhx==2a····················································1