【文档说明】山东省青岛市2022届高三上学期11月期中教学质量检测数学试题答案.doc，共(5)页，394.000 KB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-fc64f3fcee1bf8c86507d9a8e6a974aa.html

以下为本文档部分文字说明：

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=+=+sincos3sinsinsinsinsinsincoscossinBABAACAABAB+=+=++

3sinsinsinsinsinsincosBAACAAB=+=+·············3sin1cosBB=+2sin()16B−=3B=·······································5分（2

）由余弦定理2222cosbacacB=+−得：2232cos3acac=+−2()393acacac=+−=−2ac=········································8分13sin22ABC

SacB==·······················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MC

E···········································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PD

E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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=−231222322(123)nnSnn=++++−++++

令231222322nnTn=++++234121222322nnTn+=++++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+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