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高三数学答案(第1页,共6页)2023年高考适应性练习(一)数学参考答案及评分标准一、选择题BDBCCACD二、选择题9.BC10.ACD11.ACD12.ABD三、填空题13.4π14.如:12π(答案不唯一)15.2π16.1,3,1,6,4,10,2931
9nn−+四、解答题17.解:(1)由余弦定理知2222cosbacacB=+−,·······························1分所以22sin()2cos2acBbacacBac=−−=−+,所以2cos2sin=+BB,················
··························3分又因为1cossin22=+BB,所以25sin4sin0BB−=,在ABC∆中,sin0B>,所以54sin=B.··········································5分
(2)由(1)知accab56222−+=,所以22222615bacacac=−⋅++··········································7分625125acac≥−=,当且仅当ac=时等号成立.所以222bac+的最小值
为52.·······································10分18.解:(1)因为nnSnann−=−2,所以)1()1()1211−−−=−−−−nnSannn((2≥n),两式相减得22)1(1−=−−−−naannannn,·········
·····························2分化简得)2(21≥=−−naann,·········································4分所以数列}{na是以1为首项,2为公差的等差数列,所以122)1(1−=×−+=nn
an.········································6分高三数学答案(第2页,共6页)(2))121121(31)12)(12(21212121212−−−=−−=+
−+−−nnnnnnb,···························9分所以12...nnTbbb=+++)121121...121121121121(311212533−−−++−−−+−−−=+−nn)1211(3112−−=+n所以2111(1)321nnT+=−−.·
············································12分19.解:(1)证明:连接111,,OOCO,则1OO⊥平面ABC.因为1CC为母线,所以11CCOO四点共面,且11//OCOC.取CO中点N,连接1,CNMN.因为1124AB
AB==,则111ONCO==,所以四边形11ONCO为平行四边形.所以11//CNOO,所以1CN⊥平面ABC.所以1CMN∠为1CM与底面所成角,即145CMN∠=.··················································2分在1
RtCNO∆中,11CNNO==,所以12CO=,同理12CC=.在1CCO∆中,22211COCOCC=+,所以11CCCO⊥.·········································4分因为1OO⊥平面ABC,AB⊂平面ABC,所以1OOAB⊥.因为C
为AB的中点,所以ABCO⊥,又1OCOOO=,所以AB⊥平面11COOC,又1CC⊂平面11COOC,所以1CCAB⊥.·············································5分又因为11CCCO⊥,1ABCOO=,所以1
CC⊥平面1BOC;·················6分(2)以O为原点,分别以,1,OCOBOO所在的方向为,,xyz的正方向,建立空间直角坐标系Oxyz−,则(2,0,0)C,(0,0,0)O,(0
,2,0)B,1(1,0,1)C,(1,1,0)M,所以(1,1,0)BM=−,1(1,2,1)BC=−,(1,1,0)OM=,1(1,0,1)OC=.·······
7分设平面1BMC的一个法向量为1111(,,)xyz=n,由11100BMBC==nn,得11111020xyxyz−=−+=,令11x=,得111yz==,所以1(1,1,1)=n.·······························
····9分NzyxCB1O1OABC1A1M高三数学答案(第3页,共6页)设平面1OMC的一个法向量为2222(,,)xyz=n,由22100OMOC==nn,则222200xyxz+=+=.令21x=,得21y=−
,21z=−,所以2(1,1,1)=−−n,······················11分设平面1OMC与平面1BMC夹角为θ,则121111cos|cos,|||333θ−−=<>==×nn.所以平面1OMC与平面1BMC夹角的余弦值为13
.································12分20.解:(1)22()exxxfx−+′=,·································2分令()0fx′>,得02x<<,此时()fx在
(0,2)上为增函数;令()0fx′<,得0x<或2x>,此时()fx在(,0)−∞和(2,)+∞上为减函数;···········4分综上,()fx的单调增区间为(0,2),单调减区间为(,0)−∞和(2,)+∞.····
···5分(2)法一:当1x>时,1ln0x+>,所以2e(ln)xxkx−≤+1.····················7分设2()(1)e(ln)xxgxxx−=>+1,则2222lnln()e(ln)xxxxxxxgxx−−++′=+1,········8分设22()2lnlnh
xxxxxxx=−−++,则()hx′(1)(32ln)xx=−+,···········9分当1x>时,恒有32ln0x+>,()0hx′>,()hx在(1,)+∞单增,所以()(1)0hxh>=恒成立,即()0gx′>,所以()gx在(1,)+∞单增.········10分所以当1x>时,
1()(1)egxg>=−,所以k的取值范围为1(,]e−∞−.········12分法二:由1x>可得(ln)exxkxx−+1≤,即为1lne(1ln)eexxxkx+−+≤;·············8分因为1x>,所以ln1ln10ex
x++>,可得1lnee1lnexxxkx+−≥+恒成立.设()exxgx=,则1()exxgx−′=.当1x>时,()0gx′<,()gx单减.···········9分下证lnxx−>1在(1,)+∞上恒成立.令()ln
hxxx=−,11()10xhxxx−′=−=>,所以()hx在(1,)+∞上单调递增,得()(1)1hxh>=,所以1ln1xx>+>.··································10分所以()(1ln)gx
gx<+,即1ln1lneexxxx++<,所以1lne11lnexxxx+<+,高三数学答案(第4页,共6页)所以e1k−≥,可得1ek≤−,所以k的取值范围为1(,]e−∞−.···················12分21.解:(
1)由题意得列联表如下:一等品非一等品甲7525乙4832··········································1分2χ2180(75324825)4.6211235710080××−×=≈×××,···
········································2分因为0.054.6213.841x>=,依据小概率值0.05α=的独立性检验,可以认为零件是否为一等品与生产线有关联.·····························3分(2)
由已知任取一个甲生产线零件为一等品的概率为23282431004++=,任取一个乙生产线零件为一等品的概率为1517163805++=.······················4分ξ的所有可能取值为0,1,2,3,4.11224(0)4455400
Pξ==×××=,12122213223136(1)()()445554400PCCξ==×××+×××=,2222112232131323117(2)()()()()()45454455400PCCξ==×+×+
×××××=,211222323133162(3)()()455445400PCCξ==×××+×××=,223381(4)()()45400Pξ==×=,·························
·················6分所以ξ的分布列为ξ01234P44003640011740016240081400····················7分4361171628127()01234400400400400400
10Eξ=×+×+×+×+×=.····················8分高三数学答案(第5页,共6页)(3)由已知,每个零件为三等品的频率为4221118020+++=,设余下的50个零件中的三等品个数为X,则1(50,)20XB,·······
···········9分所以15()50202EX=×=.设检验费用与赔偿费用之和为Y.若不对余下的所有零件进行检验,则105120YX=×+,5()50120()501203502EYEX=+×=+×=.··································
···10分若对余下的所有零件进行检验,则检验费用605300×=元.·····················11分因为350300>,所以应对剩下零件进行检验.·······································12分22.解:(1)设221122(,),(
,)PxxQxx,因为2yx′=,所以l斜率12lkx=,·········1分所以直线OQ斜率11OQkx=−,即22221010xxxx−==−−,·····························2分所以22121211211PQxxkxxxxxx−==+=−−,
所以PQ的方程为211111()()yxxxxx−=−−,即111()1yxxx=−+,···········3分所以直线PQ过定点D(0,1);···························································4分(2)(i)12
111111||111|||||22|||2SODODxxxxxx−=+=+=,···················5分l的方程为21112()yxxxx−=−,令0y=,得1(,0)2xM,所以直线AB的方程为111()22xyxx−=−,即1124
xyx=−+,所以直线AB过定点1(0,)4N.························································6分将1124xyx=−+与2214xy+=联立,得2211115(1)04xxxx+−−=.显然0∆>,设3344(,),(,)Ax
yBxy,则134211xxxx+=+,213421154(1)xxxx=−+.所以2343434||=()4xxxxxx−+−42211211222111151615()111xxxxxxx+=+=+++,所以4211342113|||1
|216815TNDxxxxx=⋅=+−+,·····························7分高三数学答案(第6页,共6页)所以421186211123211121516151611(1)||13243124xxxxxxxxTS×==+++++,令21tx=,则
0t>,434233615161516(1)33(122)ttttTStt+++==+,··················8分设436151(()61)ftttt=++,0t>,则2276(5(1)28)()tttftt−−+−′=.令()0ft′=,得1415t+=(负根舍去),当141(0
,)5t+∈时,()0ft′>,()ft单增;当141(,)5t+∈+∞时,()0ft′<,()ft单减;所以当1415t+=时,()ft取得最大值,即2TS取得最大值,此时P的纵坐标为211415Pyx+=
=.··········9分(ii)证明:因为直线AB过定点1(0,)4N,DH平行于l,所以DHNH⊥,所以点H在以DN为直径的圆上.·······································10分设G
为DN中点,则5(0,)8G,且13||||28GHDN==,所以存在定点5(0,)8G,使得||GH为定值.·····································12分获得更多资源请扫码加入享学资源网微信公众号
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