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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,

10，29319nn−+四、解答题17.解：（1）由余弦定理知2222cosbacacB=+−,·······························1分所以22sin()2cos2acBbac

acBac=−−=−+，所以2cos2sin=+BB，··········································3分又因为1cossin22=+BB，所以25sin4sin0BB−=，在ABC∆中，sin0B>,所以54

sin=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nn

Snann−=−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−=×−+=nnan.·····································

···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(31121

2533−−−++−−−+−−−=+−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1124ABAB==，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1RtCNO∆中，11CNNO==，所以12CO=，同理1

2CC=.在1CCO∆中，22211COCOCC=+，所以11CCCO⊥.·········································4分因为1OO⊥平面ABC,AB⊂平面ABC，所以1OO

AB⊥.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1CC⊥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N

zyxCB1O1OABC1A1M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()2lnlnhxxxxxxx=−−++

，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ln1ln10exx++>，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()lnhxxx=−，11()10xhxxx−′=−=>，所以()hx在(1,)+∞上单调递增，得()(1)1hxh>=，所以1

ln1xx>+>.··································10分所以()(1ln)gxgx<+，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)4455400Pξ==×××=，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436117

1628127()0123440040040040040010Eξ=×+×+×+×+×=.····················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）12111111||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1124xyx=−+，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

(,),(,)AxyBxy，则134211xxxx+=+，213421154(1)xxxx=−+.所以2343434||=()4xxxxxx−+−42211211222111151615()111xxxxxxx+=+=+++，所以42

11342113|||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4361

51(()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.································

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