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2022届石家庄市质检一数学答案一、单选题1.C2.B3.D4.D5.A6.A7.D8.B二、多选题9.BC10.BD11.BD12.ACD三、填空题13.414.12−15.3416.40,2e四、解答题:(其他答案请参照本标准,教研组商定执行)17.解:(1)()()()
sin0=2fxx=+为偶函数..........2分()2cos=cos=sin22−−=......................3分2cos2=12sin11=0−=−...........................
5分(2)()sin++sin=sin+cos2gxxxxx==2sin+4x...........................7分2+2242kxk−+由得32244kxk−+................
........9分322,44kkkZ−+所求增区间为,.....................10分18.解:(1)5321=,8aqa=1,2q\=..................................2分211aaq\==111
12nnnaaq--骣÷ç\==?ç÷ç÷桫...........................4分()1111211S211212nnnnaqq-骣骣÷ç÷÷çç-÷÷çç÷÷琪ç÷桫-÷ç骣桫÷ç\===-?ç÷ç÷
桫--..................6分(2)答案不唯一(其它答案请酌情给分)由(1)可知S2n<,所以{}nb不妨取等比数列................8分可取12112=22nnnb--骣骣鼢珑=鼢珑鼢珑鼢桫桫.......................10分此时212121T
=41212nnn-骣骣÷ç÷÷çç?÷÷çç÷÷琪ç÷桫÷ç骣桫÷ç=-?ç÷ç÷桫-4<212nnb-骣÷ç\=÷ç÷ç÷桫符合题意........................12分19解析:(1)证明:
如图(1),设PD中点为F,连接CF,EF,因为,EF分别为PA,PD的中点,所以EFAD且12EFAD=,又因为BCAD,12BCAD=,所以EFBC....................................................
.....................................................(2分)又因为EF=BC,所以四边形BCFE为平行四边形,所以BECF.因为CF平面PCD,BE平面PCD,所以BE平
面PCD..........................................................................................(4分)(2)方法一:如图(2)取AD中点O,连接OP,
OB,因为PAD是以AD为斜边的等腰直角三角形,所以POAD⊥,因为平面PAD⊥平面ABCD,且PO平面PAD,所以PO⊥平面ABCD,且OBOD⊥.............................................
.....................(5分)则以O为坐标原点,分别以OB,OD,OP方向为x轴,y轴,z轴的正方向,建立如图所示空间直角坐标系Oxyz−,取1OB=,则则(0,0,1)P,(1,0,0)B,(0,1,0)D,图(1)图(2)(0,1,0)A−,11(0
,,)22E−(1,1,0)BD=−,(1,0,1)BP=−,11(1,,)22BE=−−,.........................................(6分)设平面PBD的法向量为(,,)xyz=m,则00B
DBP==mm,即00xyxz−+=−+=,可取1x=得:(1,1,1)=m............................................................
.(8分)设平面EBD的法向量为(,,)xyz=n,则00BDBE==nn,即011022xyxyz−+=−−+=,可取1x=得:(1,1,3)=n,.........................................
............................................(10分)则55cos,||||31133===mnmnmn,则二面角PBDE−−的余弦值为53333....
........................................................(12分)方法二:如图(3),不妨令2AD=,连接OC交BD于点F,连接PF,取AD中点O,连接OP,交DE于点H,连接FH.因为PAD是以
AD为斜边的等腰直角三角形,所以POAD⊥,因为平面PAD⊥平面ABCD,且PO平面PAD,所以PO⊥平面ABCD,所以PO⊥BD,且BDOC⊥,且OCOPO=,所以BD⊥平面POC,.........................................(6分)所以BDPF⊥,
BDFH⊥,PF平面PBD,FH平面EBD,则PFH为二面角PBDE−−的平面角.........................(8分)在PAD中,因为E是PA中点,O是AD中点,易知13OH=,则23PH=,在POF中,226HF=,62PF=,....
................(10分)则cosPFH=53333...................................................................................................(12分)法三:如图(4)
,不妨令2AD=,连接OC交BD于点F,连接PF,因为PAD是以AD为斜边的等腰直角三角形,所以POAD⊥,因为平面PAD⊥平面ABCD,且PO平面PAD,所以PO⊥平面ABCD,所以POBD⊥,因为ODCB为正方形,所以OFBD⊥,所以BD⊥平面PO
C所以PFBD⊥,所以PFO是二面角PBDA−−的平面角,22OF=,1OP=,所以tan2PFO=,.........................................(6分)取AO中点G,连接EG,取BF中点H,连接GH,EH,易知EG⊥面
ABCD,GHOF,所以GHBF⊥,所以EHBF⊥,所以EHG是二面角EBDA−−的平面角,12EG=,324GH=,所以2tan3EHG=,.............(8分)所以设二面角PBDE−−的平面角为,则2
2tantan()5PFOEHG=−=,.........................................(10分)则二面角PBDE−−的余弦值为53333........................................
..(12分)20解:(1)设甲乙比赛甲胜、乙丙比赛乙胜、丙甲比赛丙胜分别为事件,,ABC,且,,ABC相互独立,设“比赛完3局时,甲、乙、丙各胜1局”为事件M,)()()(CBAPBCAPMP+=.....................(2分)322131213132)((
)()()((+=+=CPBPAPBPCPAP))................(4分)92=......................................(5分)图(4)(2)X的可能取值为1,2...........
....................(6分)(1)()()PXPACBPABC==+()()()()()()PAPCPBPAPBPC=+21111113323236=+=...................
......................(8分)5(2)1(1)6PXPX==−==.........................................(10分)则X的分布列为X12P1656则1511()12666EX=+=......................
...................(12分)21解:(1)由2e2ca==,故2ab=,...........................(2分)椭圆方程可写为222212xybb+=,代入点2(1,)2P,解得1b=,所以椭圆C的标准方程为2212xy+=.
......................................................................(4分)(2)设点11(,)Axy,22(,)Bxy,直线l的方程为1xmy=+
与椭圆2212xy+=联立,消去x整理得()222210mymy++−=显然0成立,故12222myym+=−+,12212yym=−+........................................(6分)由椭圆定义得△1ABF的周长为442a=则△1ABF的面积136422
42S==..................................................................(7分)又由1221121||||2SFFyyyy=−=−,得2162yy−=...............................
..........(9分)从而得()22112342yyyy+−=,即222243222mmm−+=++整理得423440mm−−=,解得22m=,故2m=...................
...(11分)故直线l的方程为210xy−=............................................................................(12分)22
解:(Ⅰ)2()((1))xfxxaxae=−−−()(1)xexax=−+...................................(2分)①当1a=−时,2()(1)0xfxex=+,故()fx在R上单调递增;...............
.......(3分)②当1a−时,()fx在(,)a−递增,在(,1)a−递减,在(1,)−+递增;③当1a−时,()fx在(,1)−−递增,在(1,)a−递减,在(,)a+递增.(5分
)(Ⅱ)当1a=−时,()fx在R上递增,故当12xx时,12()()fxfx原不等式即122212()()()xxfxfxmee−−..............................................
....................(6分)从而有122212()()xxfxmefxme−−令2()()xgxfxme=−,问题等价于()gx在(0,)+上单调递减.............................(8分)2()()2xgxfx
me=−22(1)2xxxeme=+−令()0gx,则等价于2(1)2xxme+在(0,)+上恒成立......................................(10分)令2(1)()xxhxe+=,则21()
xxhxe−=,故()hx在(0,1)递增,在(1,)+递减从而只需42(1)mhe=,故m的取值范围为2[,)e+...........................................(12分)