2025年湖北省八市高三(3月)聯(lián)考數(shù)學(xué)參考答案與評(píng)分細(xì)則題號(hào)12345678答案DDBCCAAD題號(hào)910答案ADABDBCD12.1513.2π3(120)14.(1)58(2)724部分試題解析7.對(duì)于殘差圖(1)E(e)=0σ2隨著x的變化而D(e)=σ2(2)差滿足E(e)=0σ2不隨xD(e)=σ2的假設(shè).故選A8.當(dāng)a<0時(shí)，x<0f(a)=aea<a2<a2-a=g(a).當(dāng)a>0時(shí)，x>0f(x)≤g(x)即axex+a≤x2-x?axex+x+a≤x2xx?axex+x+a+x2≤x2+x2?e+ln(+x+ln(ax)≤x2+x2)x≤x2axex≤x2?a≤x

∵y=x+x為增函數(shù)，∴e+()exa≤(x=1以0<a≤1)exee綜上，a的取值范圍為(-∞,0)∪(0,1]選De10.由題知：|AF|=|1|a=2c?e=1|=|1|a=2c?e=12由題知：|AP|=|2P||AQ|=|Q|，∴ΔAPQ與ΔF2PQ的周長(zhǎng)相等A正確；又|QF|+|QF|=2a|PF|+|PF|=2a,故Δ2PQ的周長(zhǎng)為4aΔAPQ的周長(zhǎng)為4a,故B正確；21a4b∵=∴=e知M,0，23b23設(shè)Vxy(,)B|MV|2=x-b2=-1(x+b)2+16b2+(y-0)2(x∈[-b,b])339當(dāng)x=-b時(shí)，|MV|2∴|MV||MV|≤|MB|，=16b=4b293∴以點(diǎn)M為圓心，|MB|為半徑的圓與橢圓CC錯(cuò)；·1·設(shè)直線AQ的斜率為k3：kk3=-a2；b2∵直線AP的方程為：y=k(x-b)AQ的方程為：y=k(x-b)∴點(diǎn)PQ的坐標(biāo)滿足方程：[y-k(x-b)][y-k(x-b)]=0即y2+kk(x-b)2-(k1+k)(x-b)y=0y2+x=1?y2=-a22又點(diǎn)PQ橢圓上，∴點(diǎn)PQ的坐標(biāo)滿足(x2-b)a2b2b2代入上式可得：-a2(x2-b)+kk(x-b)2-(k1+k)(x-b)y=0b2∵x≠b∴-a2(x+b)+kk(x-b)-(k1+k)y=0即為直線PQ的方程，b2將M(bk3=-2a

3b2,0)代入得：k=2kk3k3≠0以k1=2k.2故DABD連接AC交BD于O接A'O,則OB=OD=6,，OC=OA=OA'=8,S=1=12×12×8=48當(dāng)A'O⊥平面BCDA'-BCD的體積最大，此時(shí)=1×48×8=128A錯(cuò)；3A'AC⊥平面ABCD，平面A'AC∩平面ABCD=AC，過A'作A'N⊥AC于NA'N⊥平面ABCD，當(dāng)DN⊥BC時(shí)，BC⊥平面A'DN時(shí)A'D⊥BC，故B正確；當(dāng)A'C=8時(shí)，ΔA'OCA'OC⊥面A'BDC作CH⊥A'O則CH=43，CM2=CH2+HM2≤49?HM≤1，即點(diǎn)M在以H為圓心，1πC正確；由前面分析知：CM與平面A'BD1，7所以CM與BD所成角α最小時(shí)，cosα=1(CM的射影與BD平行時(shí))7故D正確.BCD14.記Ai表示第i(1)P(A)=P(A)P(A|A)+P(A)PA|A1)=58×710+38×510=58(2)P(A2A|A1)=510×712=724P(A2A|A)=P(A2A3A|A)+P(AA|A12A3P(A2A|A)=P(A2A3A|A)+P(AA|A1)=510×712×914+510×512×714=724·2·P(A2A|A)=P(A2A3A4A|A)+P(AA|A)+P(AA4A|A)+P(AA|A)2A3A42A32A3A4=5×71012=724×914×16+510×712×514×916+510×512×714×916+510×512×714×716......P(A2A|A1)=724(n>2)事實(shí)上,?n∈N*P(A)=5)=58；②?n>2P(A2A|A1)=724(n∈N)；③?n>m≥2PAmA|A1)=72415.解：(1)?=2(b+c)?3asinC+acosC=b+c?3sinAsinC+sinAcosC=sinB+sinC?3sinAsinC+sinAcosC=sinAcosC+cosAsinC+sinC?3sinAsinC=sinC(cosA+1)∵C∈(0,π),sinC≠0∴3sinA-cosA=1即2sinA-π=1?????????4分6又A∈(0,π),A-π∈-π,5π=πA=ππA-??????????5分666663(2)BM=2MC?AM-AB=2(AC-AM)∴AM=1+2ABAC33|AM?AC????????????9分|2=1+22=12+42+4ABABACABAC33999又AB=c=23AC=bAM=2即|AM|=2∴4=12+4b2+4b?23cosπ

239993∴b2+3b-6=0∴b=3或b=-23(舍)故S=1bcsinA=33???????????????????????13分2216．解：(1)∵f(x)=lnx-mx2∴f'(x)=1-2mxx當(dāng)m=0時(shí)，f(x)=x,顯然x=0不是f(x)當(dāng)m≠0f(1)=-1m,f'(1)=-1m,?-m=-1m,1-2m=-1m?m2=1,,解得m=1.2m2-m-1=0???????????????????????????????????6分·3·(2)證明：g(x)=lnx+1-x-b=lnx-x+1+baaag'(x)=1-1=a-x?????????????????????????8分xaax因?yàn)閍>0以g(x)在(0,a)(a,+∞)上單調(diào)遞減，所以g(x)??????????????????????10分=g(a)=lna+bag(x)≤0g(x)≤0?b≤-alna，≤0lna+ba所以ab≤-alna????????????????????????????12分設(shè)h(x)=-xlnxx>0h'(x)=-2xlnx-x2?1x-1-1h'(x)>0?0<x<e2h'(x)<0?x>e2，=-x(2lnx+1)-1-1∴h(x)在0,ee2,+∞上單調(diào)遞減，2-1=1∴h(x)=he22e以-alna≤12e,所以ab≤12e.??????????15分17.解：(1)連接A1B交AB1于點(diǎn)FF為A1BD1F，因?yàn)槠矫鍭B1D1?平面BDC1，平面AB1D1∩平面BAC1=D1F面BDC1∩平面BAC1=C1B,以D1F?BC1所以D1為A1C1λ=12因?yàn)槠矫鍭B1D1?平面BDC1，????????????????????3分平面AB1D1∩平面ACC1A1=AD1面BDC1∩平面.ACC1A1=C1D，所以AD1?DC1AD?DC1ADC1D1為平行四邊形，所以AD=DC1=1AC以μ=122故λ=μ=1??????????????????????????????6分2(2)(i)∵AC⊥平面AB1D1B1D1?平面AB1D1∴AC⊥B1D1又AA1⊥平面A1BC1B1D1?平面A1BC1∴AA1⊥B1D1又AA1∩AC=A1AA1?平面AC1AC?平面ACC1A1，∴B1D1⊥平面ACC1A1∵D,D1分別為AC,AC1DD1?BB1DD1=BB,所以四邊形BB1D1D為平行四邊形，∴B1D1?BD∴BD⊥平面ACC1A1又AC?平面ACC1A1∴BD⊥AC????????????????????10分(ii)因?yàn)锳C⊥平面AB1D1AD1?平面AB1D1以AC⊥AD1ΔAA1D1?ΔCAA1，·4·∴AA1CA=A1D1AA1?AC=2AA1AB=2AA1以AC=AB由(i)BD⊥AC且D為AC中點(diǎn)，∴AB=BC∴AB=BC=AC????????12分設(shè)AB=2aAA1=2a，DDA為x軸，DB為y軸，DD1為z軸建立空間直角坐標(biāo)系，則A(a,0,2a),B(0,3a,0),C(-a,0,0),Ea,0,22a(-a,0,2a)，,C33∴BE,BC=(-a,-3a,2a),A=(-2a,0,-2a),???13分=a,-3a,22aC331=(x,y,z)設(shè)平面EBC1的法向量n?=0a22a3x-3ay+3z=0??BC=0-ax-3ay+2az=01不妨令x=2得z=4,y=6=(2,6,4)

又向量AC為平面BC1DAC

=(-2a,0,-2a)?????????14分|?AC|設(shè)二面角E-BC1-D的平面角為θcosθ==62a=2||?|AC|2+6+16?6a2???????????????????????????????????15分方法二：連接C1E并延長(zhǎng)交A1A于點(diǎn)HH為A1AE-BC1-D即二面角

H-BC1-D.因?yàn)锳C⊥平面AB1D1面AB1D1?平面BDC1∴AC⊥平面BDC1，因?yàn)镠D∥AC以HD⊥平面BDC1設(shè)AB=2aAA1=2aDC1=a2+(2a)2=3aDB=(2a)2-a2=3a=DC1∴ΔDC1BC1B的中點(diǎn)S接DS，則DS⊥BC1連接HS∠DSH即為二面角H-BC1-D的平面角.∵ΔBDC1DS=11DS=12C1B=6a2.又DH=12AC=6a2以DH=DS所以tan∠DSH=DHDS=1cos∠DSH=22，即二面角E-BC1-D的余弦值為21-D的余弦值為22.·5·18.解：(1)設(shè)M(x,y)(x-2)2+y2|x-1|=2???????????????2分?(x-2)2+y2=2(x-1)2?x2-y2=2所以曲線C的方程為：x2-y2=2?????????????????????4分(2)(i)由(1)知|PF|=(x1-2)2+y2=2(x1-1)??????????????6分1因?yàn)榍€C12P|PF|-|PF|=22?|PF|=22+2(x1-1)=2(x1+1)

|PF|x1+1∴=x1-1????????????????????????????9分

|PF|(ii)設(shè)點(diǎn)Q(x,y)，∵PF=λF即(2-x,-y)=λ(x2-2,y)Q2∴2-x1=λ(x2-2)-y1=λy2x1+λx2=2λ+2y1+λy2=0∵x1-y1=2,x2-y2=2∴x1-y1-λ(x2-y)=2-2λ2∴(x2-λ2x)-(y2-λ2y)=2-2λ2∴(x1-λx)(x1+λx)-(y1-λy)(y1+λy)=2-2λ2將(1)代入上式得x1-λx2=1-λx1+λx2=2λ+2聯(lián)立解得x1=λ+3???????????????????????????13分2PF15+λx1+1|PQ|=1+λ由(i)知x1-1==1+λ|PF|λPF2|PQ|?|PF||PQ|?|PF|S2=5+λ由題意PN=PF∴==????????15分2S1λ|PN|?|PF||PF|?|PF|S2+λS1+λ=5+λS2∴=+λ+1≥25+1(等號(hào)成立僅當(dāng)λ=5)+λ=5S1S1λλS2+λS1所以的最小值為25+1.????????????????????17分S1方法二:當(dāng)直線PQPQ的方程為：y=k(x-2)聯(lián)立x2-y2=2?(1-k)x2+4k2x-4k2-2=0y=k(x-2)2+2∴x(k≠±1)1+x2=4kx1x2=4k2k2-1k2-13x1-4∴-3(x1+x)+x1x2=-2?x2=22x1-3又x1+λx2=2λ+2x1=λ+32當(dāng)直線PQ.以下同解法一.19.解：(1)當(dāng)a=2,b=6時(shí)，a×33=6601=4,am=36,m=33,Sm=4+362A={4,9,16,25,36}∴T=4+9+16+25+36=90?????????????3分故T=90=3

Sm66022??????????????????????????4分am-a1b2-a2(2)因?yàn)楣頳=m-1=1m-1=1∴m=b2-a2+1·6·又A中元素的個(gè)數(shù)為b-a+1，由題意b-a+1

b2-a2+1=1????????????????????????6分100整理得b2-a2+1=100(b-a)+100?b2-a2-100(b-a)=99∴(b-a)(b+a-100)=99因?yàn)閍,bb>a所以b-a=1或b+a-100=99b-a=3或b+a-100=33b-a=9或b+a-100=b-a=b+a-100=9或b-a=33或b+a-100=3b-a=99???????????????????9分b+a-100=1解得a=99或b=100a=65或b=68a=51或b=60a=49或b=60a=35或b=68a=1b=100故滿足條件的數(shù)對(duì)有(1,100)(35,68)(49,60)(51,60)(65,68)