數(shù)學-福建省2025屆高中畢業(yè)班適應性練習卷（福建省一模）試題+答案

2025屆高中畢業(yè)班適應性練習卷1．本解答給出了一種或幾種解法供參考，如果考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)2．對計算題，當考生的解答在某一步出現(xiàn)錯誤時，如果后繼部分的解答未改變該題的內(nèi)容和難度，可視影響的程度決定后繼部分的給分，但不n2n所以{an}是以1為首項，2為公比的等比數(shù)列，································································6分n因為n∈N*，所以n=1或n=2．n同理，當n≥3且n∈N*時，bn+1<bn．·······································解法二1）同解法一．············································································因為n∈N*，所以n=1或n=2．同理，當n≥3且n∈N*時，bn+1<bn．····························································解法三1）同解法一．············································································當n≥2時，令{lbEQ\*jc3\*hps11\o\al(\s\up8(n),n)≥bEQ\*jc3\*hps11\o\al(\s\up8(n),n)EQ\*jc3\*hps11\o\al(\s\up8(1),1),,······················································································*，所以n=3．··················解法四1）同解法一．············································································（2）由（1）知a=2n?1，所以b=n2．·······································································8分nn2n?1設·····································則···················································所以f)單調(diào)遞增，在單調(diào)遞減．·····························所以b1<b2<b3>b4>b5>…，·································所以AD=2.又AP=2，PD=2，所以AP2+AD2=PD2，即AD丄AP.·········································又因為AP∩AC=A，AP,AC?平面PAC，所以AD丄平面PAC.·············又因為AD?平面ACD，所以平面PAC丄平面ACD（2）過M作ME丄AC于點E，過E作EF丄CD于點F，連接MF.所以ME丄平面ACD.·····················································P又MF?平面MEF，所以CD丄MF，即M到直線PMN因為點MMNB設AE=x，則CE=2?xBAME=3x.EME=3x.E在Rt△EFM中，MF2=ME2+EF2，即分解得x=1或（舍去）.所以AM=2，故M與點P重合.···········································13分取線段PA的中點N，連接CN，易知CN丄AP.又AD∩AP=A，AD,AP?平面PAD，所以CN丄平面PAD，zEQ\*jc3\*hps21\o\al(\s\up0(··),Z)解法二解法二1）同解法一.·······················································································由（1由（1）知平面PAC丄平面ACD，且平面PAC∩平面ACD=AC，ABCD所以AZ丄平面ACD.一BCDyxyEQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps20\o\al(\s\up8(-→),C)EQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps21\o\al(\s\up5(-),A)EQ\*jc3\*hps21\o\al(\s\up5(-),A)EQ\*jc3\*hps21\o\al(\s\up5(-),C)EQ\*jc3\*hps21\o\al(\s\up5(-),C)EQ\*jc3\*hps21\o\al(\s\up8(-),C)2=(m?2)2+(m)2=4m2?4m設點M到直線CD的距離為d，EQ\*jc3\*hps21\o\al(\s\up9(-),C)設平面PAD的法向量n=(x,y,z)，直線CM與平面PAD所成EQ\*jc3\*hps20\o\al(\s\up8(x),2)取z=1，則平面PAD的一個法向量為n=(?,0,1)，···················································P解法三1）取AC的中點Q，連接解法三1）取AC的中點Q，連接PQ，QD.BQACDCD=22，CD解得AD=2，······································又因為所以QD2=AD2+AQ2=5.又PQ=，PD=2，所以PQ2+QD2=PD2，所以PQ丄QD.······································4分所以PQ丄平面ACD.·································（2）取CD的中點R，連接QR，所以QREQ\*jc3\*hps21\o\al(\s\up8(-),Q)EQ\*jc3\*hps21\o\al(\s\up8(-→),C)EQ\*jc3\*hps21\o\al(\s\up8(-),Q)EQ\*jc3\*hps21\o\al(\s\up8(-),Q)標系Qxyz，如圖所示.·····································································zzEQ\*jc3\*hps21\o\al(\s\up5(-),A)EQ\*jc3\*hps21\o\al(\s\up5(-),C)EQ\*jc3\*hps21\o\al(\s\up5(-),C)EQ\*jc3\*hps21\o\al(\s\up5(-),A)PEQ\*jc3\*hps21\o\al(\s\up4(-),A)EQ\*jc3\*hps21\o\al(\s\up4(-),A)EQ\*jc3\*hps21\o\al(\s\up5(-),C)EQ\*jc3\*hps21\o\al(\s\up5(-),C)EQ\*jc3\*hps21\o\al(\s\up5(-),A)BAQxCRD因為點M到直線CD的距離為，BAQxCRDyEQ\*jc3\*hps20\o\al(\s\up5(-),C)·········································································設平面PAD的法向量n=(x,y,z)，直線CM與平面PAD所成的角為θ.解法四1）分別取AC,CD的中點為Q,R，連接PQ，QR，PR，則QR∥AD，且P2······················································BQCDRCDR所以QR⊥AC.又PQ⊥AC，所以∠PQR為二面角P?AC?D的平面角．················所以在△PQR中，PQ2+QR2=PR2所以平面PAC⊥平面ACD.··························································（2）同解法三．·························因為N=MN∪MN，所以P(MN)=P(N)?P(MN)=0.5?0.3=0.2．······················所以········································推送B”為事件MN.P(M∪N)=P(M)+P(N)?P(MN)·················································································9分所以P(MN)=1?P(M∪N)··························································································11分依題意，R與MNL互為對立事件，且MN與L相互獨立，···············································13分解法二1）同解法一．············································································為事件R．依題意，L與(M∪N)L互斥，M∪N與L相互獨立，·······················································11分所以P(R)=P(L∪(M∪N)L)·························································································12分=P(L)+P((M∪N)L)············································································=P(L)+P(M∪N)P(L)··················································································14分解法三1）同解法一．············································································為事件R．依題意，R=MNL∪MNL∪MNL∪MNL∪MNL∪MNL∪MNL，············································10分且MNL、MNL、MNL、MNL、MNL、MNL、MNL互斥，MN與L、MN與L、MN與L、MN與L、MN與L、MN與L、MN與L均相互獨立．························································12分又因為P(M)=0.7，P(N)=0.5，P(MN)=0.3，所以P(R)=P(MNL∪MNL∪MNL∪MNL∪MNL∪MNL∪MNL)=P(MNL)+P(MNL)+P(MNL)+P(MNL)+P(MNL)+P(MNL)+P(MNL)·····················13分=P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)························································································································································當時，f′····················································································故f(x)在(0,)單調(diào)遞增，在單調(diào)遞減．····························································8分 ττf(τ?x)=(x?)sin(τ?x)+a(τ?x)?cos(τ?x)=(x?)sinx+a(τ?x)+cosx，22所以f(τ?x)+f(x)=τa， 所以曲線y=f(x)關(guān)于點P(τ,τa)對稱 10分因為f(0)=?1，f(?τ)=?τa+1，取A(0,?1)，B(?τ,?τa+1)，·········································12分APBP，所以A,B,P不共線.··································································設A,B關(guān)于點P的對稱點分別為C,D，則|PA|=|PC|，|PB|=|PD|，所以四邊形ABCD為平行四邊形.·················································································16分因為曲線y=f(x)關(guān)于點P(τ,πa)對稱，所以C,D也在曲線y=f(x)上，22所以曲線y=f(x)上存在四個點A,B,C,D，使得四邊形ABCD為平行四邊形.·······················17分解法二1）同解法一.······························································（2）曲線y=f(x)上存在四個點，使得以這四點為頂點的四邊形是平行四邊形．··················9分所以A,B,C,D四點都在曲線y=f(x)上.········································································13分EQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps20\o\al(\s\up8(-→),C)······························EQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps20\o\al(\s\up8(-→),C)2EQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps20\o\al(\s\up8(-→),C)所以四邊形ABCD為平行四邊形．所以曲線y=f(x)上存在四個點A,B,C,D，使得四邊形ABCD為平行四邊形．······················17分解法三1）同解法一.······························································（2）曲線y=f(x)上存在四個點，使得以這四點為頂點的四邊形是平行四邊形．··················9分所以A,B,C,D四點都在曲線y=f(x)上.········································································13分ACBD，所以A,B,C,D四點不共線．······················································15分因為線段AC與線段BD的中點都是P(τ,πa)，所以|AP|=|PC|，|BP|=|PD|，22所以四邊形ABCD為平行四邊形．所以曲線y=f(x)上存在四個點A,B,C,D，使得四邊形ABCD為平行四邊形．······················17分化歸與轉(zhuǎn)化思想、分類與整合思想、數(shù)形結(jié)合思想和特殊與一般思想等，考查邏輯推理、直.|BQ|，所以y2=3(x2?1)，即（2）選擇②作為條件．··························（i）設M(x1,y1),N(x2,y2)，則T(x1,-y1.|BQ|，所以y2=3(x2?1)，即（2）選擇②作為條件．··························（i）設M(x1,y1),N(x2,y2)，則T(x1,-y1)．y/PAOBQx顯然l的斜率不為零，否則，有x2=-x1,y2=y1，l:y=y1，此時與直線TB和NB的斜率之積為6，矛盾．·······························································故可設l：x=my+t，由得(3m222x2因為直線TB和NB的斜率之積為6，所以分yM1-1)(x2-1)+y1y2=x1x2-(x1+x2)+1+y1y2AOBx3m2+9624203m2-13m2-13m2-13m23m2-13m2-13m2-13m2-1，所以DMBN是鈍角，△BMN是鈍角三角形．····················································所以M,N分別在C的兩支．不妨設M在C的右支，則x1>1，如圖．································14分EQ\*jc3\*hps20\o\al(\s\up9(-→),B)EQ\*jc3\*hps20\o\al(\s\up9(-→),R)-x1)(3-x1)+yEQ\*jc3\*hps12\o\al(\s\up4(2),1)=xEQ\*jc3\*hps12\o\al(\s\up4(2),1)-4x1+3+3xEQ\*jc3\*hps12\o\al(\s\up4(2),1)-3=4x1(x1-1)>0，···········16分y所以DBMR∈(0,)．BMRxAON因為l過點R，所以所以DBMNBMRxAON綜上可知，△BMN不可能是銳角三角形．·····································································17分BB解法二1）同解法一．·····················································································（2）選擇②作為條件．············································（i）同解法一．··································································①當M,N均在C的右支，不妨設M在x軸的上方，如圖．解法二1）同解法一．·····················································································（2）選擇②作為條件．············································（i）同解法一．··································································①當M,N均在C的右支，不妨設M在x軸的上方，如圖．yMAONx所以上MBN是鈍角，△MBN是鈍角三角形．··········································②當M,N分別在C的兩支時，不妨設M在C的右支，則x1>1，如圖．·····························14分EQ\*jc3\*hps20\o\al(\s\up9(-→),B)EQ\*jc3\*hps20\o\al(\s\up9(-→),R)-x1)(3-x1)+yEQ\*jc3\*hps11\o\al(\s\up4(2),1)=xEQ\*jc3\*hps11\o\al(\s\up4(2),1)-4x1+3+3xEQ\*jc3\*hps11\o\al(\s\up4(2),1)所以DBMR∈(0,)．因為l過點R，所以DBMN=τ-DBMR∈(,τ)，所以DBMN是鈍角，△BMN是鈍角三角形．-3=4x1(x1-1)>0，···········16分yANOBMBMRx綜上可知，△BMN不可能是銳角三角形．···································································解法三1）同解法一．············································································（2）選擇②作為條件．············································yxT=x1,yT=-y1．l:x=x1．T（i）設M(x1,y1),N(x2,y2),T(xT,yxT=x1,yT=-y1．l:x=x1．TOBMARxOBMARxN依題意解得x1=3．此時，l:x=3，過點R(3,0)下面我們證明，當l的斜率存在時，M,N,R共線．顯然直線TB斜率存在且不為零，可設直線TB：x=m1y+1(m1≠0)，同理可設直線NB：x=m2y+1，得因為直線TB和NB的斜率之積為6，所以即所以所以M,N,R共線，即l過點R(3,0)．綜上，l過定點(3,0)．·············································································EQ\*jc3\*hps21\o\al(\s\up0(12),M)1-1)(x2-1)+y1y2=(my1+2)(my2+2)+y1y2=m2y1y2+2m(y1+y2)+4+y1y2AAOBRx所以∠MBN是鈍角，△BMN是鈍角三角形．·················································所以M,N分別在C的兩支．不妨設M在C的右支，則x1>1，如圖．································14分EQ\*jc3\*hps20\o\al(\s\up9(-→),B)EQ\*jc3\*hps20\o\al(\s\up9(-→),R)-x1)(3-x1)+yEQ\*jc3\*hps12\o\al(\s\up4(2),1)=xEQ\*jc3\*hps12\o\al(\s\up4(2),1)-4x1+3+3xEQ\*jc3\*hps12\o\al(\s\up4(2),1)-3=4x1(x1-因為l過點R，所以DBMN=τ-DBMR∈(,τ)，所以DBMN是鈍角，△BMN是鈍角三角形．yTAOAORxMN綜上可知，△BMN不可能是銳角三角形．···································································解法四1）同解法一．············································································（2）選擇③作為條件．················（i）設M(x1,y1),N(x2,y2)，則T(x1,-y1)．顯然l的斜率不為零，否則x2=-x1,y2=y1，l:y=y1，因為直線TB和NA的斜率之商為2，所以故可設l：x=my+t，因為直線TB和NA的斜率之商為2，所以··················································8分因為點M在C上，所以y12=3(x12?1)，即.=3，3m2?13m2?11-1)(x2-1-1)(x2-1)+y1y2=x1x2-(x1+x2)+1+y1y23m2+962420=-++1+=<0，3m2-13m2-13m2-13m2-1所以DMBN是鈍角，△BMN是鈍角三角形．··································································3或yMMAOBN\BN\RRx3所以M,N分別在C的兩支，不妨設M在C的右支，則x1>1，如圖．································14分設R(3,0)，F(xiàn)(2,0)，則F(2,0)是以BR為直徑的圓的圓心，MF2=(x1-2)2+yEQ\*jc3\*hps12\o\al(\s\up4(2),1)=xEQ\*jc3\*hps12\o\al(\s\up4(2),1)-4x1+4+yEQ\*jc3\*hps12\o\al(\s\up4(2),1)=xEQ\*jc3\*hps12\o\al(\s\up4(2),1)-4x1+4+3xEQ\*jc3\*hps12\o\al(\s\up4(2),1)所以M在以BR為直徑的圓外，所以DBMR∈(0,)．因為l過點R(3,0)，所以DBMN=τ-DBMR∈(,τ)，所以DBMN是鈍角，△BMN是鈍角三角形．-3=4x1(x1-1)+1>1，············16分yTFANOBRxM綜上可知，△BMN不可能是銳角三角形．···································································解法五1）同解法一．············································································（2）選擇③作為條件．··································（i）設M(x1,y1),N(x2,y2),T(xT,yT)，則xT=x1,yT=-y1．當l的斜率不存在時，x2=x1,y2=-y1，l:x=x1．因為直線TB和NA的斜率之商為2，所以即，解得x1=3，此時，直線l:x=3，過點R(3,0)下面我們證明，當l的斜率存在時，M,N,R共線．顯然直線TB斜率存在且不為零，可設直線TB：x=m1y+1(m1≠0)，EQ\*jc3\*hps24\o\al(\s\up16(x),x)2EQ\*jc3\*hps24\o\al(\s\up16(m),y)EQ\*jc3\*hps24\o\al(\s\up16(1),1)EQ\*jc3\*hps13\o\al(\s\up5(2),1)2同理可設直線NA：x=m2y-1，由得因為直線TB和NA的斜率之商為2，所以即m2=2m1．所以所以M,N,R共線，即l過點R(3,0)．綜上，l過定點(3,0)．·············································································（ii）同解法二．·······················································解法六1）同解法一．············································································（2）選擇②作為條件．··································（i）設M(x1,y1),N(x2,y2)，則T(x1,-y1)．當l的斜率不存在時，x2=x1,y2=-y1，l:x=x1．又因為直線TB和NB的斜率之積為6，所此時，直線l:x=3，過點(3,0)．·····································當l的斜率存在時，設l：y=kx+m，因為直線TB和NB的斜率之積為6，所以分-3k，過點(3,0)．綜上，l過定點(3,0)．·············································································22yM1-1)(x2-1)+y1y2=x1x2-(x1+x2)+1+k2(x1-3)(x2-3)=x1x2-(x1+x2)+1+k2x1x2-3k2(x1+x2)+9k2=9k2+3-6k2+1+k2(9k2+3)-3k2.6k2+9k2AOBxk2-3k2-3k2-3k2-3=9k2+3-6k2+k2-3+k2(9k2+3)-18k4+9k2(k2-3)k2-3k2-3k2-3k2-3k2-3k2-3N=-<0，所以DMBN是鈍角，△BMN是鈍角三角形．·······························x2所以M,N分別在C的兩支，不妨設M在C的右支，則x1>1，如圖．·······························