2025年A-Level考試真題試卷考試時間：______分鐘總分：______分姓名：______第一部分1.Theequationofacurveisgivenbyy=3x^2-2x+1.a)Findthegradientofthecurveatthepointwherex=2.b)Determinethecoordinatesoftheturningpointofthecurve.c)Sketchthecurve,showingclearlythecoordinatesoftheturningpointandthey-intercept.2.Afunctionisdefinedasf(x)=x^3-4x^2+3x.a)Findthefirstderivative,f'(x).b)Solvetheequationf'(x)=0,givingyouranswerstothreesignificantfigures.c)Determinethenatureofthestationarypoint(s)youfoundinpart(b).3.Inachemicalreaction,therateofreaction,r,ismeasuredinmoldm^-3s^-1.Theconcentrationofareactant,C,ismeasuredinmoldm^-3.Therateisfoundtobeproportionaltothesquareoftheconcentrationofthereactant.WritedownanequationrelatingrandC,includingthevalueoftheconstantofproportionality,k,ifk=0.5.4.a)Define'workdone'.b)Aforceof20Nactsonanobject,causingittomove5minthedirectionoftheforce.Calculatetheworkdonebytheforce.c)Explainwhytheworkdoneinpart(b)mightnotbethetotalenergytransferrediffrictionispresent.第二部分5.Thedisplacement,smetres,ofaparticlemovinginastraightlineisgivenbytheequations=2t^3-3t^2+4t,wheretisthetimeinseconds.a)Findthevelocityoftheparticlewhent=3.b)Findtheaccelerationoftheparticlewhent=3.c)Determinethetime(s)whentheparticleisatrest.6.Evaluatethefollowingintegrals:a)∫(x^2+2x-5)dxb)∫_0^2(3e^x-2)dx7.Aboxcontains5redballs,3blueballs,and2greenballs.Aballisselectedatrandomfromthebox,itscolourisnoted,andthenitisreplaced.Thisprocessisrepeatedtwice.a)Findtheprobabilityofselectingablueballonthefirstdraw.b)Findtheprobabilityofselectingexactlyoneredballandoneblueballinthetwodraws.c)Findtheprobabilitythatthetwoballsselectedareofdifferentcolours.8.a)Define'momentofaforce'.b)Auniformbeamoflength4mandweight100Nissupportedatitsmid-pointandatoneend.Aforceof60Nactsverticallydownwardsatapoint1mfromthemid-point.Calculatetheforceexertedbythesupportatthemid-point.c)Determinetheforceexertedbythesupportattheendofthebeam.第三部分9.Solvetheequation2cos(θ)+√3=0for0°≤θ≤360°.10.Asampleofaradioactiveisotopehasamassof100grams.Thehalf-lifeoftheisotopeis5years.Writedowntheexpressionforthemass,Mgrams,oftheisotoperemainingaftertyears.a)Calculatethemassoftheisotoperemainingafter10years.b)Determinethetimetakenforthemassoftheisotopetoreduceto12.5grams.11.Describethedifferencebetweenascalarquantityandavectorquantity.Giveoneexampleofeachtype.12.Anobjectisthrownverticallyupwardsfromthegroundwithaninitialspeedof20ms^-1.Assumetheaccelerationduetogravityis9.8ms^-1downwards.a)Calculatethemaximumheightreachedbytheobject.b)Calculatethetimetakenfortheobjecttoreturntotheground.c)Determinethespeedoftheobjectwhenitisataheightof15m.13.a)Define'density'.b)Thedensityofametalblockis8gcm^-3.Theblockhasavolumeof500cm^3.Calculatethemassoftheblock.c)Explainwhatwouldhappentothedensityoftheblockifitsmassweredoubledwhileitsvolumeremainedconstant.14.Describetheprocessofosmosisinsimpleterms.Includeinyouransweradescriptionofthedirectionofwatermovementandtheconditionsrequiredforosmosistooccur.試卷答案第一部分1.a)Gradient=dy/dx=6x-2.Atx=2,Gradient=6(2)-2=12-2=10.b)dy/dx=6x-2.Setting6x-2=0givesx=1/3.y=3(1/3)^2-2(1/3)+1=1-2/3+1=2/3.Turningpoint(1/3,2/3).c)y-intercept=3(0)^2-2(0)+1=1.Thecurveisaparabolaopeningupwardswithaturningpointat(1/3,2/3)andy-interceptat(0,1).2.a)f'(x)=3x^2-8x+3.b)Solving3x^2-8x+3=0usingquadraticformula:x=[-(-8)±√((-8)^2-4*3*3)]/(2*3)=[8±√(64-36)]/6=[8±√28]/6=[8±2√7]/6=[4±√7]/3.Approximatevalues:x≈1.77orx≈0.53.c)f''(x)=6x-8.Atx=(4+√7)/3,f''(x)=6((4+√7)/3)-8=2(4+√7)-8=8+2√7-8=2√7>0.Minimum.Atx=(4-√7)/3,f''(x)=6((4-√7)/3)-8=2(4-√7)-8=8-2√7-8=-2√7<0.Maximum.Stationarypointsareamaximumatx≈0.53andaminimumatx≈1.77.3.r=kC^2.Withk=0.5,theequationisr=0.5C^2.4.a)Workdoneistheenergytransferredtoorfromanobjectbyaforceactingonitwhentheobjectmovesinthedirectionoftheforce.ItiscalculatedasWorkdone=Force×Distancemovedinthedirectionoftheforce.b)Workdone=20N×5m=100J.c)Iffrictionispresent,itactsintheoppositedirectiontothemotion,doingnegativework.Therefore,thetotalenergytransferred(theworkdonebytheappliedforce)isgreaterthantheincreaseintheobject'skineticenergy,assomeenergyistransferredtothesurroundingsasheatduetofriction.第二部分5.a)Velocityv=ds/dt=6t^2-6t+4.Att=3,v=6(3)^2-6(3)+4=54-18+4=40ms^-1.b)Accelerationa=dv/dt=12t-6.Att=3,a=12(3)-6=36-6=30ms^-2.c)Particleisatrestwhenvelocityv=0.6t^2-6t+4=0.Dividingby2:3t^2-3t+2=0.DiscriminantD=(-3)^2-4(3)(2)=9-24=-15.SinceD<0,therearenorealsolutions.Theparticleisneveratrest.6.a)∫(x^2+2x-5)dx=(x^3/3)+(x^2)-5x+C.b)∫_0^2(3e^x-2)dx=[3e^x-2x]_0^2=(3e^2-2(2))-(3e^0-2(0))=(3e^2-4)-(3-0)=3e^2-4-3=3e^2-7.7.a)P(Blue)=3/(5+3+2)=3/10.b)P(RedthenBlue)=(5/10)(3/10)=15/100.P(BluethenRed)=(3/10)(5/10)=15/100.P(exactlyoneredandoneblue)=15/100+15/100=30/100=3/10.c)P(differentcolours)=P(RedthennotRed)+P(notRedthenRed).P(RedthennotRed)=(5/10)(5/10)=25/100.P(notRedthenRed)=(7/10)(5/10)=35/100.P(differentcolours)=25/100+35/100=60/100=3/5.8.a)Themomentofaforceistheturningeffectoftheforceaboutapoint.ItiscalculatedasMoment=Force×Perpendiculardistancefromthepointofrotation.b)LetF_midbetheforceatthemid-point.Clockwisemoment=60N×1m=60Nm.Anticlockwisemoment=F_mid×2m.Forequilibrium,Clockwisemoment=Anticlockwisemoment.60=F_mid×2.F_mid=60/2=30N.Thesupportatthemid-pointexertsanupwardforceof30N.c)Sumofverticalforces=0.F_mid(upward)+F_end(upward)-100N(downward)=0.30+F_end-100=0.F_end=100-30=70N.Thesupportattheendexertsanupwardforceof70N.第三部分9.2cos(θ)+√3=0=>cos(θ)=-√3/2.Cosineisnegativeinthesecondandthirdquadrants.Referenceangleθ_ref=cos^(-1)(√3/2)=30°.Forθinthesecondquadrant:θ=180°-30°=150°.Forθinthethirdquadrant:θ=180°+30°=210°.Solutionsfor0°≤θ≤360°areθ=150°andθ=210°.10.a)UsingtheformulaM=M?(1/2)^(t/T),whereM?=100g,T=5years.M=100(1/2)^(10/5)=100(1/2)^2=100(1/4)=25grams.b)WewanttofindtwhenM=12.5grams.12.5=100(1/2)^(t/5).Divideby100:0.125=(1/2)^(t/5).Recognizing0.125=(1/2)^3:(1/2)^3=(1/2)^(t/5).Sincebasesarethesame,exponentsmustbeequal:3=t/5.t=3×5=15years.11.Ascalarquantityhasonlyamagnitude(size).Avectorquantityhasbothamagnitudeandadirection.Exampleofscalar:Mass.Exampleofvector:Force.12.a)Atmaximumheight,finalvelocityv=0.Usingv=u+at:0=20-9.8t.t=20/9.8≈2.04seconds(upwardjourney).Usings=ut+0.5at2:s=20(2.04)+0.5(-9.8)(2.04)^2=40.8-20.4=20.4m.Maximumheight≈20.4m.b)Timetoreturntoground=Timegoingup+Timegoingdown.Timegoingdownisthesameasthetimegoingup(ignoringairresistance),sotime=2.04s+2.04s=4.08seconds.c)Usings=ut+0.5at2tofindtimetakentoreach15m:15=20t-0.5(9.8)t2=>4.9t2-20t+15=0.Usingquadraticformula:t=[20±√((-20)^2-4(4.9)(15))]/(2*4.9)=[20±√(400-294)]/9.8=[20±√106]/9.8.Approximatevalues:t≈1.29sort≈1.19s.Tofindspeed,usev=u+at.Att≈1.29s:v=20-9.8(1.29)≈20-12.64=7.36ms^-1.Att≈1.19s:v=20-9.8(1.19)≈20-11.66=8.34ms^-1.Thespeedat15monthewaydownisapproximately8.3ms^-1.第四部分13.a)Densit