高考数学解题技巧Skills of solving maths problems in college entrance examination.docx
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高考数学解题技巧Skills of solving maths problems in college entrance examination.docx
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高考数学解题技巧Skillsofsolvingmathsproblemsincollegeentranceexamination
高考数学解题技巧(Skillsofsolvingmathsproblemsincollegeentranceexamination)
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//strengthisthefoundationofhighscore,strategy,techniqueandskillisthekeytogethighscore.Fortwoverystrongstudents,intheexaminationofsomeproblem-solvingskills,theuseofgoodorbad,oftenleadtothefinalscoreofthetwopeoplehaveabiggap.First,choosetheproblemsolvingstrategiesofmathematicschoicehassummedstrong,widecoverageofknowledge,compactandflexible,acomprehensiveanddepthetc.,candidatescanquickly,accuratelyandcomprehensively,andgoodsolutionchoice,becomethekeytosuccessinthecollegeentranceexamination.Thebasicrequirementsofsolutionchoiceisskilledandaccurate,flexibleandrapidmethod,surprise.Therearegenerallythreewaystosolveproblems:
oneistoconsidertheproblemfromthebeginningandtheotheristosearchfortheresult;thetwoistoconsiderthecombinationoftheproblemandthechoiceofthebranch;thethreeistoseektheconditionofsatisfyingthequestionfromthechoiceofexpenditure.Choiceiseasy(forindividualquestions,thebasicprincipleofsolvingtheintermediateproblem)is:
"cannotdoa.".1,directmethod:
involvesthemathematicstheorem,thedefinition,therule,theformulaquestion,oftenembarksfromthedesigncondition,throughtheoperationortheinference,obtainstheconclusiondirectly,andselectswiththesupporttocontrolagain.Example:
theknownfunctiony=f(x)y=g(x)inversefunctionexists,iff(3)=1,thefunctionofy=g(x1)ofthefigure:
asinthefollowingpointswillpass(B)A.(2,3)B.(0,3)C.(2,1)D.(4,1):
functiony=f(x)bytheimagepoint(3,1),itsinversefunctiony=g(x)afterimagepoint(1,3),whichcangetthefunctionofy=g(x1)afterimagepoint(0,3),theB.2,thescreeningmethod(exclusionmethod,eliminationmethod):
makefulluseofthecharacteristicsofthechoiceofradio,throughanalysis,reasoning,calculation,judgment,onebyonetoeliminatethewrongbranch,getthecorrectbranchsolution.Cases.Ifxistheminimumangleinthetriangle,thefunctionrangeisy=sinx+cosx(A)A.(1,2B.(0],
32
]
1C.[2
22
1]D.
(2)
22
]
PI
Solution:
becausexistheminimumangleinthetriangle,sox(0,3),whichcanbey=sinx+cosx>1,C,Bdebugging,DA,c..3,imagemethod(combinationofnumbersandshapes):
throughthethinkingprocessofcombinationoffiguresandshapes,bymeansofgraphicintuitive,quicklymakethechoicemethod.Casesareknown.Alphaandbetasecondquadrant,andCOSalpha>cosbeta,is(B)A..B.sin
fromthestemorselectexpenditure,byselectingspecialvalues,thespecialproblems,achieveasureornegativethreetobe"rightsmall"strategy.Thespecialvalue:
cases.AnarithmeticsequencebeforetheNand482nbeforeand60,itsformer3Nand(B)A.24B.84C.72D.36solution:
ndoesnotcontainthecorrectconclusion,independentofN,cantakethespecialvalueofN,suchasn=1atthistime,a1=48,a2=S2-S1=12,a3=a1+2d=-24,3Nandtheprevious36,D.Specialfunction:
cases.DefinitionofoddfunctiononRf(x)isadecreasingfunction,a+b=0,giventhefollowinginequality:
F(a)f(alpha)=0f(b)f(-b)=0f(a)=f(+f(b)+f(alpha)-b)andf(a)+f(b)orF(alpha)+f(b)thenumberofinequalitiesiscorrect(B)A.II.TheB.andC.oftheD..
Http:
//solution:
takeF(x)=-x,andcheckitbyitembyitem.SochooseB.
One
Theratioisgreaterthan1,thentheseries{log3an}series:
specialcases.Ifthefirstisageometricsequenceof{an}(A.)positive,isincreasingingeometricseriesB.
TheC.isadecreasinggeometricseriesisincreasingarithmeticsequenceisD.nsequencedecreasingsolution:
takean=3,namelyD.2.Specialposition:
example.Theparabolay=ax(a>0)focusFasastraightline,theparabolaisPandQtwopoints,ifthelengthoflinesegmentPFandFQispandQrespectively,thenA.2aB.
12A12A
1p1+q
Equalto(D.
1p1Q+
)
4A
C.4a,substitution
Note:
severalproblemscanalsobesolvedbyspecialpositions
Solution:
p=q=isperpendiculartotheaxisofywhenPQisperpendicular
=4a,sochooseC.)
Inspecialcases.Ffunction:
(x)x=+2(x=0)inversefunctionF1(x)imageis(.
Solution:
inF(x)x=+2(x=0)inthex=0,x=4,y=4,y=2;thatisaspecialpoint(2,0)-1-and1(4,4)intheinversefunctionofF(x)images,A,Cobservation.Cisdefinedbythedefinitiondomainofinversefunctionf(x).Thespecialcasesofequation(a>b>0).B2x2a2y2=a2b2hyperbolicasymptoteanglealpha,centrifugalratewase.
Alpha
Thencos2equals
Two
)
1C.e1D.e2
25
Two
A.eB.eofthetestsolution:
therelationshipbetweentheangleandtheeccentricityofthehyperbolicasymptote,byusingaspecialequationtosolveequationfor.
X24
-
Y21
=1,easytogetthecentrifugalratee=
52
Alpha
Cos2=
Two
Therefore,Cshouldbechosen.
Yx
Thespecialmodel:
cases.Iftherealx,y(+y=3,X2)tomeetthemaximumvalueis(
1A.2yx
)
B.
Y0X0
33
C.
32
D.
Three
Y2Y1x2x1k=
Solution:
inLenovo.=mathematicalmodel:
twolinearslopeformula,willlookintoround22(x2)+y=3andOontheoriginoftheslopeofthelineD.5,themaximumvalueestimationmethod:
byestimatingorlistofthecomplexproblemsintosimpleproblems,forajudgemethodthentheanswerinsolution.Example:
itisknownthatthecenterofthehyperbolaisattheoriginandthefocalpointisF(7,0),andtheliney=x1iscorrespondingtoMandNtwo
At2points,themidpointabscissaofMNis3,andtheequationofthishyperbolais
Http:
//A.
X2Y2x2Y2x2Y2x2Y2=1=1=1=134B.43C.52D.25x2Y2=1mn,bythedifferencemethod
Solution:
settheequationforD.note:
itisnotnecessarytosolvem,n6,inferenceanalysis:
featureanalysis:
accordingtotheinformationprovidedbythesubject,suchasnumericalcharacteristics,structurecharacteristics,locationandothercharacteristics,fastreasoning,judgmentmethod.Example:
sin=known
M3A.9M
Two
N5=m2,election
M342mm+5cosm+5m3B.|=9m|
Two
PI
0
(
Two
Then,Tan2=(D.5)
0
Two
)
1C.3
PI
Solution:
becauseoftherestrictionofsintheta+costheta=1,Misthedefinitevalue,sotanisthedefinitevalue,2 PI,theta,theta ThetaPI<<2<2,4,Tan*2>1,theD.Thelogicalanalysis: ifABAisreallytrue,otherwiseexcluded,andthereisonlyacorrectconclusioncontradiction;ifAB,AB,AandBarefalse;ifacontradiction,theremustbeareally,canCandD.cases: leta,Bsatisfyab<0therealnumber,thenB.|a+b|<|a-bC.|a-b|<|a|-|b|D.|a-b|<|a|+|b|(A.|a+b|>|a-b|)|solution: forA,Bisacontradictoryproposition,sotherewillbeatrue,thusexcludingerrorC,D.Andab<0,a=1,b=-1,intotheknowledgethatBistrue.7.validationmethod: theselectionofeachbranchintoquestiondryverification, Properorspecialvalueforinspection,orbyothermeansofverification,todeterminewhetherawrongmethod.0cases.IftheinequalityislessthanorequaltoX2-ax+a=1thesolutionsetisasingletonset,thenthevaluefora(0)(A)(B)2(C)4(D)6solution: selectB.two,fillintheblankswiththeproblemsolvingstrategiesofmultiple-choiceverificationa=2supportonebyoneintothestem,fillintheblanksisonlythebasicprincipleofproblemsolvingisthe"rightnottodo".Thebasicproblemsolvingstrategyis: theartofproblemsolving.Basicmethodsgenerallyinclude: thedirectsolvingmethod,imagemethod,structuremethodandspecialmethod(specialvalue,specialfunction,specialangle,specialseries,specialgraphiclocation,specialpoint,special,special1equationmodel),thedirectsolvingmethodisdirectlyfromthequestion: startingwiththedefinitionandtheproperties,theoremsandformulas,thecorrectconclusionobtainedbydeformation,reasoning,calculationandjudgment.Thisisthebasicmethodtofillincommonuse,shouldbegoodatgraspingtheessencethroughthephenomenon"".Striveforflexibilityandsimplicity.{an},{bn}series.Botha1=0andb1=arithmeticprogression,-4,Sk,Sk,{bn}'{an}respectivelyandthefirstitemK(kisapositiveinteger),iftheSk+Sk'=0,ak+bk=.Solution: theSk=summationformulawitharithmetic,a1+b1=+-4,Rak+bk=4.2.specialsolution: whenthefillinconclusiononlyorthevalueofthefixedvalue,weneedtotesttheparametersinthespecialvalue(orspecialfunction,specialangle,specialseries,specialgraphiclocation,specialpoint,special,specialequationmodel)instead,togettheconclusion.Inthisexample,suchas: k=2(k=1? ),soa1+a2+b1+b2=0,soa2+b2=4,ak+bk=4.WiththeknownSA,SB,SC22angleis60degrees,thedihedralangleplaneSABandplaneSACas.Solution: takingSA=SB=SCandstudyingtheprobleminthetetrahedron,itisnotdifficulttofindtheplaneSABandtheplaneSAC K(a1+ak)2K(a1+ak)2K(B1+bk)2=0,andagain Http: // 1(seearccos3.dihedralanglesforotherspecialmethodchoice)3.combinationmethod: accordingtothegeometricmeaningofthetitleandsettheconditions,drawgraphics,withintuitivegraphics,judgmentmethodquickly.Venndiagrams
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