CLASSXFORMULAE MATHS

Realnumbers:

Euclid’sdivisionlemma

Givenpositiveintegersaandb,thereexistwholenumbersqandrsatisfyinga=bq+r,0≤r<b.

Euclid’sdivisionalgorithm:

ThisisbasedonEuclid’sdivisionlemma.

Accordingtothis,theHCFofanytwopositiveintegersaandb,witha>b,isobtainedasfollows:

Step1:Applythedivisionlemmatofindqandrwherea=bq+r,0≤r<b.

Step2:Ifr=0,thenHCFisb.Ifr≠0,applyEuclid’slemmatobandr.

Step3:Continuetheprocesstilltheremainderiszero.ThedivisoratthisstagewillbeHCF(a,b).

Also,HCF(a,b)=HCF(b,r).

Thefundamentaltheoremofarithmetic

Everycompositenumbercanbeexpressed(factorised)asaproductofprimes,andthisfactorizationisunique,apartfromtheorderinwhichtheprimefactorsoccur.•Letx=p/qbearationalnumber,suchthatprimefactorisationof‘q’isoftheform2n5m,wherem,narenon-negativeintegers.Thenxhasadecimalexpansionwhichisterminating.

•Letx=p/qbearationalnumber,suchthatprimefactorizationof‘q’isnotoftheform2n5m,wherem,narenon-negativeintegers.Thenxhasadecimalexpansionwhichisnon-terminatingrepeating.

Polynomial:•Aquadraticpolynomialinxwithrealcoefficientsisoftheformax2+bx+c,wherea,b,carerealnumberswitha≠0.

•Thezeroesofapolynomialp(x)arepreciselythex-coordinatesofthepoints,wherethegraphofy=p(x)intersectsthex-axis.

•Aquadraticpolynomialcanhaveatmost2zeroesandacubicpolynomialcanhaveatmost3zeroes.

•Ifαandβarethezeroesofthequadraticpolynomialax2+bx+c,then

CLASSXFORMULAE MATHS

Sumofzeroes(α+β)=!𝒃𝒂

Productofzeroes(αβ)=𝒄𝒂

•Ifα,β,γarethezeroesofthecubicpolynomialax3+bx2+cx+d=0,then

Sumofzeroestakenoneattime(α+β+γ)=!𝒃𝒂

Sumofzeroestakentwoattime(αβ+βγ+γα)=𝒄𝒂

Productofzeroes(αβγ)=!𝒅𝒂

Pairoflinearequationsintwovariables:•Ifthelinesintersectatapoint,thenthatpointgivestheuniquesolutionofthetwoequations.Inthiscase,thepairofequationsisconsistent.

•Ifthelinescoincide,thenthereareinfinitelymanysolutions—eachpointonthelinebeingasolution.Inthiscase,thepairofequationsisdependent(consistent).

•Ifthelinesareparallel,thenthepairofequationshasnosolution.Inthiscase,thepairofequationsisinconsistent.

SimultaneouspairofLinearequation

Condition Graphicalrepresentation

Algebraicinterpretation

𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0 𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0

𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0

𝑎!𝑎!

≠𝑏!𝑏!

𝑎!𝑎!

=𝑏!𝑏!=𝑐!𝑐!

Intersectinglines;Theintersectingpointcoordinateistheonlysolution

Coincidentlines;Theanycoordinateonthelineisthesolution.

OneuniquesolutiononlyInfinitesolutions

CLASSXFORMULAE MATHS

𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0

𝑎!𝑎!

=𝑏!𝑏!≠𝑐!𝑐!

ParallelLines

Nosolution

Crossmultiplicationmethod:𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0𝑎!𝑥 + 𝑏!𝑦 + 𝑐! = 0

𝑥𝑏!𝑐! − 𝑏!𝑐!

= 𝑦

𝑎!𝑐! − 𝑎!𝑐!=

1𝑎!𝑏! − 𝑎!𝑏!

Valueofxcanbeobtainedbyusingfirstandlastexpression.

Valueofycanbeobtainedbyusingsecondandlastexpression.

Quadraticequation:

Therootsofquadraticequationax2+bx+c=0,a≠0aregivenby!!± !!!!!"!!

providedD≥0.

Discriminantofthequadraticequationax2+bx+c=0,a≠0isgivenby

D=b2-4ac

•b2-4ac>0thenwewillgettworealsolutionstothequadraticequation.

•b2-4ac=0thenwewillgettwoequalrealsolutionstothequadraticequation.

•b2-4ac>0thenwewillnorealsolutiontothequadraticequation.

•Aquadraticequationcanalsobesolvedbymethodofcompletingsquare.

(a+b)2=a2+b2+2ab

(a-b)2=a2+b2–2ab

ArithmeticProgression:•Ifa,b,careinAP,then2b=a+c

•nthtermofanarithmeticprogression:

Tn=a+(n–1)d

CLASSXFORMULAE MATHS

•Numberoftermsofanarithmeticprogression:

n= 𝒍!𝒂𝒅

+ 𝟏

Wheren=numberofterms,a=thefirstterm,l=lastterm,d=commondifference

AdditionalnotesonAPTosolvemostoftheproblemsrelatedtoAP,thetermscanbeconvenientlytakenas:

•3terms:(a–d),a,(a+d)

•4terms:(a–3d),(a–d),(a+d),(a+3d)

•5terms:(a–2d),(a–d),a,(a+d),(a+2d)

•ThenthtermofanA.Pisthedifferenceofthesumtothefirstntermsandthesumtofirst(n-1)termsofit:

Tn=Sn-Sn-1

•IfeachtermofanAPisincreased,decreased,multipliedordividedbythesamenon-zeroconstant,theresultingsequencealsowillbeinAP.

•InanAP,sumoftermsequidistantfrombeginningandendwillbeconstant.

•A.PwhichcontainfinitetermiscalledfiniteA.Pandwhichcontainsinfinitetermsiscalledinfiniteterm.

Triangles:∆ABC~∆PQR!"!"

=!"!"

=!"!"

ByA.A.testorS.A.StestorbyS.S.S.testC.P.S.T.

!!!!

=!!!!

!

Areasof~∆’sareproportionaltosquaresoftheircorrespondingsides.

A1=A2 Amediandividesa∆into2∆’swithequalarea.!!!!

=!"#$!!"#$!

=Anareaof∆’smeetingatcommonvertexandbasethroughthesamestraightlineisproportionaltotheirbases.

•Ifaperpendicularisdrawnfromthevertexoftherightangleofarighttriangletothehypotenuse,thenthetrianglesonbothsidesoftheperpendiculararesimilartothewholetriangleandalsotoeachother.

Coordinategeometry:

Distanceformula•Distance= 𝑥! − 𝑥! ! + 𝑦! − 𝑦! !.(Thesameformulaistobeusedtofindthelengthoflinesegment,sidesofatriangle,square,rectangle,parallelogrametc.)Sectionformula

CLASSXFORMULAE MATHS

Point(x,y)=!!!!! !!!!!!! !!

, !!!!! !!!!!!! !!

•IfpointP(x,y)dividesABink:1,thenthecoordinatesofpointPwillbe(kx2/k+1,ky2+y1/k+1)

•Mid-point=!!! !!

!, !!! !!

!

•Centroidofatriangle=!!! !! ! !!

!, !!! !! ! !!

!

•Toproveco-linearityofthegiventhreepointsA,B,andC,YouhavetofindlengthofAB,BC,ACthenusetheconditionAB+BC=AC.Areaoftriangle•Areaoftriangle:1/2[x1(y2–y3)+x2(y3–y1)+x3(y1–y2)]=0•Areacannotbenegativeso,weshallignorenegativesignifit’soccurringintheproblem.•Iftheareaofthetriangleiszero,thenverticesofthetrianglearecollinear.•Tofindtheareaofquadrilateralweshalldivideitintotwotrianglesbyjoiningtwooppositevertices,findtheirareasandaddthem.

Introductiontotrigonometry:

•Wherever‘Square’appearsthinkofusingtheidentities(i)Sin2θ+Cos2θ=1 (ii)Sec2θ–Tan2θ=1 (iii)Cosec2θ–Cot2θ=1•TrytoconvertallthevaluesofthegivenproblemintermsofSinθandCosθ•Cosecθmaybewrittenas1/Sinθ•Secθmaybewrittenas1/Cosθ•Cotθmaybewrittenas1/Tanθ•TanθmaybewrittenasSinθ/Cosθ•Whereverfractionalpartsappearsthenthinkoftakingtheir‘LCM’•Thinkofusing(a+b)2,(a–b)2,(a+b)3,(a–b)3formulaeetc.

CLASSXFORMULAE MATHS

•Rationalisethedenominator[Ifa+b,(or)a–bformatisgiveninthedenominator]•Youmayseparatethedenominator

ForEx:!"# ! ! !"# !

!"# !as

!"# !!"# !

+!"# !!"# !

=1+Cotθ•Sin(90–θ)=Cosθ : Cos(90–θ)=Sinθ•Sec(90–θ)=Cosecθ: Cosec(90–θ)=Secθ•Tan(90–θ)=Cotθ : Cot(90–θ)=Tanθ

Circles:

•Thetangenttoacircleisperpendiculartotheradiusthroughthepointofcontact.

•Thelengthsofthetwotangentsfromanexternalpointtoacircleareequal.

Arearelatedtocircle:•AreaofaCircle =πr2•PerimeterofaCircle =2πr•Areaofsector =θ/360°(πr2)•Lengthofanarc =θ/360°(2πr)•Areaofring =π(R2–r2)•Areaofsegment=Areaofthecorrespondingsector–Areaofthecorrespondingtriangle

=!!

!( !"!"#!

− 𝑠𝑖𝑛𝜃)Where,𝜃 isthecentralangleindegrees.•Distancemovedbyawheelinonerevolution=Circumferenceofthewheel.•Angledescribedbyminutehandin60minutes=3600

•Angledescribedbyminutehandin1minute=!"#!

!"!=60

•Numberofrevolutions=!"#$% !"#$%&'( !"#$%

!"#$%&'(#()$( !" !"# !"##$

Surfaceareasandvolume:

CylinderVolumeofacylinder =πr2hCurvedsurfacearea =2πrhTotalsurfacearea =2πrh+2πr2=2πr(h+r)Volumeofhollowcylinder=πR2h–πr2h=π(R2–r2)hTSAofhollowcylinder =OuterCSA+InnerCSA+2Areaofring=2πRh+2 πrh+2[πR2–πr2]

Cone

VolumeofaCone=!!πr2h

CSAofaCone =πrℓ(Here‘ℓ’refersto‘SlantHeight’)[whereℓ= ℎ! + 𝑟! ] TSAofaCone =πrℓ+πr2=πr(ℓ+r)

CLASSXFORMULAE MATHS

Frustum

Volumeofafrustum=!!πh[R2+r2+Rr]

CSAofafrustum=πℓ[R +r](Here‘ℓ’refersto‘Slantheight’)[whereℓ= ℎ! + (𝑅 − 𝑟)! ] TSAofafrustum=πℓ(R+r)+πr2+πR2=

SphereSurfaceareaofaSphere=4 πr2(IncaseofSphere,CSA=TSAi.e.theyaresame)

Volumeofhemisphere=!!πr3 [Takehalfthevolumeofasphere]

CSAofhemisphere=2πr2[TakehalftheSAofasphere]TSAofhemisphere=2πr2+πr2=3πr2

Volumeofasphere =!!πr3

Volumeofsphericalshell=Outervolume–Innervolume=!!π(R3–r3)

Whilesolvingtheproblemsbasedoncombinationofsolidsitwouldbebetterifyoutakecommon.•T.S.Aofcombinedsolid=C.S.Aofsolid1+C.S.Aofsolid2+C.S.Aofsolid3•Ifasolidismeltedand,recastintonumberofothersmallsolids,thenVolumeofthelargersolid=NoofsmallsolidsxVolumeofthesmallersolidForEx:Acylinderismeltedandcastintosmallerspheres.FindthenumberofspheresVolumeofCylinder=Noofsphere×Volumeofsphere•Ifan‘icecreamconewithhemisphericaltop’isgiventhenyouhavetotake(a)TotalVolume=VolumeofCone+VolumeofHemisphere(b)Surfacearea=CSAofCone+CSAofhemisphere

Statistics:

MeanThemeanforgroupeddatacanbefoundby:(i)Thedirectmethod:𝑋 = !"#"

!"

(ii)Theassumedmeanmethod:𝑋 = 𝑎 + !!!!!"

,where 𝑑! = 𝑥! − 𝑎

(iii)Thestepdeviationmethod: 𝑋 = 𝑎 + !!!!!"×ℎ, where𝑢! =

!!!!!

ModeThemodeforthegroupeddatacanbefoundbyusingtheformula:Mode=𝑙 + !!!!!

!!!!!!!!!×ℎ

l=lowerlimitofthemodalclass.f! = frequency of the modal class.𝑓! =frequencyoftheprecedingclassofthemodalclass.𝑓! =frequencyofthesucceedingclassofthemodalclass.h=sizeoftheclassinterval.

CLASSXFORMULAE MATHS

MedianThemedianforthegroupeddatacanbefoundbyusingtheformula:

Median= 𝑙 + !!!!"

! ×ℎ

l=lowerlimitofthemedianclass.n=numberofobservations.cf=cumulativefrequencyofclassintervalprecedingthemedianclass.f=frequencyofmedianclass.h=classsize.•EmpiricalFormula:Mode=3median–2mean.

Probability:

Probabilityofanevent:P(event)=!"#$%& !" !"#$%"&'( !"#$!%&'!"#$% !"#$%& !" !"#$!%&'

Inadeckofplayingcards,therearefourtypesofcards:♠(SpadesinBlackcolour)havingA,2,3,4,5,6,7,8,9,10,J,K,andQtotal13cards♣(ClubsinBlackcolour)havingA,2,3,4,5,6,7,8,9,10,J,K,andQtotal13cards♥(HeartsinRedcolour)havingA,2,3,4,5,6,7,8,9,10,J,K,andQtotal13cards♦(DiamondinRedcolour)havingA,2,3,4,5,6,7,8,9,10,J,K,andQtotal13cards

52cards •Jack,KingandQueenareknownas‘FaceCards’,asthesecardsarehavingsomepicturesonit.•AlwaysrememberAceisnotafacecardasitdoesn’tcarryanyfaceonit.