ncert solution
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CBSEClass–12Mathematics
NCERTsolution
Chapter-1
Relations&Functions-Exercise1.1
1.Determinewhethereachofthefollowingrelationsarereflexive,symmetricand
transitive:
(i)RelationRinthesetA={1,2,3,………..13,14}definedasR= .
(ii)RelationRinthesetNofnaturalnumbersdefinedasR={(x,y):y=x+5andx<4}
(iii)RelationRinthesetA={1,2,3,4,5,6}asR=
(iv)RelationRinthesetZofallintegersdefinedasR=
(v)RelationRinthesetAofhumanbeingsinatownataparticulartimegivenby:
(a)R=
(b)R=
(c)R=
(d)R=
(e)R=
Ans.(i)R= inA={1,2,3,4,5,6,……13,14}
ClearlyR={(1,3),(2,6),(3,9),(4,12)}
Since, R, Risnotreflexive.
Again Rbut R Risnotsymmetric.
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Also(1,3) Rand(3,9) Rbut(1,9) R, Risnottransitive.
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
(ii)R= insetNofnaturalnumbers.
ClearlyR={(1,6),(2,7),(3,8)}
Now R, Risnotreflexive.
Again Rbut R Risnotsymmetric.
Also(1,6) Rand(2,7) Rbut(1,7) R, Risnottransitive.
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
(iii)R= inA={1,2,3,4,5,6}
ClearlyR=(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(5,5),(6,6)
Now i.e.,(1,1),(2,2)and(3,3) R Risreflexive.
Again i.e.,(1,2) Rbut R Risnotsymmetric.
Also(1,4) Rand(4,4) Rand(1,4) R, Ristransitive.
Therefore,Risreflexiveandtransitivebutnotsymmetric.
(iv)R= insetZofallintegers.
Now i.e.,(1,1)=1–1=0 Z Risreflexive.
Again Rand R,i.e., and areaninteger Rissymmetric.
Also Zand Zand
R, Ristransitive.
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Therefore,Risreflexive,symmetricandtransitive.
(v)RelationRinthesetAofhumanbeinginatownataparticulartime.
(a)R=
Since R,because and workatthesameplace. Risreflexive.
Now,if Rand R,since and workatthesameplaceand and
workatthesameplace. Rissymmetric.
Now,if Rand R R. Ristransitive
Therefore,Risreflexive,symmetricandtransitive.
(b)R=
Since R,because and liveinthesamelocality. Risreflexive.
Also R Rbecause and liveinsamelocalityand and also
liveinsamelocality. Rissymmetric.
Again Rand R R Ristransitive.
Therefore,Risreflexive,symmetricandtransitive.
(c)R=
isnotexactly7cmtallerthan ,so R Risnotreflexive.
Also isexactly7cmtallerthan but isnot7cmtallerthan ,so Rbut
R Risnotsymmetric.
Now isexactly7cmtallerthan and isexactly7cmtallerthan thenitdoesnot
implythat isexactly7cmtallerthan Risnottransitive.
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
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(d)R=
isnotwifeof ,so R Risnotreflexive.
Also iswifeof but isnotwifeof ,so Rbut R Risnot
symmetric.
Also Rand Rthen R Risnottransitive.
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
(e)R=
isnotfatherof ,so R Risnotreflexive.
Also isfatherof but isnotfatherof ,
so Rbut R Risnotsymmetric.
Also Rand Rthen R Risnottransitive.
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
2.ShowthattherelationRinthesetRofrealnumbersdefinedasR=
isneitherreflexivenorsymmetricnortransitive.
Ans.R= ,RelationRisdefinedasthesetofrealnumbers.
(i)Whether R,then whichisfalse. Risnotreflexive.
(ii)Whether ,then and ,itisfalse. Risnotsymmetric.
(iii)Now , ,whichisfalse. Risnottransitive
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
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3.CheckwhethertherelationRdefinedintheset{1,2,3,4,5,6}asR=
isreflexive,symmetricortransitive.
Ans.R= R
NowR={(1,2),(2,3),(3,4),(4,5),(5,6)}and
(i)When whichisfalse,so R, Risnotreflexive.
(ii)Whether ,then and ,false Risnotsymmetric.
(iii)Nowif R, R R
(iv)Then and whichisfalse. Risnottransitive.
Therefore,Risneitherreflexive,norsymmetricandnortransitive.
4.ShowthattherelationRinRdefinedasR= ,isreflexiveand
transitivebutnotsymmetric.
Ans.(i) whichistrue,so R, Risreflexive.
(ii) but whichisfalse. Risnotsymmetric.
(iii) and whichistrue. Ristransitive.
Therefore,Risreflexiveandtransitivebutnotsymmetric.
5.CheckwhethertherelationRinRdefinedbyR= isreflexive,
symmetricortransitive.
Ans.(i)For whichisfalse. Risnotreflexive.
(ii)For and whichisfalse. Risnotsymmetric.
(iii)For and , whichisfalse. Risnottransitive.
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Therefore,Risneitherreflexive,norsymmetricandnortransitive.
6.ShowthattherelationinthesetA={1,2,3}givenbyR={(1,2),(2,1)}issymmetric
butneitherreflexivenortransitive.
Ans.R={(1,2),(2,1)},sofor ,(1,1) R. Risnotreflexive.
Alsoif then R Rissymmetric.
Now Rand thendoesnotimply R Risnottransitive..
Therefore,Rissymmetricbutneitherreflexivenortransitive.
7.ShowthattherelationRinthesetAofallthebooksinalibraryofacollege,givenby
R= isanequivalencerelation.
Ans.Books and havesamenumberofpages R Risreflexive.
If R R,so Rissymmetric.
Nowif R, R R Ristransitive.
SinceRisreflexive,symmetricandtransistive,therefore,Risanequivalencerelation.
8.ShowthattherelationRinthesetA={1,2,3,4,5}givenbyR=
isanequivalencerelation.Showthatalltheelementsof{1,3,
5}arerelatedtoeachotherandalltheelementsof{2,4}arerelatedtoeachother.But
noelementof{1,3,5}isrelatedtoanyelementof{2,4}.
Ans.A={1,2,3,4,5}andR= ,thenR={(1,3),(1,5),(3,5),(2,4)}
(a)For whichiseven. Risreflexive.
If iseven,then isalsoeven. Rissymmetric.
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Now,if and iseventhen isalsoeven. Ris
transitive.
Therefore,Risanequivalencerelation.
(b)Elementsof{1,3,5}arerelatedtoeachother.
Since allareevennumbers
Elementsof{1,3,5}arerelatedtoeachother.
Similarlyelementsof(2,4)arerelatedtoeachother.
Since anevennumber,thennoelementoftheset{1,3,5}isrelatedtoany
elementof(2,4).
Hencenoelementof{1,3,5}isrelatedtoanyelementof{2,4}.
9.ShowthateachoftherelationRinthesetA= givenby:
(i)R=
(ii)R= isanequivalencerelation.Findthesetofallelementsrelatedto1in
eachcase.
Ans.A={x∈Z:0≤x≤12}={0,1,2,3,4,5,6,7,8,9,10,11,12}
(i)R={(a,b):|a-b|isamultipleof4}
Foranyelementa∈A,wehave(a,a)∈Ras|a-a=0|isamultipleof4.
∴Risreflexive.Now,let(a,b)∈R⇒|a-b|isamultipleof4.
⇒|-(a-b)|⇒|b-a|isamultipleof4.
⇒(b,a)∈R∴Rissymmetric.
Now,let(a,b),(b,c)∈R.⇒|(a-b)|isamultipleof4and|(b-c)|isamultipleof4.
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⇒(a-b)isamultipleof4and(b-c)isamultipleof4.
⇒(a-c)=(a–b)+(b–c)isamultipleof4.
⇒|a-c|isamultipleof4.
⇒(a,c)∈R∴Ristransitive.Hence,Risanequivalencerelation.
Thesetofelementsrelatedto1is{1,5,9}since
|1-1|=0isamultipleof4,
|5-1|=4isamultipleof4,and
|9-1|=8isamultipleof4.
(ii)R={(a,b):a=b}
Foranyelementa∈A,wehave(a,a)∈R,sincea=a.∴Risreflexive.Now,let(a,b)∈R.⇒a=b⇒b=a⇒(b,a)∈R∴Rissymmetric.
Now,let(a,b)∈Rand(b,c)∈R.⇒a=bandb=c⇒a=c⇒(a,c)∈R∴Ristransitive.
Hence,Risanequivalencerelation.
TheelementsinRthatarerelatedto1willbethoseelementsfromsetAwhichareequalto1.
Hence,thesetofelementsrelatedto1is{1}.
10.Giveanexampleofarelation,whichis:
(i)Symmetricbutneitherreflexivenortransitive.
(ii)Transitivebutneitherreflexivenorsymmetric.
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(iii)Reflexiveandsymmetricbutnottransitive.
(iv)Reflexiveandtransitivebutnotsymmetric.
(v)Symmetricandtransitivebutnotreflexive.
Ans.(i)Therelation“isperpendicularto” isnotperpendicularto
If then ,howeverif and then isnotperpendicularto
SoitisclearthatR“isperpendicularto”isasymmetricbutneitherreflexivenor
transitive.
(ii)RelationR=
Weknowthat isfalse.Also but isfalseandif , this
implies
Therefore,Ristransitive,butneitherreflexivenorsymmetric.
(iii)“isfriendof”R=
Itisclearthat isfriendof Risreflexive.
Also isfriendof and isfriendof Rissymmetric.
Alsoif isfriendof and isfriendof then
cannotbefriendof Risnottransitive.
Therefore,Risreflexiveandsymmetricbutnottransitive.
(iv)“isgreaterorequalto”R=
Itisclearthat Risreflexive.
And doesnotimply Risnotsymmetric.
But , Ristransitive.
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Therefore,Risreflexiveandtransitivebutnotsymmetric.
(v)“isbrotherof”R=
Itisclearthat isnotthebrotherof Risnotreflexive.
Also isbrotherof and isbrotherof Rissymmetric.
Alsoif isbrotherof and isbrotherof then
canbebrotherof Ristransitive.
Therefore,Rissymmetricandtransitivebutnotreflexive.
11.ShowthattherelationRinthesetAofpointsinaplanegivenbyR={(P.Q):
distanceofthepointPfromtheoriginissameasthedistanceofthepointQfromthe
origin},isanequivalencerelation.Further,showthatthesetofallpointsrelatedtoa
pointP (0,0)isthecirclepassingthroughPwithoriginascentre.
Ans.PartI:R={(P,Q):distanceofthepointPfromtheoriginisthesameasthe
distanceofthepointQfromtheorigin}
LetP andQ andO(0,0).
OP=OQ = =
Now,For(P,P),OP=OP Risreflexive.
AlsoOP=OQandOQ=OP (P,Q)=(Q,P) R Rissymmetric.
AlsoOP=OQandOQ=OR OP=OQ Ristransitive.
Therefore,Risanequivalentrelation.
PartII:As = = (let) whichrepresentsacirclewith
centre(0,0)andradius
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12.ShowthattherelationRdefinedinthesetAofalltrianglesasR={(T1,T2):T1is
similartoT2},isanequivalencerelation.ConsiderthreerightangletrianglesT1with
sides3,4,5,T2withsides5,12,13andT3withsides6,8,10.WhichtrianglesamongT1,
T2andT3arerelated?
Ans.PartI:R={(T1,T2):T1issimilartoT2}andT1,T2aretriangle.
Weknowthateachtrianglesimilartoitselfandthus(T1,T2) R Risreflexive.
Alsotwotrianglesaresimilar,thenT1 T2 T1 T2 Rissymmetric.
Again,ifthenT1 T2andthenT2 T3 thenT1 T3 Ristransitive.
Therefore,Risanequivalentrelation.
PartII:ItisgiventhatT1,T2andT3arerightangledtriangles.
T1withsides3,4,5,T2withsides5,12,13and
T3withsides6,8,10
Since,twotrianglesaresimilarifcorrespondingsidesareproportional.
Therefore,
Therefore,T1andT3arerelated.
13.ShowthattherelationRdefinedinthesetAofallpolygonsasR={(P1,P2):P1and
P2havesamenumberofsides},isanequivalencerelation.Whatisthesetofall
elementsinArelatedtotherightangletriangleTwithsides3,4,and5?
Ans.PartI:R={(P1,P2):P1andP2havesamenumberofsides}
(i)Considertheelement(P1,P2),itshowsP1andP2havesamenumberofsides.
Therefore,Risreflexive.
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(ii)If(P1,P2) Rthenalso(P2,P1) R
(P1,P2)=(P2,P1)asP1andP2havesamenumberofsides,therefore,Rissymmetric.
(iii)If(P1,P2) Rand(P2,P3) Rthenalso(P1,P3) RasP1,P2andP3havesame
numberofsides,therefore,Ristransitive.
Therefore,Risanequivalentrelation.
PartII:weknowthatif3,4,5arethesidesofatriangle,thenthetriangleisrightangled
triangle.Therefore,thesetAisthesetofrightangledtriangle.
14.LetLbethesetofalllinesinXYplaneandRbetherelationinLdefinedasR={(L1,
L2):L1isparalleltoL2}.ShowthatRisanequivalencerelation.Findthesetofalllines
relatedtotheline
Ans.PartI:R={(L1,L2):L1isparalleltoL2}
(i)ItisclearthatL1 L1i.e.,(L1,L1) R Risreflexive.
(ii)IfL1 L2andL2 L1then(L1,L2) R Rissymmetric.
(iii)IfL1 L2andL2 L3 L1 L3 Ristransitive.
Therefore,Risanequivalentrelation.
PartII:Allthelinesrelatedtotheline and where isareal
number.
15.LetRbetherelationintheset{1,2,3,4}givenbyR={(1,2),(2,2),(1,1),(4,4),(1,3),
(3,3),(3,2)}.Choosethecorrectanswer:
(A)Risreflexiveandsymmetricbutnottransitive.
(B)Risreflexiveandtransitivebutnotsymmetric.
(C)Rissymmetricandtransitivebutnotreflexive.
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(D)Risanequivalencerelation.
Ans.LetRbetherelationintheset{1,2,3,4}isgivenby
R={(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}
(a)(1,1),(2,2),(3,3),(4,4) R Risreflexive.
(b)(1,2) Rbut(2,1) R Risnotsymmetric.
(c)If(1,3) Rand(3,2) Rthen(1,2) R Ristransitive.
Therefore,option(B)iscorrect.
16.LetRbetherelationinthesetNgivenbyR= Choosethe
correctanswer:
(A)(2,4) R
(B)(3,8) R
(C)(6,8) R
(D)(8,7) R
Ans.Given:
(A) ,Here isnottrue,therefore,thisoptionisincorrect.
(B) and 3=6,whichisfalse.
Therefore,thisoptionisincorrect.
(C) and 6=6,whichistrue.
Therefore,option(C)iscorrect.
(D) and 8=5,whichisfalse.
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CBSEClass–12Mathematics
NCERTsolution
Chapter-1
Relations&Functions-Exercise1.2
1. Show that the function defined by is one-one and onto,
where isthesetofallnon-zerorealnumbers.Istheresulttrue,ifthedomain is
replacedbyNwithco-domainbeingsameas ?
Ans.
PartI: and
If then
isone-one.
isonto.
Part II: When domain R is replaced by N, co-domain R remaining the same, then,
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If
where N
isone-one.
Buteveryrealnumberbelongingtoco-domainmaynothaveapre-imageinN.
e.g. N isnotonto.
2.Checktheinjectivityandsurjectivityofthefollowingfunctions:
(i) givenby
(ii) givenby
(iii) givenby
(iv) givenby
(v) givenby
Ans.(i) givenby
If then
isinjective.
Therearesuchnumbersofco-domainwhichhavenoimageindomainN.
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e.g.3 co-domainN,butthereisnopre-imageindomainof
therefore isnotonto. isnotsurjective.
(ii) givenby
Since,Z= therefore,
and1havesameimage. isnotinjective.
Therearesuchnumbersofco-domainwhichhavenoimageindomainZ.
e.g.3 co-domain,but domainof isnotsurjective.
(iii) givenby
As
and1havesameimage. isnotinjective.
e.g. co-domain,but domainRof isnotsurjective.
(iv) givenby
If then
i.e.,forevery N,hasauniqueimageinitsco-domain. isinjective.
There are many such members of co-domain of which do not have pre-image in its
domaine.g.,2,3,etc.
Therefore isnotonto. isnotsurjective.
(v) givenby
If then
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i.e.,forevery Z,hasauniqueimageinitsco-domain. isinjective.
There are many such members of co-domain of which do not have pre-image in its
domain.
Therefore isnotonto. isnotsurjective.
3. Prove that the Greatest integer Function , given by is
neitherone-onenoronto,where denotesthegreatestintegerlessthanorequalto
Ans.Function ,givenby
and
isnotone-one.
Alltheimagesof Rbelongtoitsdomainhaveintegersastheimagesinco-domain.But
no fraction proper or improper belonging to co-domain of has any pre-image in its
domain.
Therefore, isnotonto.
4.Showthat theModulusFunction ,givenby isneitherone-
onenoronto,whereis if ispositiveor0and is if isnegative.
Ans.ModulusFunction ,givenby
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Now
contains
Thusnegativeintegersarenotimagesofanyelement. isnotone-one.
AlsosecondsetRcontainssomenegativenumberswhicharenotimagesofanyrealnumber.
isnotonto.
5. Show that the Signum Function , given by is
neitherone-onenoronto.
Ans.SignumFunction ,givenby
for
isnotone-one.
Except noothermembersofco-domainof hasanypre-imageitsdomain.
isnotonto.
Therefore, isneitherone-onenoronto.
6.LetA={1,2,3},B={4,5,6,7}andlet ={(1,4),(2,5),(3,6)}beafunctionfromAto
B.Showthat isone-one.
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Ans.A={1,2,3},B={4,5,6,7}and ={(1,4),(2,5),(3,6)}
Here, and
Here,alsodistinctelementsofAhavedistinctimagesinB.
Therefore, isaone-onefunction.
7. In each of the following cases, state whether the function is one-one, onto or
bijective.Justifyyouranswer.
(i) definedby
(ii) definedby
Ans.(i) definedby
Now,if R,then and
Andif ,then isone-one.
Again,ifeveryelementofY(–R)isimageofsomeelementofX(R)under i.e.,forevery
Y,thereexistsanelement inXsuchthat
Now
isontoorbijectivefunction.
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(ii) definedby
Now,if R,then and
Andif ,then
isnotone-one.
Again,ifeveryelementofY(–R)isimageofsomeelementofX(R)under i.e.,forevery
Y,thereexistsanelement inXsuchthat
Now,
isnotonto.
Therefore, isnotbijective.
8.LetAandBbesets.Showthat :AxB BxAsuch that isa
bijectivefunction.
Ans.Injectivity:Let and AxBsuchthat
and
=
= forall AxB
So, isinjective.
Surjectivity:Let beanarbitraryelementofBxA.Then Band A.
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AxB
Thus,forall BxA,theirexists AxBsuchthat
So, AxB BxAisanontofunction,therefore isbijective.
9.Let bedefinedby forall N.
Statewhetherthefunction isbijective.Justifyyouranswer.
Ans. bedefinedby
(a) and
Theelements1,2,belongingtodomainof havethesameimage1initsco-domain.
So, isnotone-one,therefore, isnotinjective.
(b)Everynumberofco-domainhaspre-imageinitsdomaine.g.,1hastwopre-images1and
2.
So, isonto,therefore, isnotbijective.
10. Let A = R – {3} and B = R – {1}. Consider the function : A B defined by
Is one-oneandonto?Justifyyouranswer.
Ans.A=R–{3}andB=R–{1}and
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Let A,then and
Now,for
isone-onefunction.
Now
=
Therefore, isanontofunction.
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11.Let bedefinedas Choosethecorrectanswer:
(A) isone-oneonto
(B) ismany-oneonto
(C) isone-onebutnotonto
(D) isneitherone-onenoronto
Ans. and
Let ,then and
Therefore, isnotone-onefunction.
Now,
and
Therefore, isnotontofunction.
Therefore,option(D)iscorrect.
12.Let bedefinedas Choosethecorrectanswer:
(A) isone-oneonto
(B) ismany-oneonto
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(C) isone-onebutnotonto
(D) isneitherone-onenoronto
Ans.Let Rsuchthat
Therefore, isone-onefunction.
Now,consider R(co-domainof )certainly R(domainof )
Thusforall R(co-domainof )thereexists R(domainof )suchthat
Therefore, isontofunction.
Therefore,option(A)iscorrect.
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CBSEClass–12MathematicsNCERTsolutionChapter-1
Relations&Functions-Exercise1.3
1.Let :{1,3,4} {1,2,5}and :{1,2,5} {1,3}begivenby ={(1,2),(3,5),(4,1)}and ={(1,3),(2,3),(5,1)}.Writedown
Ans. ={(1,2),(3,5),(4,1)}and ={(1,3),(2,3),(5,1)}
Now, and
and
Hence, {(1,3),(3,1),(4,3)}
2.Let and befunctionsfromR R.Showthat:
Ans.(a)Toprove:
L.H.S.= = = =R.H.S.
(b)Toprove:
L.H.S.= = = =R.H.S.
3.Find and ,if:
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(i) and
(ii) and
Ans.Tofind: and
(i) and
and = =
(ii) and
and =
4. If show that for all What is the
inverseof
Ans.Given:
L.H.S.= = = =
= =R.H.S.
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Now,
Henceinverseof
5.Statewithreasonwhetherfollowingfunctionshaveinverse:
(i) :{1,2,3,4} {10}with ={(1,10),(2,10),(3,10),(4,10)}
(ii) :{5,6,7,8} {1,2,3,4}with ={(5,4),(6,3),(7,4),(8,2)}
(iii) :{2,3,4,5} {7,9,11,13}with ={(2,7),(3,9),(4,11),(5,13)}
Ans.(i) ={(1,10),(2,10),(3,10),(4,10)}
Itismany-onefunction,therefore hasnoinverse.
(ii) ={(5,4),(6,3),(7,4),(8,2)}
Itismany-onefunction,therefore hasnoinverse.
(iii) ={(2,7),(3,9),(4,11),(5,13)}
isone-oneontofunction,therefore, hasaninverse.
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6.Showthat Rgivenby isone-one.Findtheinverse
ofthefunction Range
Ans.PartI: Rgivenby
Let ,then and
When then
isone-one.
PartII:Let Rangeof
forsome in
As
isonto.
Therefore,
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7.Consider :R Rgivenby Showthat isinvertible.Findtheinverseof [Hint: ]
Ans.Consider :R Rgivenby
Let R,then and
Now,for ,then isone-one.
Let Rangeof
isonto.
Therefore, isinvertibleandhence, .
8.Consider givenby Showthat is invertiblewiththeinverse of givenby where isthesetofallnon-negativerealnumbers.
Ans.Consider and
Let R ,then and
isone-one.
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Now
as
isonto.
Therefore, isinvertibleand .
9. Consider given by Show that is
invertiblewith
Ans.Consider and
Let R ,then and
Now, then
isone-one.
Now,again
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= = =
=
= isonto.
Therefore, isinvertibleand .
10.Let beaninvertiblefunction.Showthat hasuniqueinverse.
(Hint: Suppose and are two inverses of Then for allUseone-onenessof ).
Ans.Given: beaninvertiblefunction.
Thus is1–1andontoandtherefore exists.
Let and betwoinversesof Thenforall Y,
Theinverseisuniqueandhence hasauniqueinverse.
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11.Consider :{1,2,3} givenby and
Find andshowthat
Ans. ,thenitisclearthat is1–1andontoandtherefore
exists.
Also, and
Hence,
12.Let beaninvertiblefunction.Showthattheinverseof is
,i.e.,
Ans.Let beaninvertiblefunction.
Then isone-oneandonto
Xwhere isalsoone-oneandontosuchthat
and
Now, and
13.If :R Rgivenby then is:
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(A)
(B)
(C)
(D)
Ans. :R Rand
=
= = =
Therefore,option(C)iscorrect.
14.Let :R– Rbeafunctiondefinedas Theinverseof
isthemap :Rangeof givenby:
(A)
(B)
(C)
(D)
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Ans.Given: :R– Rand
Now,Rangeof
Let
f-1(y)=g(y)=
Therefore,option(B)iscorrect.
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CBSEClass–12Mathematics
NCERTsolution
Chapter-1
Relations&Functions-Exercise1.4
1. Determine whether or not each of the definition of * given below gives a binary
operation.Intheeventthat*isnotabinaryoperation,givenjustificationforthis.
(i)On define*by
(ii)On define*by
(iii)OnR,define*by
(iv)On define*by
(v)On define*by
Ans.(i)On ={1,2,3,…..},
Let
Therefore,operation*isnotabinaryoperationon .
(ii)On ={1,2,3,…..},
Let
Therefore,operation*isabinaryoperationon .
(iii)onR(setofrealnumbers)
Let R
Therefore,operation*isabinaryoperationonR.
(iv)On ={1,2,3,…..},
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Let
Therefore,operation*isabinaryoperationon .
(v)On ={1,2,3,…..},
Let
Therefore,operation*isabinaryoperationon .
2.Foreachbinaryoperation*definedbelow,determinewhether*iscommutativeor
associative:
(i)On define
(ii)OnQ,define
(iii)OnQ,define
(iv)On define
(v)On define
(vi)OnR–{–1},define
Ans.(i)Forcommutativity: and =
Forassociativity: =
Also, =
Therefore,theoperation*isneithercommutativenorassociative.
(ii)Forcommutativity: and
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Forassociativity: =
Also, =
Therefore,theoperation*iscommutativebutnotassociative.
(iii)Forcommutativity: and =
Forassociativity: =
Also, =
Therefore,theoperation*iscommutativeandassociative.
(iv)Forcommutativity: and
Forassociativity: =
Also, =
Therefore,theoperation*iscommutativebutnotassociative.
(v)Forcommutativity: and
Forassociativity: =
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Also, =
Therefore,theoperation*isneithercommutativenorassociative.
(vi)Forcommutativity: and
Forassociativity: =
Also, =
Therefore,theoperation*isneithercommutativenorassociative.
3. Consider the binary operation on the set {1, 2, 3, 4, 5} defined by
Writetheoperationtableoftheoperation
Ans.LetA={1,2,3,4,5}definedby i.e.,minimumof and
1 2 3 4 5
1 1 1 1 1 1
2 1 2 2 2 2
3 1 2 3 3 3
4 1 2 3 4 4
5 1 2 3 4 5
4. Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following
multiplicationtable(table1.2).
(i)Compute(2*3)*4and2*(3*4)
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(ii)Is*commutative?
(iii)Compute(2*3)*(4*5)
(Hint:Usethefollowingtable)
Table1.2
* 1 2 3 4 5
1 1 1 1 1 1
2 1 2 1 2 1
3 1 1 3 1 1
4 1 2 1 4 1
5 1 1 1 1 5
Ans.(i)2*3=1and3*4=1
Now(2*3)*4=1*4=1and2*(3*4)=2*1=1
(ii)2*3=1and3*4=1
2*3=3*2andotherelementofthegivenset.
Hencetheoperationiscommutative.
(iii)(2*3)*(4*5)=1*1=1
5.Let*’bethebinaryoperationontheset{1,2,3,4,5}definedby H.C.F.of
and Istheoperation*’sameastheoperation*definedinExercise4above?Justify
youranswer.
Ans.LetA={1,2,3,4,5}and H.C.F.of and
*’ 1 2 3 4 5
1 1 1 1 1 1
2 1 2 1 2 1
3 1 1 3 1 1
4 1 2 1 4 1
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5 1 1 1 1 5
Weobservethattheoperation*'isthesameastheoperation*inExp.4.
6.Let*bethebinaryoperationonNgivenby L.C.M.of and Find:
(i)5*7,20*16
(ii)Is*commutative?
(iii)Is*associative?
(iv)Findtheidentityof*inN.
(v)WhichelementsofNareinvertiblefortheoperation*?
Ans. L.C.M.of and
(i)5*7=L.C.M.of5and7=35
20*16=L.C.M.of20and16=80
(ii) L.C.M.of and =L.C.M.of and =
Therefore,operation*iscommutative.
(iii) =
=
Similarly,
Thus,
Therefore,theoperationisassociative.
(iv)Identityof*inN=1because =L.C.M.of and1=
(v)Onlytheelement1inNisinvertiblefortheoperation*because
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7.Is*definedontheset{1,2,3,4,5}by L.C.M.of and abinaryoperation?
Justifyyouranswer.
Ans.LetA={1,2,3,4,5}and L.C.M.of and
* 1 2 3 4 5
1 1 2 3 4 5
2 2 2 6 4 10
3 3 x 3 12 15
4 4 4 12 4 20
5 5 x 15 20 5
Here,2*3=6 A
Therefore,theoperation*isnotabinaryoperation.
8. Let * be the binary operation on N defined by H.C.F. of and Is *
commutative?Is*associative?DoesthereexistidentityforthisbinaryoperationonN?
Ans. H.C.F.of and
(i) H.C.F.of and =H.C.F.of and =
Therefore,operation*iscommutative.
(ii) =
=
Therefore,theoperationisassociative.
Therefore,theredoesnotexistanyidentityelement.
9.Let*beabinaryoperationonthesetQofrationalnumbersasfollows:
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(i) (ii)
(iii) (iv)
(v) (vi)
Findwhichofthebinaryoperationsarecommutativeandwhichareassociative.
Ans.(i) operation*isnotcommutative.
And
Here, operation*isnotassociative.
(ii) operation*iscommutative.
And
Here, operation*isnotassociative.
(iii) and
Therefore,operation*isnotcommutative.
And
Here, operation*isnotassociative.
(iv) operation*iscommutative.
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And
Here, operation*isnotassociative.
(v) operation*iscommutative.
And
Here, operation*isassociative.
(vi) and operation*isnotcommutative.
And
Here, operation*isnotassociative.
10.Showthatnoneoftheoperationsgivenabovetheidentity.
Ans.LettheidentitybeI.
(i)
(ii)
(iii)
(iv)
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(v)
(vi)
Therefore,noneoftheoperationsgivenabovehasidentity.
11. Let A = N x N and * be the binary operation on A defined by
Showthat*iscommutativeandassociative.Findtheidentityelementfor*onA,ifany.
Ans.A=NxNand*isabinaryoperationdefinedonA.
The operation is
commutative
Again,
And
Here, Theoperationisassociative.
Letidentityfunctionbe ,then
Foridentityfunction
Andfor
As0 N,therefore,identity-elementdoesnotexist.
12.Statewhetherthefollowingstatementsaretrueorfalse.Justify:
(i)Foranarbitrarybinaryoperation*onasetN,
(ii)If*isacommutativebinaryoperationonN,then
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Ans.(i)*beingabinaryoperationonN,isdefinedas
Henceoperation*isnotdefined,therefore,thegivenstatementisfalse.
(ii)*beingabinaryoperationonN.
Thus, ,thereforethegivenstatementistrue.
13.Considerabinaryoperation *onNdefinedas .Choose the correct
answer:
(A)Is*bothassociativeandcommutative?
(B)Is*commutativebutnotassociative?
(C)Is*associativebutnotcommutative?
(D)Is*neithercommutativenorassociative?
Ans. Theoperation*iscommutative.
Again,
And
Theoperation*isnotassociative.
Therefore,option(B)iscorrect.
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CBSEClass–12Mathematics
NCERTsolution
Chapter-1
Relations&Functions-MiscellaneousExercise
1. Let be defined as Find the function
suchthat
Ans.Given:
Now and
2.Let bedefinedas if isoddand if is
even.Showthat isinvertible.Findtheinverseof Here,Wisthesetofallwhole
numbers.
Ans.Given: definedas
Injectivity:Let beanytwooddrealnumbers,then
Again,let beanytwoevenwholenumbers,then
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Is isevenand isodd,then
Also,if oddand iseven,then
Hence,
isaninjectivemapping.
Surjectivity:Let beanarbitrarywholenumber.
If isanoddnumber,thenthereexistsanevenwholenumber suchthat
If isanevennumber,thenthereexistsanoddwholenumber suchthat
Therefore,every Whasitspre-imageinW.
So, isasurjective.Thus isinvertibleand exists.
For :
and
Hence,
3.If isdefinedby find
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Ans.Given:
=
4. Show that the function defined by
Risone-oneandontofunction.
Ans.Itisgiventhatf: isdefinedasf(x)=
Forone-one
Supposef(x)=f(y),wherex,y R
Itcanbeobservedthatifxispositiveandyisnegative.
Then,wehave
Since,xispositiveandyisnegative,
x>y ,but,2xyisnegative
Then,2xy x-y.
Thus,thecaseofxbeingpositiveandybeingnegativecanberuledout.
Underasimilarargument,xbeingnegativeandybeingpositivecanalsoberuledout.
xandyhavetobeeitherpositiveornegative.
Whenxandyarebothpositive,wehave
f(x)=f(y)
Whenxandyarebothnegative,wehave
f(x)=f(y)
fisone-one
ForOnto
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Now,lety Rsuchthat-1<y<1.
Ifyisnegativethen,thereexistsx= suchthat
f(x)=
Ifyispositive,then,thereexistsx= suchthat
f(x)=
Therefore,fisonto.Hence,fisone-oneandonto.
5.Showthatthefunction givenby isinjective.
Ans.Let Rbesuchthat
Therefore, isone-onefunction,hence isinjective.
6. Give examples of two functions and such that is
injectivebut isnotinjective.
(Hint:Consider and )
Ans.Given:twofunctions and
Let and
Therefore, isinjectivebut isnotinjective.
7. Give examples of two functions and such that is
ontobut isnotonto.
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(Hint:Consider and )
Ans.Let
Thesearetwoexamplesinwhich isontobut isnotonto.
8.GivenanonemptysetX,considerP(X)whichisthesetofallsubsetsofX.
DefinetherelationRinP(X)asfollows:
ForsubsetsA,BinP(X),ARBifandonlyifA B.IsRanequivalencerelationonP(X)?
Justifyyouranswer.
Ans.(i)A A Risreflexive.
(ii)A B B A Risnotcommutative.
(iii)IfA B,B C,thenA C Ristransitive.
Therefore,Risnotequivalentrelation.
9.Givenanon-emptysetX,considerthebinaryoperation*:P(X)xP(X) P(X)given
byA*B=A B A,BinP(X),whereP(X)isthepowersetofX.ShowthatXisthe
identityelement for thisoperationandX is theonly invertibleelement inP (X)with
respecttotheoperation*.
Ans.LetSbeanon-emptysetandP(S)beitspowerset.LetanytwosubsetsAandBofS.
A B S
A B P(S)
Therefore, isanbinaryoperationonP(S).
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Similarly,ifA,B P(S)andA–B P(S),thentheintersectionofsets anddifferenceof
setsarealsobinaryoperationonP(S)andA S=A=S AforeverysubsetAofsets
A S=A=S AforallA P(S)
Sistheidentityelementforintersection onP(S).
10.Findthenumberofallontofunctionsfromtheset{1,2,3,……., }toitself.
Ans.The number of onto functions that can be defined from a finite set A containing
elementsontoafinitesetBcontaining elements=
11.LetS= andT={1,2,3}.Find ofthefollowingfunctionsFfromStoT,if
itexists.
(i)F=
(ii)F=
Ans.S= andT={1,2,3}
(i)F=
(ii)
Fisnotone-onefunction,sinceelement and havethesameimage1.
Therefore,Fisnotone-onefunction.
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12. Consider the binary operation * : R x R R and o = R x R R defined as
and R.Showthat*iscommutativebutnotassociative,
o is associative but not commutative. Further, show that R,
[Ifitisso,wesaythattheoperation*distributesoverthe
operationo].Doesodistributeover*?Justifyyouranswer.
Ans.PartI: also operation*iscommutative.
Now,
And
Here, operation*isnotassociative.
PartII: R
And,
operation isnotcommutative.
Now and
Here operation isassociative.
PartIII:L.H.S. =
R.H.S. = =L.H.S.Proved.
Now,anotherdistributionlaw:
L.H.S.
R.H.S.
AsL.H.S. R.H.S.
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Therefore,theoperation doesnotdistributeover.
13.Givenanon-emptysetX,let*:P(X)xP(X) P(X)bedefinedasA*B=(A–B)
(B–A), A,B P(X).Showthattheemptyset istheidentityfortheoperation*
andalltheelementsAofP(X)areinvertiblewithA-1=A.(Hint:
and )
Ans.ForeveryA P(X),wehave
=
And =
istheidentityelementfortheoperation*onP(X).
AlsoA*A=(A–A) (A–A)=
EveryelementAofP(X)isinvertiblewith =A.
14. Define binary operation * on the set {0, 1, 2, 3, 4, 5} as
Show that zero is the identity for this operation and each element of the set is
invertiblewith beingtheinverseof
Ans.Abinaryoperation(orcomposition)*ona(non-empty)setisafunction*:AxA A.
Wedenote by foreveryorderedpair AxA.
Abinaryoperationonano-emptysetAisarulethatassociateswitheveryorderedpair
ofelements (distinctorequal)ofAsomeuniqueelement ofA.
* 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
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2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 0 1 2 3 4
Forall A,wehave (mod6)=0
And and
0istheidentityelementfortheoperation.
Alsoon0=0–0=0*
2*1=3=1*2
15.LetA={–1,0,1,2},B={–4,–2,0,2}and bethefunctionsdefinedby
Aand A.Are and equal?Justifyyour
answer.
(Hint: One may note that two functions and such that
A,arecalledequalfunctions).
Ans.When then and
At and
At and
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At and
Thusforeach A,
Therefore, and areequalfunction.
16. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are
reflexiveandsymmetricbutnottransitiveis:
(A)1
(B)2
(C)3
(D)4
Ans.Itisclearthat1isreflexiveandsymmetricbutnottransitive.
Therefore,option(A)iscorrect.
17.LetA={1,2,3}.Thennumberofequivalencerelationscontaining(1,2)is:
(A)1
(B)2
(C)3
(D)4
Ans.2
Therefore,option(B)iscorrect.
18.Let betheSignumFunctiondefinedas and
betheGreatestFunctiongivenby where isgreatestintegerlessthan
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orequalto Then,does and coincidein(0,1)?
Ans.Itisclearthat and
Consider whichlieon(0,#1)
Now,
And
in(0,1]
No,fogandgofdon'tcoincidein(0,1].
19.Numberofbinaryoperationontheset are:
(A)10
(b)16
(C)20
(D)8
Ans.A=
AxA=
=4
Numberofsubsets= =16
Hencenumberofbinaryoperationis16.
Therefore,option(B)iscorrect.
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CBSEClass–12Mathematics
NCERTsolution
Chapter-2
InverseTrigonometricFunctions-Exercise2.1
Findtheprincipalvaluesofthefollowing:
1.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
2.
Ans.Let
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Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
3.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
4.
Ans.Let
Since,theprincipalvaluebranchof is
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Therefore,Principalvalueof is
5.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
6.
Ans.Let
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Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
7.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
8.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
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9.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
10.
Ans.Let
Since,theprincipalvaluebranchof is
Therefore,Principalvalueof is
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Findthevalueofthefollowing:
11.
Ans.
=
=
= =
12.
Ans.
=
=
=
13.If then:
A)
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(B)
(C)
(D)
Ans.Bydefinitionofprincipalvaluefor
Therefore,option(B)iscorrect.
14. isequalto:
(A)
(B)
(C)
(D)
Ans.
=
=
=
Therefore,option(B)iscorrect.
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CBSEClass–12Mathematics
NCERTsolution
Chapter-2
InverseTrigonometricFunctions-Exercise2.2
Provethefollowing:
1.
Ans.Weknowthat:
Putting
Putting ,
Proved.
2.
Ans.Weknowthat:
Putting
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Putting ,
Proved.
3.
Ans.L.H.S.=
=
=
=
=
=R.H.S.
Proved.
4.
Ans.L.H.S.=
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=
=
=
=
= =R.H.S.
Proved.
Writethefollowingfunctionsinthesimplestform:
5.
Ans.Putting sothat
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=
=
=
=
=
=
=
6.
Ans.Putting sothat
=
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Ans.
Dividingthenumeratoranddenominatorby
=
=
=
=
9.
Ans.Putting sothat
=
=
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=
=
=
10.
Ans.
= [Dividingnumeratoranddenominatorby ]
Putting sothat
=
=
= =
Findthevaluesofeachofthefollowing:
11.
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Ans.
=
=
=
=
=
=
12.
Ans.
=
13.
Ans.Putting and
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=
=
=
=
=
=
14.If thenfindthevalueof
Ans.Given:
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15.If thenfindthevalueof
Ans.Given:
FindthevaluesofeachoftheexpressionsinExercises16to18.
16.
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Ans.
=
=
=
=
17.
Ans.
=
=
=
=
=
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18.
Ans.Putting and sothat and
Now, =
And
and
=
= =
= =
19. isequalto:
(A)
(B)
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(C)
(D)
Ans.
=
=
Therefore,option(B)iscorrect.
20. isequalto:
(A)
(B)
(C)
(D)1
Ans.
=
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=
=
=
Therefore,option(D)iscorrect.
21. isequalto:
(A)
(B)
(C)
(D)
Ans.
=
=
=
=
=
=
Therefore,option(B)iscorrect.
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CBSEClass–12Mathematics
NCERTsolution
Chapter-2
InverseTrigonometricFunctions-MiscellaneousExercise
Findthevalueofthefollowing:
1.
Ans.
=
=
=
= =
2.
Ans.
=
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=
=
= =
3.Provethat:
Ans.Let sothat
Since
= = =
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4.Provethat:
Ans.Let sothat
Again,Let sothat
Since
=
=
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5.Provethat:
Ans.Let sothat
Again,Let sothat
Since =
=
6.Provethat:
Ans.Let sothat
Again,Let sothat
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Since =
=
7.Provethat:
Ans.Let sothat
Again,Let sothat
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Since =
=
8.Provethat:
Ans.L.H.S.=
=
=
=
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=
=
=
R.H.S.
9.Provethat:
Ans.Let sothat
=
=
=
=
10.Provethat:
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Ans.Weknowthat =
Again,
=
L.H.S.=
=
=
=
11.Provethat:
Ans.Putting sothat
L.H.S.=
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=
=
=
Dividingeverytermby
=
=
= =
= =R.H.S.
12.Provethat:
Ans.L.H.S.=
=
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= …(i)
Now,let sothat
Fromeq.(i), =R.H.S.
13.Solvetheequation:
Ans.
14.Solvetheequation:
Ans.Putting
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(B)
(C)
(D)
Ans.Let where sothat
Putting
Therefore,option(D)iscorrect.
16. then isequalto:
(A)
(B)
(C)0
(D)
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Ans.Putting sothat
or
or
But doesnotsatisfythegivenequation.
Therefore,option(C)iscorrect.
17. isequalto:
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(A)
(B)
(C)
(D)
Ans.
=
=
=
=
=
=
Therefore,option(C)iscorrect.