Sets
Question1
LetS={1,2,3,...,10}.SupposeMisthesetofallthesubsetsofS,then
therelationR={(A,B):A∩B≠φ;A,B∈M}is:
[27-Jan-2024Shift1]
Options:
A.
symmetricandreflexiveonly
B.
reflexiveonly
C.
symmetricandtransitiveonly
D.
symmetriconly
Answer:D
Solution:
Question2
LetRbearelationonZ×Zdefinedby(a,b)R(c,d)ifandonlyifad-
bcisdivisibleby5.ThenRis
[29-Jan-2024Shift1]
Options:
A.
Reflexiveandsymmetricbutnottransitive
B.
Reflexivebutneithersymmetricnottransitive
C.
Reflexive,symmetricandtransitive
D.
Reflexiveandtransitivebutnotsymmetric
Answer:A
Solution:
Question3
IfRisthesmallestequivalencerelationontheset{1,2,3,4}such
that{(1,2),(1,3)}⊂R,thenthenumberofelementsinRis
[29-Jan-2024Shift2]
Options:
A.
10
B.
12
C.
8
D.
15
Answer:A
Solution:
Question4
Agroupof40studentsappearedinanexaminationof3subjects-
Givenset{1,2,3,4}
Minimumorderpairsare
(1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),(1,3),(1,2)
Thusno,ofelements=10
Mathematics,Physics&Chemistry.Itwasfoundthatallstudentspassed
inatleastoneofthesubjects,20studentspassedinMathematics,25
studentspassedinPhysics,16studentspassedinChemistry,atmost11
studentspassedinbothMathematicsandPhysics,atmost15students
passedinbothPhysicsandChemistry,atmost15studentspassedin
bothMathematicsandChemistry.Themaximumnumberofstudents
passedinallthethreesubjectsis____
[30-Jan-2024Shift1]
Answer:10
Solution:
Question5
Thenumberofsymmetricrelationsdefinedontheset{1,2,3,4}which
arenotreflexiveis____
[30-Jan-2024Shift2]
Answer:960
Solution:
Question6
LetA={1,2,3,4}andR={(1,2),(2,3),(1,4)}bearelationonA.
LetSbetheequivalencerelationonAsuchthatR⊂Sandthenumberof
elementsinSisn.Then,theminimumvalueofnis____
[31-Jan-2024Shift1]
Answer:16
Solution:
Allelementsareincluded
Answeris16
Question7
LetA={1,2,3,........100}.LetRbearelationonAdefinedby(x,y)∈
Rifandonlyif2x=3y.LetR1beasymmetricrelationonAsuch
thatR⊂R1andthenumberofelementsinR1isn.Then,theminimum
valueofnis
[31-Jan-2024Shift2]
Answer:66
Solution:
Question8
LetA={1,2,3,...20}.LetR1andR2tworelationonAsuchthat
R1={(a,b):bisdivisiblebya}
R2={(a,b):aisanintegralmultipleofb}.
Then,numberofelementsinR1−R2isequalto_____
[1-Feb-2024Shift1]
Answer:46
Solution:
Question9
Thenumberofelementsintheset
S={(x,y,z):x,y,z∈Z,x+2y+3z=42,x,y,z≥0}equals____
[1-Feb-2024Shift1]
Answer:169
Solution:
Question10
ConsidertherelationsR1andR2definedasaR1b⇔a2+b2=1forall
a,b,∈Rand(a,b)R2(c,d)⇔a+d=b+cforall(a,b),(c,d)∈N×N.
Then
[1-Feb-2024Shift2]
Options:
A.
OnlyR1isanequivalencerelation
B.
OnlyR2isanequivalencerelation
C.
R1andR2bothareequivalencerelations
D.
NeitherR1norR2isanequivalencerelation
Answer:B
Solution:
Question11
Answer:13
Solution:
Question12
Inagroupof100persons75speakEnglishand40speakHindi.Each
personspeaksatleastoneofthetwolanguages.Ifthenumberof
persons,whospeakonlyEnglishisαandthenumberofpersonswho
speakonlyHindiisβ,thentheeccentricityoftheellipse25(β2x2+α2y2)
=α2β2is:
[6-Apr-2023shift2]
Options:
A.
B.
Theminimumnumberofelementsthatmustbeaddedtotherelation
R= {(a,b), (b,c), (b,d)}ontheset{a,b,c,d}sothatitisan
equivalencerelation,is__
[24-Jan-2023Shift2]
GivenR= {(a,b), (b,c), (b,d)}
Inordertomakeitequivalencerelationaspergivenset,Rmustbe
{ (a,a), (b,b), (c,c), (d,d), (a,b), (b,a), (b,c), (c,b),(b,d), (d,b), (a,c), (a,d), (c,d), (d,c), (c,a), (d,a) }
Therealreadygivenso13moretobeadded.
C.
D.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question13
LetthenumberofelementsinsetsAandBbefiveandtworespectively.
ThenthenumberofsubsetsofA×Beachhavingatleast3andatmost
6elementsis:
[8-Apr-2023shift1]
Options:
A.
752
B.
772
C.
782
D.
792
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question14
Thenumberofelementsintheset{n∈Z:|n2−10n+19|<6}is
[10-Apr-2023shift1]
Answer:6
Solution:
Question15
ThenumberofelementsinthesetS={θ∈[0,2π]:3cos4θ−5cos2θ−
2sin6θ+2=0}is:
[11-Apr-2023shift1]
Options:
A.
10
B.
9
C.
8
D.
12
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question16
Anorganizationawarded48medalsinevent'A',25inevent'B'and
18inevent'C'.Ifthesemedalswenttototal60menandonlyfivemen
gotmedalsinallthethreeevents,then,howmanyreceivedmedalsin
exactlytwoofthreeevents?
[11-Apr-2023shift1]
Options:
A.
15
B.
9
C.
21
D.
10
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question17
Thenumberoftherelations,onthe
set{1,2,3}containing(1,2)and(2,3),whicharereflexiveandtransitive
butnotsymmetric,is________.
[12-Apr-2023shift1]
Answer:3
Solution:
Question18
Thenumberofelementsintheset{n∈N:10≤n≤100.and3n−3isa
multipleof7}is________
[15-Apr-2023shift1]
Answer:15
Solution:
Question19
LetA = {z∈C:1≤ | z− (1+i) | ≤2}andB = {z∈A: | z− (1−i) | = 1}.
Then,B:
[24-Jun-2022-Shift-1]
Options:
A.isanemptyset
B.containsexactlytwoelements
C.containsexactlythreeelements
D.isaninfiniteset
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question20
Options:
A.A −B= (−1,1)
B.B −A=R− (−3,1)
C.A ∩B= (−3, −1]
D.A ∪B=R− [1,3)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question21
LetA = { n∈N :H.C.F.(n,45) = 1}and
LetB = {2k :k∈ {1,2, ......, 100}}.Thenthesumofalltheelementsof
A∩Bis______
[26-Jun-2022-Shift-1]
LetA = {x∈R: | x+1| <2}andB = {x∈R: | x−1| ≥2}.Thenwhich
oneofthefollowingstatementsisNOTtrue?
[25-Jun-2022-Shift-2]
Answer:5264
Solution:
Solution:
SumofallelementsofA∩B=2[Sumofnaturalnumbersupto100whichareneitherdivisibleby3norby5]
Question22
Options:
A.reflexive,symmetricbutnottransitive.
B.reflexive,transitivebutnotsymmetric.
C.reflexivebutnotsymmetricandtransitive.
D.anequivalencerelation.
Answer:D
Solution:
LetasetA =A1∪A2∪ ..... ∪ Ak,whereAi∩Aj=φfori ≠j,1≤j,j≤k.
DefinetherelationRfromAtoAbyR = { (x,y) : y∈Ai.ifandonlyif
.x∈Ai,1≤i≤k}.Then,Ris:
[29-Jun-2022-Shift-1]
©
R= {(x,y) : y∈Ai,iff x ∈Ai1≤i≥k}
(1)Reflexive
(a,a) ⇒ a∈Aiiffa∈Ai
(2)Symmetric
(a,b) ⇒ a∈Aiiffb∈Ai
(b,a)∈Rasb∈Aiiffa∈Ai
(3)Transitive
(a,b) ∈ R&(b,c) ∈ R
⇒a∈Aiiffb∈Ai&b∈Aiiffc∈Ai
⇒a∈Aiiffc∈Ai
⇒(a,c) ∈ R.
⇒RElationisequivalnece.
Question23
Answer:107
Solution:
Question24
Answer:112
Solution:
LetA = {1,2,3,4,5,6,7}.DefineB = { T⊆A:either1 ∉T or2 ∈T}
andC = { T⊆A:T thesumofalltheelementsofT isaprimenumber
}.ThenthenumberofelementsinthesetB ∪Cis____
[25-Jul-2022-Shift-2]
∵(B∪C)′=B′∩C′
B′isasetcontainingsubsetsofAcontainingelement1andnotcontaining2.
AndC′isasetcontainingsubsetsofAwhosesumofelementsisnotprime.
So,weneedtocalculatenumberofsubsetsof{3,4,5,6,7}whosesumofelementsplus1iscomposite.
Numberofsuch5elementssubset = 1
Numberofsuch4elementssubset = 3(exceptselecting3or7)
Numberofsuch3elementssubset = 6(exceptselecting{3,4,5}, {3,6,7}, {4,5,7}or{5,6,7})
Numberofsuch2elementssubset = 7(exceptselecting{3,7}, {4,6}, {5,7})
Numberofsuch1elementssubset = 3(exceptselecting{4}or{6})
Numberofsuch0elementssubset = 1
n(B′∩C′) = 21 ⇒n(B∪C) = 27−21 =107
LetA = {1,2,3,4,5,6,7}andB = {3,6,7,9}.Thenthenumberof
elementsintheset{C⊆A:C∩B≠φ}is_______.
[26-Jul-2022-Shift-2]
Question25
Options:
A.R1isanequivalencerelationbutnotR2
B.R2isanequivalencerelationbutnotR1
C.bothR1andR2areequivalencerelations
D.neitherR1norR2isanequivalencerelation
Answer:D
Solution:
Question26
Options:
LetR1andR2betworelationsdefinedonℝbyaR1b⇔ab ≥0and
aR2b⇔a≥bThen,
[27-Jul-2022-Shift-1]
©
R1= {xy ≥0,x,y∈R}
Forreflexivex×x≥0whichistrue.
Forsymmetric
Ifxy ≥0⇒yx ≥0
Ifx=2,y=0andz= −2
Thenx⋅y≥0&y⋅z≥0butx⋅z≥0isnottrue
⇒nottransitiverelation.
⇒R1isnotequivalence
R2ifa≥bitdoesnotimpliesb≥a
⇒R2isnotequivalencerelation
Forα ∈N ,considerarelationRonN givenbyR. = { (x,y) : 3x +αyisa
multipleof7}.TherelationRisanequivalencerelationifandonlyif:
[28-Jul-2022-Shift-1]
©
A.α =14
B.αisamultipleof4
C.4istheremainderwhenαisdividedby10
D.4istheremainderwhenαisdividedby7
Answer:D
Solution:
Question27
Options:
A.600
B.660
C.540
D.720
Answer:B
Solution:
R= { (x,y) : 3x +αyismultipleof7},nowRtobeanequivalencerelation
(1)Rshouldbereflexive:(a,a) ∈ R∀a∈N
∴3a +aα =7k
∴(3+α)a=7k
∴3+α=7k1⇒α=7k1−3
=7k1+4
(2)Rshouldbesymmetric:aRb ⇔bRa
aRb :3a + (7k −3)b=7m
⇒3(a−b) + 7kb =7m
⇒3(b−a) + 7ka =7m
So,aRb ⇒bRa
∴Rwillbesymmetricfora=7k1−3
(3)Transitive:Let(a,b) ∈ R, (b,c) ∈ R
⇒3a + (7k −3)b=7k1and
3b + (7k2−3)c=7k3
Adding3a +7kb + (7k2−3)c=7(k1+k3)
3a + (7k2−3)c=7m
∴(a,c) ∈ R
∴Ristransitive
∴α=7k −3=7k +4
LetRbearelationfromtheset{1,2,3, ..., 60}toitselfsuchthat
R= { (a,b) : b=pq,wherep,q≥3areprimenumbers}.Then,the
numberofelementsinRis:
[29-Jul-2022-Shift-1]
©
Question28
Answer:5
Solution:
Question29
Options:
A.PandQ
bcantakeitsvaluesas9,15,21,33,39,51,57,25,35,55,49
bcantakethese11valuesandacantakeanyof60values
So,numberofelementsinR=60 ×11 =660
LetA = { n∈N:nisa3-digitnumber}B = {9k +2:k∈N}
andC = {9k +I:k∈N}forsomeI (0<1<9)
IfthesumofalltheelementsofthesetA ∩ (B∪C)is274 ×400,thenl
isequalto
[2021,24Feb.Shift-1]
Given,A= { n∈N:nisa3-digitnumber}
B= {9k +2:k∈N}
C= {9k +1:k∈N}
∵3digitnumberoftheform3k +2are{101,109, .. . 992}
⇒Sum =100
2[101 +992] = 100 ×1093
2
Similarly,3-digitnumberoftheform9k +5is
100
2[104 +995] = 100 ×1099
2
[ ∵numbersare104,113, ..., 995 ]Theirsum = 100 ×1093
2+100 ×1099
2
=100 ×1096 =400 ×274
Hence,wecansaythevalueofI=5asthesecondseriesofnumbersobtainedbysetCisoftheform9k +5.
∴RequiredvalueofI=5
Inaschool,therearethreetypesofgamestobeplayed.Someofthe
studentsplaytwotypesofgames,butnoneplayallthethreegames.
WhichVenndiagramcanjustifytheabovestatement?
[2021,17MarchShift-1]
B.PandR
C.Noneofthese
D.0andR
Answer:C
Solution:
Solution:
Question30
Answer:832
Solution:
Solution:
LetA = {n∈N∣n2≤n+10,000},
B= {3k +1∣k∈N}andC = {2k ∣k∈N},thenthesumofallthe
elementsofthesetA ∩ (B−C)isequalto
[2021,27JulyShift-II]
LetA= {n∈N∣n2≤n+10000}
n2≤n+10000
n2−n≤10000
⇒n(n−1) ≤ 100 ×100
⇒A= {1,2,3, ......, 100}
Now,B= {3k +1∣k∈N}
Question31
Answer:256
Solution:
Question32
B= {4,7,10,13, ...}
andC= {2k ∣k∈N}
C= {2,4,6,8, ...}
So, B−C= {7,13,19, ......, 97, ...}
So,A∩ (B−C) = {7,13,19, ......, 97}
ThisformanAPwithcommondifference
(d=6)
⇒97 =7+ (n−1)6
n=97 −7
6+1=16 [∵an=a+ (n−1)d]
Hence,sum =16
2[7+97]
=832 ∵Sn=n
2(a+1)
{ }
IfA = {x∈R: | x−2| >1},
B=x∈R:x2−3>1 and
C= {x∈R: | x−4| ≥2}andZ istheset
ofallintegers,thenthenumberof
subsetsoftheset(A∩B∩C)C∩Z
is
[2021,27Aug.Shift-I]
{√}
A= {x∈R:|x−2| >1}
⇒A= (−∞,1) ∪ (3,∞)
B=x∈R:x2−3>1
⇒B= (−∞, −2) ∪ (2,∞)
C= {x∈R:|x−4| ≥2}
⇒C= (−∞,2] ∪ [6,∞)
⇒A∩B∩C= (−∞, −2) ∪ [6,∞)
⇒ (A∩B∩C)C= [−2,6)
∴(A∩B∩C)C∩Z= {−2, −1,0,1,2,3,4,5}
Numberofsubsetsof (A∩B∩C)C∩Z
=28=256
{√}
Outofallthepatientsinahospital89%arefoundtobesufferingfrom
heartailmentand98%aresufferingfromlungsinfection.IfK %of
themaresufferingfrombothailments,thenK cannotbelongtotheset
[2021,26Aug.Shift-1]
Options:
A.{80,83,86,89}
B.{84,86,88,90}
C.{79,81,83,85}
D.{84,87,90,93}
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question33
Answer:29
Solution:
Question34
©
LetA= PatientsufferingfromheartailmentandB= Setofpatientsufferingfromlungsinfection
Given,n(A) = 89%andn(B) = 98%
n(A∪B) ≥ n(A) + n(B) − n(A∩B)
⇒100 ≥89 +98 −n(A∩B)
⇒87 ≤n(A∩B)
Also, n(A∩B) = min{n(A), n(B)}
⇒n(A∩B) ≤ 89
∴87 ≤n(A∩B) ≤ 89
So,n(A∩B) ∉ {79,81,83,85}.
LetX = {n∈N:1≤n≤50}.IfA = { n∈X:nisamultipleof2}and
B= { n∈X:nisamultipleof7},thenthenumberofelementsinthe
smallestsubsetofX containingbothAandBis
[Jan.7,2020(II)]
Fromthegivenconditions,n(A) = 25,n(B) = 7 and n(A∩B) = 3
n(A∪B) = n(A) + n(B) − n(A∩B)
=25 +7−3=29
SetAhasmelementsandsetBhasnelements.Ifthetotalnumberof
subsetsofAis112morethanthetotalnumberofsubsetsofB,thenthe
valueofm ⋅nis
[Sep.06,2020(I)]
Answer:28
Solution:
Question35
Options:
A.63
B.36
C.54
D.38
Answer:B
Solution:
Question36
Options:
A.29
2m=112 +2n⇒2m−2n=112
⇒2n(2m−n−1) = 24(23−1)
∴m=7,n=4⇒mn =28
Asurveyshowsthat73%ofthepersonsworkinginanofficelikecoffee,
whereas65%liketea.Ifxdenotesthepercentageofthem,wholike
bothcoffeeandtea,thenxcannotbe:
[Sep.05,2020(I)]
©
Given,n(C) = 73,n(T) = 65,n(C∩T) = x
∴65 ≥n(C∩T) ≥ 65 +73 −100
⇒65 ≥x≥38 ⇒x≠36
Asurveyshowsthat63%ofthepeopleinacityreadnewspaperA
whereas76%readnewspaperB.Ifx%ofthepeoplereadboththe
newspapers,thenapossiblevalueofxcanbe:
[Sep.04,2020(I)]
©
B.37
C.65
D.55
Answer:D
Solution:
Question37
Options:
A.15
B.50
C.45
D.30
Answer:D
Solution:
Question38
Letn(U) = 100,thenn(A) = 63,n(B) = 76n(A∩B) = x
Now,n(A∪B) = n(A) + n(B) − n(A∩B) ≤ 100
=63 +76 −x≤100
⇒x≥139 −100 ⇒x≥39
∵n(A∩B) ≤ n(A)
⇒x≤63
∴39 ≤x≤63
Let⋃i=1
50 Xi= ⋃i=1
nYi=T,whereeachX icontains10elementsand
eachY icontains5elements.IfeachelementofthesetT isanelement
ofexactly20ofsetsX i
′sandexactly6ofsetsY i
′s,thennisequalto
[Sep.04,2020(II)]
©
⋃i=1
50 Xi= ⋃i=1
nYi=T
∵n(Xi) = 10,n(Yi) = 5
So,⋃i=1
50 Xi=500, ⋃i=1
nYi=5n
⇒500
20 =5n
6⇒n=30
LetS = {1,2,3, ..., 100}.Thenumberofnon-emptysubsetsAofSsuch
thattheproductofelementsinAisevenis:
[Jan.12,2019(I)]
Options:
A.2100 −1
B.250(250 −1)
C.250 −1
D.250 +1
Answer:B
Solution:
Question39
Options:
A.215
B.218
C.212
D.210
Answer:A
Solution:
©
∵Productoftwoevennumberisalwaysevenandproductoftwooddnumbersisalwaysodd.
∴Numberofrequiredsubsets
=Totalnumberofsubsets−Totalnumberofsubsetshavingonlyoddnumbers
=2100 −250 =250(250 −1)
LetZ bethesetofintegers.IfA =x∈Z:2(x+2)(x2−5x +6)=1 and
B= {x∈Z: −3<2x −1<9},thenthenumberofsubsetsoftheset
A×B,is:
[Jan.12,2019(II)]
©
{ }
(a)Letx∈A,then
∵2(x+2)(x2−5x +6)=1⇒ (x+2)(x−2)(x−3) = 0
x= −2,2,3
A= {−2,2,3}
Then,n(A) = 3
Letx∈B,then
−3<2x −1<9
Question40
Options:
A.102
B.42
C.1
D.38
Answer:D
Solution:
−1<x<5andx∈Z
∴B= {0,1,2,3,4}
n(B) = 5
n(A×B) = 3×5=15
Hence,NumberofsubsetsofA×B=215
Inaclassof140studentsnumbered1to140,allevennumbered
studentsoptedMathematicscourse,thosewhosenumberisdivisibleby
3optedPhysicscourseandthosewhosenumberisdivisibleby5opted
Chemistrycourse.Thenthenumberofstudentswhodidnotoptforany
ofthethreecoursesis:
[Jan.10,2019(II)]
P= {30,60,90,120}
⇒n(P) = 4
Q= {6n :n∈N,1≤n≤23} − P
⇒n(Q) = 19
R= {15n :n∈N,1≤n≤9} − P
⇒n(R) = 5
S= {10n :n∈N,1≤n≤14} − P
⇒n(S) = 10
n(T) = 70 −n(P) − n(Q) − n(S) = 70 −33 =37
n(V) = 46 −n(P) − n(Q) − n(R) = 46 −28 =18
n(W) = 28 −n(P) − n(R) − n(S) = 28 −19 =9
⇒Numberofrequiredstudents = 140 − (4+19 +5+10 +37 +18 +9)
=140 −102 =38
Question41
Options:
A.B ∩C≠φ
B.If(A−B) ⊆ C,thenA ⊆C
C.(C∪A) ∩ (C∪B) = C
D.If(A−C) ⊆ B,thenA ⊆B
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question42
Options:
A.13.9
B.12.8
C.13
D.13.5
Answer:A
Solution:
Solution:
LetA,BandCbesetssuchthatφ ≠A∩B⊆C.Thenwhichofthe
followingstatementsisnottrue?
[April12,2019(II)]
©
(1),(2)and(4)arealwayscorrect
In(3)option,
IfA=CthenA−C=φ
Clearly,φ⊂eqBbutA⊂eqBisnotalwaystrue.
TwonewspapersAandBarepublishedinacity.Itisknownthat25%of
thecitypopulationreadsAand20%readsBwhile8%readsbothAand
B.Further,30%ofthosewhoreadAbutnotBlookintoadvertisements
and40%ofthosewhoreadBbutnotAalsolookintoadvertisements,
while50%ofthosewhoreadbothAandBlookintoadvertisements.
Thenthepercentageofthepopulationwholookintoadvertisementsis:
[April.09,2019(II)]
-------------------------------------------------------------------------------------------------
Question43
Options:
A.containsexactlyoneelement.
B.containsexactlytwoelements.
C.containsexactlyfourelements.
D.isanemptyset
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question44
%ofpeoplewhoreadsAonly = 25 −8=17%
%ofpeoplewhoreadBonly = 20 −8=12%
%ofpeoplefromAonlywhoreadadvertisement = 17 ×0.3 =5.1%
%ofpeoplefromBonlywhoreadadvertisement = 12 ×0.4 =4.8%
%ofpeoplefromA&Bbothwhoreadadvertisement = 8×0.5 =4%
∴total%ofpeoplewhoreadadvertisement = 5.1 +4.8 +4=13.9%
LetS = { x∈R:x≥0and2 | √x−3| +√x(√x−6) + 6=0.ThenS
[2018]
©
Case-I:x∈ [0,9]
2(3− √x) + x−6√x+6=0
⇒x−8√x+12 =0⇒ √x=4,2
⇒x=16,4
Sincex∈ [0,9]
∴x=4
Case-II:x∈ [9,∞]
2(√x−3) + x−6√x+6=0
⇒x−4√x=0⇒x=16,0
Sincex∈ [9,∞]
∴x=16
Hence,x=4&16
Iff (x) + 2f 1
x=3x,x≠0andS = {x∈R:f(x) = f(−x)};thenS
[2016]
©
( )
Options:
A.containsexactlytwoelements.
B.containsmorethantwoelements.
C.isanemptyset.
D.containsexactlyoneelement.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question45
Options:
A.P ⊂QandQ −P≠φ
B.Q not ⊂P
C.P =Q
D.P not ⊂Q
Answer:C
Solution:
Solution:
f(x) + 2f 1
x=3x .... . (1)
f1
x+2f (x) = 3
x
Adding(1)and(2)
⇒f(x) + f1
x=x+1
x
Substracting(1)from(2)
⇒f(x) − f1
x=3
x−3x...
Onadding(3)and(4)
⇒f(x) = 2
x−x
f(x) = f(−x) ⇒ 2
x−x=−2
x+x⇒x=2
x
x2=2or x= √2, −√2
( )
( )
( )
( )
LetP = {θ:sin θ −cos θ = √2cos θ}andQ = { θ:sin θ+
cos θ = √2sin θ }betwosets.Then:
[OnlineApril10,2016]
©
sin θ −cos θ = √2 cos θ
⇒sin θ =cos θ + √2 cos θ
-------------------------------------------------------------------------------------------------
Question46
Options:
A.Only(A)and(C)arecorrect.
B.Only(B)and(C)arecorrect.
C.Al l (A), (B)and(C)arecorrect.
D.Only(A)and(B)arecorrect.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question47
Options:
= (√2+1)cos θ =2−1
√2−1cos θ
⇒(√2−1)sin θ =cos θ
⇒sin θ +cos θ = √2 sin θ
∴P=Q
( )
Inacertaintown,25%ofthefamiliesownaphoneand15%ownacar;
65%familiesownneitheraphonenoracarand2,000familiesownboth
acarandaphone.Considerthefollowingthreestatements:
(A)5%familiesownbothacarandaphone
(B)35%familiesowneitheracaroraphone
(C)40,000familiesliveinthetown
Then,
[OnlineApril10,2015]
©
n(P) = 25%
n(C) = 15%
n(P′∪C′) = 65%
⇒n(P∪C)′=65%
n(P∪C) = 35%
n(P∩C) = n(P) + n(C) − n(P∪C)
25 +15 −35 =5%
x×5% = 2000
x=40,000
ArelationonthesetA = {x: | x| <3,x∈Z}whereZ isthesetof
integersisdefinedbyR = {(x,y) : y= | x| ,x≠ −1}.Thenthenumberof
elementsinthepowersetofRis:
[OnlineApril12,2014]
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A.32
B.16
C.8
D.64
Answer:B
Solution:
Solution:
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Question48
Options:
A.52
B.35
C.25
D.53
Answer:B
Solution:
Solution:
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Question49
(b)A= {x:|x| <3,x∈Z}
A= {−2, −1,0,1,2}
R= {(x,y) : y= | x| ,x≠ −1}
R= {(−2,2), (0,0), (1,1), (2,2)}
RhasfourelementsNumberofelementsinthepowersetofR
=24=16
LetX = {1,2,3,4,5}.Thenumberofdifferentorderedpairs(Y,Z)
thatcanformedsuchthatY ⊂eqX ,Z⊂eqX andY ∩Z isemptyis:
[2012]
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LetX= {1,2,3,4,5}n(x) = 5
Eachelementofxhas3options.EitherinsetYorsetZornone.
(∵Y∩Z=φ)
So,numberoforderedpairs = 35
IfA,BandCarethreesetssuchthatA ∩B=A∩CandA ∪B=A∪C,
then
[2009]
Options:
A.A =C
B.B =C
C.A ∩B=φ
D.A =B
Answer:B
Solution:
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Findingthevalue:
A∪B=A∪C
⇒(A∪B) ∩ C= (A∪C) ∩ C
⇒(A∩C) ∪ (B∩C) = C
⇒(A∩B) ∪ (B∩C) = C... . (i) (∵A∩C=A∩B)
⇒A∪B=A∪C
⇒(A∪B) ∩ B= (A∪C) ∩ B
⇒B= (A∩B) ∪ (C∩B)
= (A∩B) ∪ (B∩C) ⋅⋅⋅⋅⋅⋅ ⋅ (ii)
From(i)and(ii)
B=C