CBSE Class 11 Mathematics - Vidya Champ

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MaximumMarks:80

TimeAllowed:3hours

GeneralInstructions:

i. Allthequestionsarecompulsory.

ii. Thequestionpaperconsistsof36questionsdividedinto4sectionsA,B,C,andD.

iii. SectionAcomprisesof20questionsof1markeach.SectionBcomprisesof6questionsof

2markseach.SectionCcomprisesof6questionsof4markseach.SectionDcomprisesof

4questionsof6markseach.

iv. Thereisnooverallchoice.However,aninternalchoicehasbeenprovidedinthree

questionsof1markeach,twoquestionsof2markseach,twoquestionsof4markseach,

andtwoquestionsof6markseach.Youhavetoattemptonlyoneofthealternativesinall

suchquestions.

v. Useofcalculatorsisnotpermitted.

SectionA

1. Twofinitesetshavemandnelements.Thenumberoelementsinthepowersetofthe

firstis48morethanthetotalnumberofelementsinthepowersetofthesecond.

Thenthevaluesofmandnare

a. 6,4

b. 6,3

c. 3,7

d. 7,6

2. Thenumberofwaysinwhichntiescanbeselectedfromarackdisplaying3n

CBSE Class 11 Mathematics

Sample Papers 07 (2019-20)

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differenttiesis

a. noneofthese

b.

c.

d.

3. Ifthecoefficientsof and intheexpansionof areequalthenn=

a. 45

b. 55

c. 56

d. 15

4. Afairdiceisrolledntimes.Thenumberofallthepossibleoutcomesis

a. 6n

b.

c.

d. noneofthese

5. Iff: andg: aregivenbyf(x0= andg(x)=[x]foreachx R,

then{x R}:g(f(x0) f(g(x))]=

a. ZU(- ,0)

b.

c. R

d. Z

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6. If isdivisibleby64foralln N,thentheleastnegativeintegral

valueof is

a. -1

b. -3

c. -4

d. -2

7. Acoinistossedonce.Ifaheadcomesup,thenitistossedagainandifatailcomesup,

adiceisthrown.Thenumberofpointsinthesamplespaceofexperimentis

a. 4

b. 12

c. 8

d. 24

8. Alinemakingangles withthepositivedirectionsoftheaxisofxandy

makeswiththepositivedirectionofZ-axis,anangleof

a.

b.

c.

d.

9. Fromeachofthefourmarriedcouples,oneofthepartnersisselectedatrandom.The

probabilitythatthoseselectedareofthesamesexis

a.

b.

c.

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d.

10. Theexponentofxoccurringinthe termofexpansionof is

a. -5

b. 3

c. 5

d. -3

11. Fillintheblanks:

LetAandBbeanytwonon-emptyfinitesetscontainingmandn

elementsrespectively,then,thetotalnumberofsubsetsof(A B)is________.

12. Fillintheblanks:

Thestructurewhichisusedtounderstandandrememberthecoefficientsofvariables

inanyexpansion,looklikeatrianglewith1atthetopvertexandrunningdownthe

twoslantingsidesiscalled________.

13. Fillintheblanks:

ThevaluesofP(15,3)is________.

14. Fillintheblanks:

ListhefootofperpendiculardrawnfromthepointP(3,4,5)onzx-planes.The

coordinatesofLare________

OR

Fillintheblanks:

Theequationx=brepresentsaplaneparallelto________plane.

15. Fillintheblanks:

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Thederivativeofcosxis________.

OR

Fillintheblanks:

Thederivativeofxatx=1is________.

16. Describe{x:x Zand }setinRosterform.

17. Findthenumberofchordsthatcanbedrawnthrough16pointsonacircle.

18. Expressthecomplexnumbersi9+i19

OR

Showthatin+in+1+in+2+in+3=0, .

19. IfN={1,2,3},thenfindtherelation

R={(x,y):x N,y Nand2x+y=10}inN N.

20. If .find .

21. LetAandBbetwosets.Provethat:(A-B) B=AifandonlyifB A.

OR

DescribethefollowingsetsinRosterform:

i. Thesetofallvowelsintheword'EQUATION'

ii. Thesetofall-naturalnumberslessthan7.

22. AletterischosenatrandomfromthewordASSASSINATIONfindtheprobabilitythat

letteris

(i)avowel

(ii)aconsonant

23. Expand

24. Findtheequationoftheperpendicularbisectorofthelinesegmentjoiningthepoints

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(1,1)and(2,3).

OR

Checkwhetherthepoints(1,-1),(5,2)and(9,5)arecollinearornot.

25. Checkthevalidityofthestatement:

p:100isamultipleof4and5.

26. Solve:2cos2x+3sinx=0

27. Inagroupof400peopleinUSA,250canspeakSpanishand200canspeakEnglish.

HowmanypeoplecanspeakbothSpanishandEnglish?

28. If findthevaluesofxandy.

OR

LetA={1,2}andB={3,4}write .Howmanysubsetswill have?List

them

29. Evaluate

30. Solve

31. Solvethefollowinginequation:

OR

Solvetheinequalitiesgraphicallyintwo-dimensionalplane:

32. Provethefollowingbyusingtheprincipleofmathematicalinductionforall

isamultipleof27.

33. Provethat:cos40°cos80°cos160°= .

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OR

Ifcos +cos +cos = ,thenprovethatcos +cos +cos =

sin +sin +sin =0.

34. Findthesumofntermsofseries

35. Findtheequationofthecirclewhichiscircumscribedaboutthetriangle,whose

verticesare(-2,3),(5,2)and(6,-1).

OR

Findtheequationofthecirclewhichpassesthroughthecentreofthecirclex2+y2+

8x+10y-7=0andisconcentricwiththecircle2x2+2y2-8x-12y-9=0

36. Themeasurementsofthediameters(inmm)oftheheadsof107screwsaregiven

below:

Diameter(inmm) 33-35 36-38 39-41 42-44 45-47

No.ofscrews 17 19 23 21 27

Calculatethestandarddeviation.

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Solution

SectionA

1. (a)6,4

Explanation:LetAhasmelementsandBgasnelements.Then,no.ofelementsin

P(A)=2mandno.ofelementsinP(B)=2n.]

Bythequestion,

2m=2n+48

2m-2n=48

Thisispossible,if2m=64,2n=16.(As64-16=48)

Also,

2. (d)

Explanation:

Thenumberofselectionsofrobjectsfromthegivennobjectsisdenotedby

andwehave

Nowntiescanbeselectedfromarackdisplaying3ndifferenttiesin

differentways

3. (b)55

CBSE Class 11 Mathematics

Sample Papers 07

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Explanation:Wehavethegeneraltermintheexpansionof isgivenby

Now

and

Given

4=55

4. (c)

Explanation:

eachtimethereare6possibilities,thereforeforntimesthereare6npossibilities.

5. (c)R

Explanation:

Wehave,

now, forsome

Wherenisapositiveinteger

DomainofF=R

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6. (a)-1

Explanation:

Whenn=1wehavethevalueoftheexpressionas65.Giventhattheexpressionis

divisiblebe64.Hencethevalueis-1.

7. (c)8

Explanation:

SampleSpaceis

S={HH,HT,T1,T2,T3,T4,T5,T6}

sonumberofoutcomesinsamplespaceis8

8. (a)

Explanation:

cos2 +cos2 +cos2 =1

putthevaluesinaboveequation

( )2+(1/2)2+cos2 =1

cos

9. (a)

Explanation:

Here,s={(MMMM),(FFFF),......}

Clearly,n(s)=16

Requiredproability=P[(MMMM)or(FFFF)]

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=P[(MMMM)+(FFFF)]

10. (d)-3

Explanation:Wehavethegeneraltermof is

Nowconsider

heren=9andr+1=7 r=6

Also and

Hencetheexponentofx=-3

11. 2mn

12. Pascal'striangle

13. 2730

14. (3,0,5)

OR

yz-plane

15. -sinx

OR

1

16. Wefindthatxisanintegersatisfying

and, =0,1,2

x=0,±1,±2

So,xcantakevalues-2,-1,0,1,2.

{x:x Zand }={-2,-1,0,1,2}

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17. Since,thepointsliesonthecircumferenceofthecircle.So,nothreeofthemare

collinear.

Thus,numberofchordsformedby16pointsbytaking2attime=16C2

=120

18.

=i-i=0

OR

Given,LHS

=in(1+i-1-i)[ i2=-1,i3=i2.i=-i]

=in(0)=0=RHS

Henceproved.

19. Here,R={(x,y):x N,y Vand2x+y=10}in

N N.

R={ }

Domainof={ }

RangeofR={ }

20. Here

21. First,letusconsiderthat,(A-B) B=A.

Then,wehavetoprovethatB A.

WeknowthatA-BreferstothoseelementsofAwhicharenotpresentinB,thatisA-

B=A B'.....(i)

Now,(A-B) B=A

......[from(i)]

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Theaboveconditionisonlypossiblewhen,

Conversely,letB A.Then,wehavetoprovethat(A-B) B=A.

Now,(A-B) B=(A B') B

=(A B) (B' B)

NowasweknowthatB A

=A

OR

IntheRoasterformalltheelementsofthesetarelistedinside"{}"bracketsandare

seperatedbycommas.

i. Thevowelsintheword'EQUATION'areA,E,I,O,U

So,therequiredsetcanbedescribedasfollows:{A,E,I,O,U}

ii. Naturalnumberslessthan7are1,2,3,4,5,6.

Hence,therequiredsetcanbedescribedasfollows:{1,2,3,4,5,6}.

22. Thereare13lettersinthewordASSASSINATIONofwhich6vowelsand7consonants.

Oneletterisselectedoutof13lettersin ways

(i)Outof6vowels,1vowelcanbeselectedin6ways

P(1vowelselected)

(ii)Outof7consonants,1consonantcanbeselectedin7ways.

P(1consonantselected)t

23. =4Co +4C1 +4C2 +4C3

+4C4

=1 +4 +6 +4 +1

[using4Co=4C4=1,

4C3=4C1=4and

4C2= =6]

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= - +6- +

24. LetPbethemid-pointofthelinesegmentjoiningpointsA(1,1)andB(2,3).Then,

thecoordinatesofPare .

LetmbetheslopeoftheperpendicularbisectorofAB.

Then,

m SlopeofAB=-1

m =-1

m=

Clearly,theperpendicularbisectorofABpassesthroughP andhasslopem=

- .So,itsequationis

y-2=- or,2x+4y-11=0.

OR

LetA=(1,-1),B=(5,2)andC=(9,5)

Now,distancebetweenAandB,

DistancebetweenBandC,

DistancebetweenAandC,

Clearly,AC=AB+BC

Hence,A,BandCarecollinearpoints.

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25. Thestatementis:

"100ismultipleof4and5".

Weknowthat100isamultipleof4aswellas5.Thus,pisatruestatement.

Hence,thestatementistruei.e.thestatement"p"isavalidstatement.

26.

2sinx(sinx-2)+1(sinx-2)=0

(sinx-2)(2sinx+1)=0

2sinx+1=0

27. LetSbethesetofpeoplewhospeakSpanish,andEbethesetofpeoplewhospeak

English

∴n(S∪E)=400,n(S)=250,n(E)=200n(S∩E)=?Weknowthat:

n(S∪E)=n(S)+n(E)–n(S∩E)∴400=250+200–n(S∩E)⇒400=450–n(S∩E)⇒n(S∩E)=450–400∴n(S∩E)=50Thus,50peoplecanspeakbothSpanishandEnglish.

28. Here

and

and

and

x=2andy=1

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OR

HereA={1,2}andB={3,4}

={(1,3},(1,4),(2,3),(2,4)}

Numberofelementsin

Numberofsubsetsof

Thesubsetare:

{(1,3)},{(1,4)},{(2,3)},{(2,4)},{(1,3),(1,4)},{(1,3),(2,3)},{(1,3),(2,4)},{(1,4),(2,

3)},{(1,4),(2,4)},{(2,3),(2,4)},{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},{(1,3),(2,3),(2,

4)}{(1,4),(2,3),(2,4)},{(1,3),(1,4),(2,3),(2,4)}

29. Here

Puttingx+1=y,as

30. Here

Comparingthegivenquadraticequationwithax2+bx+c=0,wehave

a=3,b=-4and

Thus and

31. Here

Multiplyingbothsidesby12

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Dividingbothsidesby-10

Thussolutionsetofgiveninequationis

OR

Thegiveninequalityis

Drawthegraphoftheline2x+y=6

Tableofvaluessatisfyingtheequation2x+y=6

X 1 2

Y 4 2

Putting(0,0)inthegiveninequation,wehave

whichisfalse.

Halfplaneof isalwaysfromorigin

32. LetP(n)=41n-14nisamultipleof27

Forn=1

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P(1)=411-141isamultipleof27 27isamultipleof27

P(1)istrue

LetP(n)betrueforn=k

isamultipleof27 ....(i)

Forn=k+1

P(k+1)41k+1-14k+1isamultipleof27

Now

[Using(i)]

isamultipleof27

P(k+1)istrue

ThusP(k)istrue P(k+1)istrue

Hencebyprincipleofmathematicalinduction,P(n)istrueforall .

33. cos40°cos80°cos160°=

LHS=cos40°cos80°cos160°

=cos80°cos40°cos160°

Multiplyinganddividingby2

{cos80° (2cos40°cos160°)}

Because2cosAcosB=cos(A+B)+cos(A-B)

cos80°[cos(40°+160°)+cos(40°-160°)]

cos80°[cos200+cos(-120)]

cos80°[cos200+cos120]

cos80°{cos(180°+20°)+cos(180°-60°)}

cos80°(-cos20°-cos60°)

cos80°cos20°- cos80°cos60°

(2cos80°cos20°)- cos80°

[2cos80°cos20°+cos80°]

[cos(80°+20°)+cos(80°-20°)+cos80°]

[cos100°+cos60°+cos80°]

[cos(180°-80°)+cos60°+cos80°]

[-cos80°+cos60°+cos80°]

cos60°

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=RHS

OR

Given,

+cos =

2[cos +cos +cos ]=-3

[ ]

+2cos cos +2cos cos ]+[2sin sin +2sin sin

+2sin sin ]+3=0

[2cos cos +2cos cos +2cos cos ]+[2sin sin +2sin sin

+2sin sin ]+(cos2 +sin2 )+(cos2 +sin2 )+(cos2 +sin2 )=0

[ ]

[cos2 +cos2 +cos2 +2cos cos +2cos cos +2cos cos ]+[sin2

+sin2 +sin2 +2sin sin +2sin sin +2sin sin ]=0

Weknowthat,sumoftwopositivetermswillbezero,ifbothareequaltozero.

and

Henceproved.

34. LetthegivenseriesbeS=

Then,nthtermTn=

Now,wewillsplitthedenominatorofthenthtermintotwopartsorwewillwrite

Tnasthedifferenceoftwoterms.

Tn=

=

Onputtingn=1,2,3,4,...successively,weget

T1=

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T2=

T3= ................

Tn=

Onaddingalltheseterms,weget

S=T1+T2+T3+...+Tn

=

=1-

S=

35. Thecirclewhichiscircumscribedaboutthetriangle,whoseverticesare(-2,3),(5,2)

and(6,-1)meansthecirclepassesthroughthesethreepoints.

Lettheequationofcirclebe

(x-h)2+(y-k)2=r2...(i)

Since,equation(i)passesthroughthepoints(-2,3),

(-2-h)2+(3-k)2=r2

h2+4h+4+k2-6k+9=r2...(ii)

alsoequation(i)passesthroughthepoint(5,2)

=>(5-h)2+(2-k)2=r2

h2-10h+25+k2-4k+4=r2...(iii)

againequation(i)passesthroughthepoint(6,-1)=>(6-h)2+(-1-k)2=r2

h2-12h+36+k2+2k+1=r2...(iv)

Nowsubtractingequation(iii)fromequation(ii),weget

14h-21-2k+5=0

i.e.,14h-2k=16

=>7h-k=8...(v)

againsubtractingequation(iv)fromequation(iii),weget

2h-11-6k+3=0

i.e.,2h-6k=8

=>h-3k=4...(vi)

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Onsolvingequation(v)andequation(vi),weget

h=1andk=-1

Onputtingthevaluesofh=1andk=-1inequation(ii),weget

1+4+4+1+6+9=r2

r2=25

r=5

Nowputtingh=1,k=-1andr=5inequation(i)weget

(x-1)2+(y+1)2=25

x2+y2-2x+2y-23=0

whichistherequiredequationofthecircle.

OR

Wehavetofindtheequationofcircle(C2)whichpassesthroughthecentreofcircle

(C1)andisconcentricwithcircle(C3).

Wehave,equationofcircle(C1),

x2+y2+8x+10y-7=0...(i)

Oncomparingitwithx2+y2+2gx+2fy+c=0,weget

g=4,f=5andc=-7

CentreofC1isO1=(-g,-f)

O1=(-4,-5)

Now,equationofcircle(C2)whichisconcentricwithgivencircle(C3)havingequation

2x2+2y2-8x-12y-9=0is

2x2+2y2-8x-12y+k=0...(ii)

Since,circle(C2)passesthroughO1(-4,-5)

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2(-4)2+2(-5)2-8(-4)-12(-5)+k=0

32+50+32+60+k=0

k=-174

OnputtingthevalueofkinEq.(ii),weget

2x2+2y2-8x-12y-174=0

x2+y2-4x-6y-87=0[dividingbothsidesby2]

whichisrequiredequationofcircle(C2).

36. Heretheclassintervalsareformedbytheinclusivemethod.But,themid-pointsof

class-intervalsremainthesamewhethertheyareformedbytheinclusivemethodor

exclusivemethod.Sothereisnoneedtoconvertthemintoanexclusiveseries.

CalculationofStandardDeviation

Diameter(in

mm)

Mid-values,

xi

No.ofscrews,

fi

ui= fiui fiui2

33-35 34 17 -2 -34 68

36-38 37 19 -1 -19 19

39-41 40 23 0 0 0

42-44 43 21 1 21 21

45-47 46 27 2 54 108

fi=107 fiui=

22

fiui2=

216

HereN= fi=107, fiui=22, fiui2=216,A=40and,h=3

Var(X)=h2 =9

Var(X)=9(2.0187-0.0420)=9 1.9767=17.7903

S.D.= =4.2178