Answer Key
QA #1 Factors and Multipless
QuestionsΒ andΒ AnswersΒ SheetΒ 1
Β
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β FactorsΒ andΒ MultiplesΒ
Β
QuestionΒ #1Β
WhatΒ isΒ Β
β
βπ₯
π
(π₯
π
!)
Β Β whereΒ Β
π₯
π
Β Β is aΒ discreteΒ variable?Β
DoΒ youΒ considerΒ Β
(π₯
π
!)Β =Β (π₯
π
)(π₯
π
βΒ 1)Β β―Β 1
Β andΒ doΒ productΒ ruleΒ onΒ eachΒ term,Β orΒ somethingΒ else?Β
Answer:Β
StepΒ 1Β
TheΒ derivativeΒ ofΒ aΒ functionΒ ofΒ aΒ discreteΒ variableΒ doesn'tΒ really makeΒ senseΒ inΒ theΒ typicalΒ calculus setting.Β
However,Β thereΒ is aΒ continuous variantΒ ofΒ theΒ factorialΒ functionΒ calledΒ theΒ Β
πΎ
Β Β function,Β forΒ whichΒ youΒ canΒ
takeΒ derivatives andΒ evaluateΒ theΒ derivativeΒ atΒ integerΒ values.Β
InΒ particular,Β sinceΒ Β
π!Β =Β πΎ(πΒ +Β 1)
Β ,Β thereΒ is aΒ niceΒ formulaΒ forΒ Β
Ξ³
β²
Β Β atΒ integerΒ values:Β
Β
Ξ³
β²
(πΒ +Β 1)Β =Β π!Β (βΞ³Β +Β β
1
π
π
π=1
)
Β Β
whereΒ Β
πΎ
Β Β is theΒ Euler-MascheroniΒ constant.Β
StepΒ 2Β
As has beenΒ mentioned,Β theΒ GammaΒ functionΒ Β \gamma(x)Β Β is theΒ way toΒ go.Β
IntegrationΒ by partsΒ yieldsΒ
Β
πΎ(π₯)Β =Β β«Β π
βπ‘
β
0
π‘
π₯β1
ππ‘
Β Β
Β
=Β (π₯Β βΒ 1)Β β«Β π
βπ‘
β
0
π‘
π₯β2
ππ‘
Β Β
Β
=Β (π₯Β βΒ 1)πΎ(π₯Β βΒ 1)
Β Β
TakingΒ theΒ derivativeΒ ofΒ theΒ logarithmΒ ofΒ Β
πΎ(π₯)
Β Β givesΒ
Β
Ξ³
β²
(π₯)
Ξ³(π₯)
=
1
π₯β1
+
Ξ³
β²
(π₯β1)
Ξ³(π₯β1)
Β Β
BecauseΒ Β
πΎ(π₯)
Β Β isΒ log-connvexΒ andΒ
Β
lim
π₯ββ
Ξ³
β²
(π₯)
Ξ³(π₯)
βΒ log(π₯)Β =Β 0
Β Β
weΒ getΒ thatΒ
Β
Ξ³
β²
(π₯)
Ξ³(π₯)
=Β βπΎΒ +Β β
(
1
π
β
1
π+π₯β1
)
β
π=1
Β Β
ForΒ integerΒ Β
π,Β β!Β =Β πΎ(πΒ +Β 1)
,Β soΒ theΒ derivativeΒ isΒ
Β
πΎ(πΒ +Β 1)Β =Β πΎ(πΒ +Β 1)Β (βΞ³Β +Β β
π
π(π+π)
β
π=1
)
Β Β
Β
=Β π!Β (βΞ³Β +Β π»
π
)
Β Β
whereΒ Β
π»
π
Β Β is theΒ Β
π
π‘β
Β Β Harmonic NumberΒ (withΒ theΒ conventionΒ thatΒ Β
π»
0
=Β 0
)
Β
QuestionΒ #2Β
HowΒ doΒ youΒ evaluateΒ Β
7πΆ
2
Β Β ?
Β
Answer:Β
Explanation:Β
Β
ππΆπΒ =Β ((π),Β (π))Β =
π!
π!(πβπ)!
Β Β
Β
ππΆπΒ =Β ((7),Β (2))Β =
7!
2!(7β2)!
=
7!
2!β 5!
=
5040
2β 120
=
5040
2040
=Β 21
Β
Β
QuestionΒ #3Β
HowΒ doΒ youΒ factorΒ Β
π¦Β =Β π₯
3
βΒ 2π₯
2
+Β π₯Β βΒ 2
Β ?
Β
Answer:Β
Β
(π₯Β βΒ 2)(π₯Β +Β π)(π₯Β βΒ π)
Β Β
Explanation:Β
factorΒ theΒ terms by groupingΒ
Β
=Β π₯
2
(π₯Β βΒ 2)Β +Β 1(π₯Β βΒ 2)
Β Β
takeΒ outΒ theΒ commonΒ factorΒ Β
(π₯Β βΒ 2)
Β Β
Β
=Β (π₯Β βΒ 2)(π₯
2
+Β 1)
Β Β
Β
π₯
2
+Β 1
Β Β canΒ beΒ factoredΒ usingΒ differenceΒ ofΒ squaresΒ
Β
π
2
βΒ π
2
=Β (πΒ βΒ π)(πΒ +Β π)
Β
withΒ Β
πΒ =Β π₯
Β Β andΒ
πΒ =Β πΒ βΒ (πΒ =Β ββ1)
Β Β
Β
=Β (π₯Β βΒ 2)(π₯Β +Β π)(π₯Β βΒ π)Β βΒ inΒ factoredΒ form
Β
Β
QuestionΒ #4Β
HowΒ doΒ youΒ factorΒ Β
20π₯
5
βΒ 5
Β
Β
Answer:Β
Β
20π₯
5
βΒ 5Β =Β 5(4π₯
2
βΒ 1)Β =Β 5(2π₯Β βΒ 1)(2π₯Β +Β 1)
Β Β
Explanation:Β
SoΒ weΒ haveΒ Β
20π₯
2
βΒ 5Β =Β 5(4π₯
2
βΒ 1)
Β ;Β dividingΒ thruΒ by 5Β
Β
5(4π₯
2
βΒ 1)
Β Β
AndΒ Β
4π₯
2
βΒ 1
Β Β isΒ theΒ differenceΒ betweenΒ 2Β squares,Β Β
2π₯Β
Β andΒ 1Β
Β
(4π₯
2
βΒ 1)Β =Β (2π₯Β βΒ 1)(2π₯Β +Β 1)
Β
PuttingΒ itΒ allΒ together:Β
Β
20π₯
2
βΒ 5Β =Β 5(2π₯Β βΒ 1)(2π₯Β +Β 1)
Β Β
SoΒ thereΒ areΒ rootsΒ atΒ Β
Β±
1
2
Β
Β
QuestionΒ #5Β
WhatΒ isΒ theΒ nextΒ numberΒ inΒ theΒ sequenceΒ Β
1,2,6,24,120,Β β―Β ?
Β
Β
Answer:Β
TheseΒ areΒ theΒ firstΒ 5Β factorialsΒ (startingΒ atΒ 1!)Β
Β
1!Β =Β 1
Β Β
Β
2!Β =Β 2Β Β ΓΒ 1Β =Β 2
Β Β Β
Β
3!Β =Β 3Β Β ΓΒ 2Β Β ΓΒ 1Β =Β 6
Β Β Β
Β
4!Β =Β 4Β ΓΒ 3Β Β ΓΒ 2Β Β ΓΒ 1Β Β =Β Β 24
Β Β Β
Β
5!Β =Β 5Β Β ΓΒ 4Β Β ΓΒ 3Β Β ΓΒ 2Β Β ΓΒ 1Β =Β 120
Β Β Β
Β
6!Β =Β 6Β Β ΓΒ 5Β Β ΓΒ 4Β Β ΓΒ 3Β Β ΓΒ 2Β Β ΓΒ 1Β =Β 720
Β
Β
QuestionΒ #6Β
UseΒ LagrangeΒ multipliers toΒ findΒ theΒ maximumΒ andΒ minimumΒ values ofΒ theΒ functionΒ subjectΒ toΒ theΒ givenΒ
constraint.Β Β
π(π₯,Β π¦)Β =Β 3π₯Β +Β π¦;Β π₯
2
+Β π¦
2
=Β 10
Β
Β
Answer:Β
StepΒ 1Β
WeΒ firstΒ findΒ ourΒ gradientsΒ ofΒ fΒ andΒ gΒ
Β
βπΒ =<Β 3,1Β >
Β Β
Β
βπΒ =<Β 2π₯,Β 2π¦Β >
Β Β Β
StepΒ 2Β Β
WeΒ thenΒ setΒ ourΒ gradientsΒ equalΒ toΒ eachΒ otherΒ usingΒ aΒ LaGrangeΒ multiplier.Β WeΒ solveΒ forΒ xΒ
Β 3Β =Β π2π₯
Β Β Β
Β
3
2π₯
=
Β Β Β
Β
1Β =Β π2π¦
Β Β Β
1
2π¦
=
Β Β
Β
3
2π₯
=
1
2π¦
Β Β Β
6y=2xΒ Β
3y=xΒ Β
StepΒ 3Β Β
WeΒ plugΒ 3yΒ inΒ forΒ xΒ inΒ ourΒ constraintΒ function.Β This determines whatΒ y willΒ beΒ equalΒ too.Β WeΒ findΒ thatΒ Β
π¦Β =
Β±1
Β Β .Β WeΒ canΒ plugΒ this backΒ intoΒ ourΒ LaGrangeΒ equation.This yields theΒ pointsΒ (-3,-1)Β andΒ (3,1).Β TheΒ signsΒ
mustΒ beΒ theΒ sameΒ dueΒ toΒ ourΒ LaGrangeΒ equation.Β
Β (
3π¦)^2Β +Β π¦^2Β =Β 10Β
Β Β
Β
10π¦
2
=Β 10
Β Β
Β
π¦Β =Β Β±1
Β Β Β
@y=-1Β :Β 3(-1)=-3=xΒ Β
@y=1Β :Β 3(1)=3=xΒ Β
StepΒ 4Β Β
WeΒ plugΒ theseΒ intoΒ ourΒ f(x,y)Β toΒ findΒ that:Β Β
(3,1)Β is aΒ maximumΒ Β
(-3,-1)Β is aΒ minimumΒ Β
f(-3,-1)=3(-3)-1=-10Β Β
f(3,1)=3(3)+1=10Β Β
resultΒ Β
Max:Β 10Β atΒ (3,1)Β Β
Min:Β -10Β atΒ (-3,-1)Β
QuestionΒ #7Β
FactorΒ
Β
π₯
3
βΒ 10π₯
2
+Β 29π₯
Β
Β
Answer:Β
FactorΒ theΒ following:Β
Β
π₯
3
βΒ 10π₯
2
+Β 29π₯
Β Β
Hint:Β FactorΒ commonΒ terms outΒ ofΒ Β
Β
π₯
3
βΒ 10π₯
2
+Β 29π₯.
Β Β
FactorΒ xΒ outΒ ofΒ Β
Β
π₯
3
βΒ 10π₯
2
+Β 29π₯
:Β Β
Answer:Β Β
Β
π₯(π₯
2
βΒ 10π₯Β +Β 29)
Β
Β
QuestionΒ #8Β
OfΒ theΒ multiplesΒ ofΒ 2Β betweenΒ 1Β andΒ 99,Β whatΒ fractionΒ areΒ alsoΒ multiples ofΒ 3?Β
π΄)
16
99
Β
π΅)
16
49
Β
πΆ)
1
3
Β
π·)
33
49
Β
49
99
Β
Answer:Β
StepΒ 1Β
theΒ multiplesΒ ofΒ 2Β betweenΒ 1Β andΒ 99Β areΒ theΒ evenΒ numbers .Β
Therefore,Β thereΒ areΒ 49Β evenΒ numbers fromΒ 1Β toΒ 99Β
TheΒ numbers thatΒ areΒ multipleΒ ofΒ 2Β areΒ 2,4,6,8,10,12,...49Β
outΒ ofΒ thoseΒ evenΒ numbers,Β theΒ numbers thatΒ areΒ alsoΒ theΒ multiplesΒ ofΒ 3Β areΒ theΒ numbers thatΒ areΒ divisibleΒ
by 3.Β
thereforeΒ fromΒ theΒ setΒ 2,4,6,8,10,12,...49Β every thirdΒ numberΒ isΒ multipleΒ ofΒ 3.Β
therefore,Β thereΒ areΒ 16Β suchΒ numbersΒ
StepΒ 2Β
therefore,Β theΒ requiredΒ fractionΒ isΒ givenΒ byΒ 1649Β
theΒ answerΒ isΒ 1649
Β
QuestionΒ #9
Β
TimΒ always eatsΒ strawberriesΒ inΒ multiplesΒ ofΒ three.Β ToddΒ eatsΒ themΒ inΒ multiplesΒ ofΒ fourΒ andΒ RickΒ inΒ
multiples ofΒ six.Β GavinΒ is notΒ fussy aboutΒ itΒ butΒ likes creamΒ toΒ goΒ withΒ them.Β UncleΒ SamΒ wantsΒ eachΒ ofΒ theΒ
fourΒ kids toΒ eatΒ theΒ sameΒ numberΒ ofΒ strawberries.Β WhatΒ is theΒ minimumΒ numberΒ ofΒ strawberriesΒ thatΒ heΒ
has toΒ giveΒ toΒ eachΒ kid?Β
SelectΒ oneΒ answer:Β
Β
π΄)24Β π΅)6Β πΆ)3Β π·)1Β πΈ)12
Β
Β
Answer:
Β
StepΒ 1Β
TimΒ eatsΒ multiples ofΒ threeΒ
ToddΒ eatsΒ multiplesΒ ofΒ fourΒ
Rick eatsΒ multiplesΒ ofΒ six.Β
GavinΒ isΒ notΒ bussy.Β
Thus weΒ haveΒ numbers toΒ work withΒ 3,4,6.Β
StepΒ 2Β
Therefore,Β sinceΒ uncleΒ SamΒ wantΒ toΒ giveΒ
sameΒ numberΒ ofΒ strawberries.Β
ThenΒ theΒ numberΒ Β
=Β 2ππ(3,4,6)
Β Β
Β
=Β 12
Β Β
Therefore,Β uncleΒ SamΒ shouldΒ giveΒ 12Β strawberriesΒ toΒ eachΒ ofΒ kid.Β
OptionΒ EΒ isΒ correct.
Β
QuestionΒ #10Β
ExerciseΒ 1:Β CountingΒ binary strings.Β
CountΒ theΒ numberΒ ofΒ binary strings ofΒ lengthΒ 10Β subjectΒ toΒ eachΒ ofΒ theΒ followingΒ restrictions.Β
ThereΒ isΒ only oneΒ binary stringΒ ofΒ lengthΒ tenΒ withΒ noΒ 1's:Β 00000000000.Β ThereΒ areΒ Β
2
10
Β Β binary strings ofΒ
lengthΒ ten.Β ThereforeΒ theΒ numberΒ ofΒ binary strings ofΒ lengthΒ tenΒ withΒ atΒ leastΒ oneΒ 1Β isΒ Β
2
10
βΒ 1
Β .Β
(b)Β
TheΒ stringΒ has atΒ leastΒ oneΒ 1Β andΒ atΒ leastΒ oneΒ 0.Β
(c)Β
TheΒ stringΒ containsΒ exactly fiveΒ 1's orΒ itΒ beginsΒ withΒ aΒ 0.Β Β
ExerciseΒ 2:Β CountingΒ integerΒ multiples.Β
(b)Β
HowΒ many integersΒ inΒ theΒ rangeΒ 1Β throughΒ 140Β areΒ integerΒ multiples ofΒ 2,Β 5,Β orΒ 7?
Β
Answer:Β
StepΒ 1Β
TheΒ stringΒ has atΒ leastΒ oneΒ 1Β andΒ atΒ leastΒ oneΒ 0.Β
This removes theΒ possibilityΒ ofΒ allΒ oneΒ andΒ allΒ zerosΒ
Thus numberΒ ofΒ binary stringsΒ is Β
2
10
βΒ 2
Β Β
StepΒ 2Β
TheΒ stringΒ containsΒ exactly fiveΒ 1's orΒ itΒ beginsΒ withΒ aΒ 0.Β
FirstΒ digitΒ is fixedΒ as zeroΒ Β
InΒ remainingΒ 9Β places LetsΒ chooseΒ 5Β places andΒ placeΒ 1s thereΒ Β
NumberΒ ofΒ ways areΒ Β
πΆ
5
9
Β Β
126Β
StepΒ 3Β
HowΒ many integersΒ inΒ theΒ rangeΒ 1Β throughΒ 140Β areΒ integerΒ multiples ofΒ 2,Β 5,Β orΒ 7Β
ThereΒ areΒ 70Β multiplesΒ ofΒ 2Β
23Β multiplesΒ ofΒ 3Β leftΒ Β
weΒ haveΒ 9Β multiples ofΒ 5Β thatΒ haveΒ notΒ yetΒ beenΒ included.Β
TotalΒ is 102
Β
QuestionΒ #11Β
Β
1Β =Β ππ₯Β +Β ππ¦
Β ,Β forΒ allΒ integersΒ xΒ andΒ yΒ
Β
1Β =Β ππ₯Β +Β ππ¦
Β ,Β forΒ someΒ integers xΒ andΒ yΒ
Β 1Β =Β ππ₯Β +Β ππ¦
Β ,Β forΒ uniqueΒ integers xΒ andΒ yΒ
All theΒ above
Β
Answer:Β Β
StepΒ 1Β
TwoΒ integers areΒ relativelyΒ 'Β whenΒ thereΒ areΒ noΒ commonΒ factors otherΒ thanΒ 1.Β This means thatΒ noΒ otherΒ
integerΒ couldΒ divideΒ bothΒ numbers.Β
TwoΒ integers a,bΒ areΒ calledΒ relatively 'Β toΒ eachΒ otherΒ ifΒ Β
Β πππ(π,Β π)Β =Β 1
Β .Β
StepΒ 2Β
ForΒ anyΒ positiveΒ integersΒ aΒ andΒ b,Β thereΒ existΒ integers xΒ andy suchΒ thatΒ Β axΒ + by = gcd(a,Β b)Β .Β Furthermore,Β
as xΒ andΒ yΒ vary overΒ allΒ integers aΒ
π₯Β Β +Β Β ππ¦
Β Β attainsΒ allΒ multiples andΒ only multiplesΒ ofΒ gcd(a,Β b).Β
HereΒ Β
πππ(π,Β π)Β =Β 1
Β Β
SoΒ thereΒ existΒ integers xΒ andΒ yΒ suchΒ thatΒ Β
ππ₯Β Β +Β Β ππ¦Β =Β Β 1
Β Β
SoΒ theΒ correctΒ choiceΒ isΒ b.Β
Β
1Β =Β Β ππ₯Β Β +Β Β ππ¦Β
,Β forΒ someΒ integers xΒ andΒ y.
Β
QuestionΒ #12Β
ConsiderΒ theΒ functionΒ
π(π₯)Β =Β 4Β sin(Ο2(π₯Β βΒ 3))Β +Β 8
.Β StateΒ theΒ amplitudeΒ A,Β periodΒ P,Β andΒ midline.Β StateΒ theΒ
phaseΒ shiftΒ andΒ verticalΒ translation.Β InΒ theΒ fullΒ periodΒ [0,Β P],Β stateΒ theΒ maximumΒ andΒ minimumΒ y-valuesΒ andΒ
theirΒ correspondingΒ x-values.Β
EnterΒ theΒ exactΒ answers.Β
Amplitude:Β Β
π΄Β =
Β Β Β
Period:Β Β
πΒ =
Β Β
Midline:Β Β
π¦Β =
Β Β
TheΒ phaseΒ shiftΒ is _____Β
TheΒ verticalΒ translationΒ is_______Β
HintsΒ forΒ theΒ maximumΒ andΒ minimumΒ values ofΒ f(x):Β
TheΒ maximumΒ valueΒ ofΒ Β
π¦Β =Β sin(π₯)
Β Β is Β
π¦Β =Β 1
Β Β andΒ theΒ correspondingΒ xΒ values areΒ Β
π₯Β =Β π2
Β Β andΒ multiplesΒ
ofΒ Β
2π
Β Β less thanΒ andΒ moreΒ thanΒ this xΒ value.Β YouΒ may wantΒ toΒ solveΒ Β
π2(π₯Β βΒ 3)Β =Β π2
Β .Β
TheΒ minimumΒ valueΒ ofΒ Β
π¦Β =Β sin(π₯)
Β Β isΒ Β
π¦Β =Β β1
Β Β andΒ theΒ correspondingΒ xΒ valuesΒ areΒ Β
π₯Β =Β 3Β π
Β 2Β Β andΒ
multiples ofΒ Β
2π
Β Β lessΒ thanΒ andΒ moreΒ thanΒ thisΒ xΒ value.Β YouΒ mayΒ wantΒ toΒ solveΒ Β
π2(π₯Β βΒ 3)Β =Β 3π2
Β .Β
IfΒ youΒ getΒ aΒ valueΒ forΒ xΒ thatΒ isΒ less thanΒ 0,Β youΒ couldΒ addΒ multiples ofΒ PΒ toΒ getΒ intoΒ theΒ nextΒ cycles.Β
IfΒ youΒ getΒ aΒ valueΒ forΒ xΒ thatΒ is moreΒ thanΒ P,Β youΒ couldΒ subtractΒ multiples ofΒ PΒ toΒ getΒ intoΒ theΒ previousΒ
cycles.Β
ForΒ xΒ inΒ theΒ intervalΒ [0,Β P],Β theΒ maximumΒ y-valueΒ andΒ correspondingΒ x-valueΒ isΒ at:Β
Β
πΒ =
Β Β
Β
πΒ =
Β Β
ForΒ xΒ inΒ theΒ intervalΒ [0,Β P],Β theΒ minimumΒ y-valueΒ andΒ correspondingΒ x-valueΒ is at:Β
Β
πΒ =
Β Β
Β
πΒ =
Β
Β
Answer:Β
StepΒ 1Β
weΒ haveΒ Β
π(π₯)Β =Β 4Β sin(2Ο(π₯Β βΒ 3))Β +Β 8
Β Β
AmplitudeΒ Β
π΄Β =
Β cofficientΒ ofΒ Β
sin(2Ο(π₯Β βΒ 3))Β =Β 4
Β periodΒ Β
πΒ =Β 2Β π
Β Β cofficientΒ ofΒ Β
(π₯Β βΒ 3)Β =Β 2π2πΒ =Β 1
Β Β
maximumΒ valueΒ andΒ minimumΒ valueΒ Β
β1Β β€Β sin(2Ο(π₯Β βΒ 3))Β β€Β 1Β βΒ 4Β β€Β 4Β sin(2Ο(π₯Β βΒ 3))Β β€Β 4Β βΒ 4Β +Β 8Β β€
4Β sin(2Ο(π₯Β βΒ 3))Β +Β 8Β β€Β 4Β +Β 84Β β€Β 4Β sin(2Ο(π₯Β βΒ 3))Β +Β 8Β β€Β 12
Β Β maximumΒ valueΒ is 12Β andΒ minimumΒ valueΒ isΒ 4.Β
StepΒ 2Β
inΒ intervalΒ [0,1]Β accordingΒ toΒ graphΒ ofΒ functionΒ Β
π(π₯)Β =Β 4Β sin(2Ο(π₯Β βΒ 3))Β +Β 8
Β Β maximumΒ valueΒ Β
π₯Β =Β 0.25
Β Β
andΒ minimumΒ valueΒ atΒ Β
π₯Β =Β 0.75
Β Β
Β
QuestionΒ #13Β
ConsiderΒ theΒ functionΒ Β
π(π₯)Β =Β 2Β sin(Ο2(π₯Β βΒ 3))Β +Β 4
Β .Β
StateΒ theΒ amplitudeΒ A,Β periodΒ P,Β andΒ midline.Β StateΒ theΒ phaseΒ shiftΒ andΒ verticalΒ translation.Β InΒ theΒ fullΒ
periodΒ [0,Β P],Β stateΒ theΒ maximumΒ andΒ minimumΒ y-valuesΒ andΒ theirΒ correspondingΒ x-values.Β
HintsΒ forΒ theΒ maximumΒ andΒ minimumΒ values ofΒ f(x):Β
TheΒ maximumΒ valueΒ ofΒ Β
π¦Β =Β sin(π₯)
Β Β is Β
π¦Β =Β 1Β
Β andΒ theΒ correspondingΒ xΒ valuesΒ areΒ Β
π₯Β =Β π2
Β Β andΒ multiples ofΒ Β
2\piΒ Β less thanΒ andΒ moreΒ thanΒ thisΒ xΒ value.Β YouΒ mayΒ wantΒ toΒ solveΒ Β
π2(π₯Β βΒ 3)Β =Β π2
Β .Β
TheΒ minimumΒ valueΒ ofΒ Β
π¦Β =Β sin(π₯)
Β Β isΒ Β y=-1Β Β andΒ theΒ correspondingΒ xΒ values areΒ Β
π₯Β =Β 3π2
Β Β andΒ multiplesΒ ofΒ Β
2π
Β Β lessΒ thanΒ andΒ moreΒ thanΒ this xΒ value.Β YouΒ may wantΒ toΒ solveΒ Β
π2(π₯Β βΒ 3)Β =Β 3π2
Β .Β
IfΒ youΒ getΒ aΒ valueΒ forΒ xΒ thatΒ isΒ less thanΒ 0,Β youΒ couldΒ addΒ multiples ofΒ PΒ toΒ getΒ intoΒ theΒ nextΒ cycles.Β
IfΒ youΒ getΒ aΒ valueΒ forΒ xΒ thatΒ is moreΒ thanΒ P,Β youΒ couldΒ subtractΒ multiples ofΒ PΒ toΒ getΒ intoΒ theΒ previousΒ
cycles.Β
ForΒ xΒ inΒ theΒ intervalΒ [0,Β P],Β theΒ maximumΒ y-valueΒ andΒ correspondingΒ x-valueΒ isΒ at:Β
Β
π₯Β =
Β Β
Β
π¦Β =
Β Β
ForΒ xΒ inΒ theΒ intervalΒ [0,Β P],Β theΒ minimumΒ y-valueΒ andΒ correspondingΒ x-valueΒ is at:Β
Β
π₯Β =
Β Β
Β
π¦Β =
Β
Β
Answer:Β
StepΒ 1Β
ConsiderΒ aΒ givenΒ functionΒ
Β
π(π₯)Β =Β 2Β sinΒ [
Ο
2
(π₯Β βΒ 3)Β +Β 4]
Β .Β
StateΒ theΒ amplitudeΒ A,Β periodΒ PΒ andΒ mid-line.Β
StateΒ theΒ phaseΒ shiftΒ andΒ verticalΒ translation.Β
StateΒ theΒ minimumΒ andΒ maximumΒ y-valuesΒ andΒ theirΒ corrospondingΒ x-valuesΒ inΒ theΒ intervalΒ [0,P].Β
StepΒ 2Β
a)Β ConsiderΒ aΒ givenΒ functionΒ
Β
π(π₯)Β =Β 2Β sinΒ [
Ο
2
(π₯Β βΒ 3)]Β +Β 4
Β Β
TheΒ amplitudeΒ ofΒ theΒ functionΒ fΒ isΒ 2.Β Hence,Β Β
π΄Β =Β 2
Β .Β
TheΒ periodΒ ofΒ theΒ functionΒ fΒ is Β
2Ο
(
Ο
2
)
=Β 4
Β .Β Hence,Β Β
πΒ =Β 4
Β .Β
TheΒ mid-lineΒ ofΒ theΒ trigonometric functionΒ is aΒ pointΒ whereΒ theΒ functionΒ attainΒ itsΒ maximumΒ values.Β
Now,Β
Β
Ο
2
(π₯Β βΒ 3)Β =
(2πβ1)Ο
2
Β Β
Β
π₯Β βΒ 3Β =Β 2πΒ βΒ 1
Β Β
Β
βΒ π₯Β =Β 2πΒ +Β 2
Β Β forΒ allΒ Β
πβΒ βΒ π
Β Β
Hence,Β theΒ mid-lineΒ ofΒ theΒ functionΒ fΒ isΒ Β
π₯Β =Β 2πΒ +Β 2
Β Β forΒ allΒ Β
πΒ Β β
Β
π
Β Β
StepΒ 3Β
b)Β ConsiderΒ aΒ givenΒ functionΒ
Β
π(π₯)Β =Β 2Β sinΒ [
Ο
2
(π₯Β βΒ 3)]Β +Β 4
Β
TheΒ functionΒ fΒ is shiftedΒ rightΒ sideΒ byΒ 3Β units.Β Hence,Β itsΒ phaseΒ shiftΒ is 3.Β
TheΒ functionΒ fΒ is shiftedΒ verticallyΒ upwardΒ by 4Β units.Β Hence,Β itsΒ vertically shiftΒ isΒ 4Β units.Β
StepΒ 4Β
c)Β
TheΒ rangeΒ ofΒ theΒ trigonometric functionΒ is Β
[β1,1]
Β .Β
ItΒ implies thatΒ Β
β1Β β€Β sin(π₯)Β β€Β 1
Β .Β
Hence,Β
Β
β1Β β€Β sinΒ (
Ο
2
(π₯Β βΒ 3))Β β€Β 1
Β Β
Β
β2Β β€Β 2Β sinΒ (
Ο
2
(π₯Β βΒ 3))Β β€Β 2
Β Β
Β
β2Β +Β 4Β β€Β 2Β sinΒ (
Ο
2
(π₯Β βΒ 3))Β +Β 4Β β€Β 2Β +Β 4
Β Β
Β
2Β β€Β π(π₯)Β β€Β 6
Β Β
Hence,Β minimumΒ valueΒ ofΒ theΒ functionΒ fΒ is 2Β andΒ theΒ maximumΒ valueΒ ofΒ theΒ functionΒ fΒ isΒ 6.Β
ItΒ implies thatΒ Β
π
πππ
=Β 2
Β Β andΒ Β
π
πππ₯
=Β 6
Β .Β
StepΒ 5Β
Further,Β
MaximumΒ attainΒ atΒ Β
π₯Β =Β 2πΒ +Β 2
Β Β forΒ allΒ Β
πΒ Β βΒ π
Β .Β
Hence,Β theΒ pointΒ ofΒ maximaΒ is Β
π₯Β =Β 0,2,4Β
.Β
Β
sinΒ (
Ο
2
(π₯Β βΒ 3))Β =Β 0
Β Β
Β
Ο
2
(π₯Β βΒ 3)Β =Β π
Β Β
Β
π₯Β βΒ 3Β =Β 2π
Β Β
Β
βΒ π₯Β =Β 2πΒ +Β 3
Β Β forΒ allΒ Β
πΒ Β βΒ π
Β Β
Hence,Β theΒ pointΒ ofΒ minimaΒ is 1,3.
Β
QuestionΒ #14Β
FiftyΒ raffleΒ ticketsΒ areΒ numberedΒ 1Β throughΒ 50,Β andΒ oneΒ ofΒ themΒ is drawnΒ atΒ random.Β WhatΒ is theΒ
probabilityΒ thatΒ theΒ numberΒ is aΒ multipleΒ ofΒ 5Β orΒ 7?Β ConsiderΒ theΒ followingΒ ''solution'':Β SinceΒ 10Β ticketsΒ bearΒ
numbers thatΒ areΒ multiples ofΒ 5Β andΒ 7Β ticketsΒ bearΒ numbers thatΒ areΒ multiplesΒ ofΒ 7,Β weΒ concludedΒ theΒ
requiredΒ probabilityΒ is as follows.Β
10
50
+
7
50
=
17
50
Β
IsΒ thisΒ theΒ correctΒ answer?Β (IfΒ so,Β enterΒ yes.Β IfΒ not,Β enterΒ theΒ correctΒ answer.)
Β
Answer:Β
StepΒ 1Β
FiftyΒ raffleΒ ticketsΒ areΒ numberedΒ 1Β throughΒ 50,Β andΒ oneΒ ofΒ themΒ is drawnΒ atΒ random.Β WhatΒ is theΒ
probabilityΒ thatΒ theΒ numberΒ is aΒ multipleΒ ofΒ 5Β orΒ 7?Β ConsiderΒ theΒ followingΒ ''solution'':Β SinceΒ 10Β ticketsΒ bearΒ
numbers thatΒ areΒ multiples ofΒ 5Β andΒ 7Β ticketsΒ bearΒ numbers thatΒ areΒ multiplesΒ ofΒ 7,Β weΒ concludedΒ theΒ
requiredΒ probabilityΒ is as follows.Β
StepΒ 2Β
LetΒ AΒ beΒ theΒ eventΒ thatΒ theΒ numberΒ is multipleΒ ofΒ 5Β
BΒ beΒ theΒ eventΒ thatΒ theΒ numberΒ isΒ multipleΒ ofΒ 7Β
FindΒ Β
π(π΄Β βͺΒ π΅)
Β Β
Therefore,Β Β
π(π΄Β βͺΒ π΅)Β =Β π(π΄)Β +Β π(π΅)Β βΒ π(π΄Β β©Β π΅)
Β Β
NumberΒ ofΒ ticketsΒ thatΒ areΒ multipliesΒ ofΒ bothΒ 5Β andΒ 7Β areΒ Β
50
5Γ7
=
50
35
=Β 1
Β Β
Β
π(π΄Β β©Β π΅)Β =
1
50
Β Β
Β
π(π΄Β βͺΒ π΅)Β =
10
50
+
7
50
β
1
50
Β Β
Β
=
17
50
β
1
50
Β Β
Β
=
16
50
Β Β
TheΒ probabilityΒ thatΒ theΒ numberΒ ofΒ ticketsΒ areΒ multipliesΒ ofΒ 5Β orΒ 7Β is Β
16
50
Β
QuestionΒ #15Β
ConsiderΒ theΒ functionΒ Β
π(π₯)Β =Β 3Β sinΒ (
Ο
3
(π₯Β βΒ 4))Β +Β 5
Β .Β
StateΒ theΒ amplitudeΒ A,Β periodΒ P,Β andΒ midline.Β StateΒ theΒ phaseΒ shiftΒ andΒ verticalΒ translation.Β InΒ theΒ fullΒ
periodΒ [0,Β P],Β stateΒ theΒ maximumΒ andΒ minimumΒ y-valuesΒ andΒ theirΒ correspondingΒ x-values.Β
EnterΒ theΒ exactΒ answers.Β
Amplitude:Β Β
π΄Β =
Β Β
Period:Β Β
πΒ =
Β Β
Midline:Β Β
π¦Β =
Β Β Β
TheΒ phaseΒ shiftΒ is:Β Β
a.Β upΒ 4Β unitsΒ
b.Β 4Β unitsΒ toΒ theΒ rightΒ
c.Β 4Β unitsΒ toΒ theΒ leftΒ
d.Β 5Β unitsΒ toΒ theΒ leftΒ
TheΒ verticalΒ translationΒ is:Β Β
a.Β upΒ 4Β unitsΒ
b.Β downΒ 5Β unitsΒ
c.Β upΒ 5Β unitsΒ
d.Β downΒ 4Β unitsΒ
HintsΒ forΒ theΒ maximumΒ andΒ minimumΒ values ofΒ f(x):Β
TheΒ maximumΒ valueΒ ofΒ Β
π¦Β =Β sin(π₯)
Β Β is Β
π¦Β =Β 1
Β Β andΒ theΒ correspondingΒ xΒ values areΒ Β
π₯Β =
Ο
2
Β Β andΒ multiplesΒ ofΒ Β
2π
Β Β lessΒ thanΒ andΒ moreΒ thanΒ this xΒ value.Β YouΒ may wantΒ toΒ solveΒ Β
Ο
3
(π₯Β βΒ 4)Β =
Ο
2
Β .Β
TheΒ minimumΒ valueΒ ofΒ Β
π¦Β =Β sin(π₯)
Β Β isΒ Β
π¦Β =Β β1
Β Β andΒ theΒ correspondingΒ xΒ valuesΒ areΒ Β
π₯Β =
3Ο
2
Β Β andΒ multiplesΒ
ofΒ Β
2π
Β less thanΒ andΒ moreΒ thanΒ thisΒ xΒ value.Β YouΒ mayΒ wantΒ toΒ solveΒ Β
Ο
3
(π₯Β βΒ 4)Β =
3Ο
2
Β .Β
IfΒ youΒ getΒ aΒ valueΒ forΒ xΒ thatΒ isΒ less thanΒ 0,Β youΒ couldΒ addΒ multiples ofΒ PΒ toΒ getΒ intoΒ theΒ nextΒ cycles.Β
IfΒ youΒ getΒ aΒ valueΒ forΒ xΒ thatΒ is moreΒ thanΒ P,Β youΒ couldΒ subtractΒ multiples ofΒ PΒ toΒ getΒ intoΒ theΒ previousΒ
cycles.Β
ForΒ xΒ inΒ theΒ intervalΒ [0,Β P],Β theΒ maximumΒ y-valueΒ andΒ correspondingΒ x-valueΒ isΒ at:Β
x=Β
y=Β
ForΒ xΒ inΒ theΒ intervalΒ [0,Β P],Β theΒ minimumΒ y-valueΒ andΒ correspondingΒ x-valueΒ is at:Β
Β
π₯Β =
Β Β
Β
π¦Β =
Β
Β
Answer:Β
StepΒ 1Β
ConsiderΒ theΒ providedΒ question,Β
GivenΒ function,Β Β
π(π₯)Β =Β 3Β sinΒ (
Ο
3
(π₯Β βΒ 4))Β +Β 5
Β Β
FirstΒ findΒ Amplitude,Β PeriodΒ andΒ Midline.Β
Amplitude=coefficientΒ ofΒ Β
sinΒ (
Ο
3
(π₯Β βΒ 4))Β =Β 3
Β Β
PeriodΒ Β
=
2Ο
πππππππππππ‘ππΒ π₯
=
2Ο
Ο
3
=Β 6
Β Β
Midline:Β (y=verticalΒ displacement):Β Β
π¦Β =Β 5
Β Β
StepΒ 2Β
DrawΒ theΒ graphΒ ofΒ theΒ givenΒ function,Β
Β
StepΒ 3Β Β
FromΒ theΒ aboveΒ diagramΒ itΒ isΒ clearΒ that,Β Β
phaseΒ shiftΒ is 4Β unitsΒ toΒ theΒ right.Β Β
So,Β theΒ correctΒ answerΒ is optionΒ (b).Β Β
AndΒ theΒ verticalΒ translationΒ is 5Β units upΒ Β
(2Β +
8β2
2
=Β 5)
Β Β Β
So,Β theΒ correctΒ answerΒ is optionΒ (c).Β
StepΒ 4Β
Now,Β inΒ theΒ intervalΒ Β
[0,Β π]Β =Β [0,6]
Β theΒ maximumΒ valueΒ ofΒ yΒ correspondingΒ toΒ xΒ is,Β Β
FromΒ theΒ aboveΒ graphΒ inΒ theΒ intervalΒ [0,Β 6],Β Β
TheΒ maximumΒ valueΒ ofΒ Β
π¦Β =Β 8Β
Β correspondingΒ toΒ Β
π₯Β =Β 5.5
Β Β Β
So,Β Β
π₯Β Β =Β Β 5.5
Β Β
Β
π¦Β Β =Β Β 8
Β Β
StepΒ 5Β
InΒ theΒ intervalΒ Β [0,P]=[0,6]Β Β theΒ minimumΒ valueΒ ofΒ y correspondingΒ toΒ xΒ is,Β Β
FromΒ theΒ aboveΒ graphΒ inΒ theΒ intervalΒ [0,Β 6],Β Β
TheΒ minimumΒ valueΒ ofΒ Β
π¦Β Β =Β Β 2
Β Β correspondingΒ toΒ Β
π₯Β =Β 2.5
Β Β Β
So,Β Β
π₯Β Β =Β Β 2.5
Β Β Β
Β
π¦Β Β =Β Β 2
Β
Β
QuestionΒ #16Β
TheΒ UniversalΒ Set,Β U,Β consistsΒ ofΒ theΒ naturalΒ numbers fromΒ 20Β toΒ 60Β incluiveΒ
a.Β DefineΒ orΒ describeΒ inΒ words theΒ followingΒ threeΒ (3)Β sets:Β factorsΒ ofΒ 64,Β 'Β numbers,Β andΒ multiples ofΒ 3.Β
b.Β ListΒ theΒ elementsΒ inΒ eachΒ ofΒ yourΒ sets:Β
Β
π΄Β Β =Β Β Β {
Β Β
Β
π΅Β Β =Β Β Β {
Β Β
Β
πΆΒ Β =Β Β Β {
Β Β
c.Β DetermineΒ theΒ probabilityΒ ofΒ eachΒ ofΒ theΒ following:Β
πΌ.Β π(πΆ)
Β Β
Β
πΌπΌ.Β π(π΄Β βͺΒ π΅)
Β Β
Β
πΌπΌπΌ.Β π(π΄Β β©Β π΅Β β©Β πΆ)
Β Β
Β
πΌπ.Β π(π΅\πΆ)
Β Β
Β
π.Β π((π΄\π΅)Β β©Β πΆ)
Β
Β
Answer:Β
StepΒ 1:Β Given,Β
TheΒ universalΒ setΒ U, consistΒ ofΒ theΒ naturalΒ numbers fromΒ 20Β toΒ 60Β inclusive.Β WeΒ haveΒ toΒ answerΒ theΒ
following...Β
StepΒ 2:Β ExplanationΒ
SolutionΒ (a).Β
IfΒ weΒ haveΒ 3Β sets,Β Β
Β
π
1
=
Β factors ofΒ 64Β i.e.Β 1,2,4,8,16,32,64Β
Β
π
2
=
Β Β 'Β numbersΒ
Β
π
3
=
Β Β multipleΒ ofΒ 3Β
NowΒ describeΒ theΒ setsΒ inΒ words,Β Β
Β
π
1
=Β Β Β {π₯Β βΆΒ Β π₯Β
Β is aΒ factorΒ ofΒ 64Β andΒ Β
20Β Β β€Β π₯Β Β β€Β 60}Β
Β
Β
π
2
=Β Β Β {π₯Β βΆΒ Β π₯
Β Β isΒ 'Β andΒ Β
20Β Β β€Β π₯Β Β β€Β 60}
Β Β
Β
π
3
=Β Β Β {π₯Β βΆΒ Β π₯
Β Β is multipleΒ ofΒ 3Β andΒ Β
20Β Β β€Β π₯Β Β β€Β 60}
Β Β
SolutionΒ (b).Β Β
ListΒ ofΒ theΒ elementsΒ
Β
π
1
=Β 32
Β Β
Β
π
2
=Β 23,29,31,37,41,43,47,53,59
Β Β
Β
π
3
=Β 21,24,27,30,Β β¦Β ,57,60
Β Β
SolutionΒ (c).Β
DetermineΒ theΒ probabilityΒ
SinceΒ weΒ haveΒ fromΒ theΒ bΒ part,Β Β
π
1
βͺΒ π
2
=Β 32,23,29,31,37,41,43,47,53,59
Β Β
Β
π
1
β©Β π
2
β©Β π
3
=
Β Β
Β
πΌ.Β π(π
3
)Β =
14
40
=
7
20
Β Β
Β
πΌπΌ.Β π(π
1
βͺΒ π
2
)Β =
10
40
=
1
4
Β Β
Β
πΌπΌπΌ.Β π(π
1
β©Β π
2
β©Β π
3
)Β =
0
40
=Β 0
Β
Β
QuestionΒ #17Β
Values forΒ theΒ functionΒ f(x)Β areΒ shownΒ inΒ theΒ table.Β
Β
0
1
2
3
4
5
(Β )
2
4
8Β 16
32
64
x
fΒ x
Β
WhichΒ statementΒ proves thatΒ f(x)Β isΒ anΒ exponentialΒ function?Β
A)Β AllΒ ofΒ theΒ values ofΒ f(x)Β areΒ oddΒ numbers.Β
B)Β AllΒ ofΒ theΒ values ofΒ f(x)Β areΒ multiples ofΒ 2.Β
C)Β TheΒ functionΒ f(x)Β grows by equalΒ factors overΒ equalΒ intervals.Β
D)Β TheΒ functionΒ f(x)Β grows by equalΒ differences overΒ equalΒ intervals.
Β
Answer:Β
StepΒ 1Β
GivenΒ aΒ tableΒ ofΒ valuesΒ ofΒ f(x)Β atΒ differentΒ points.Β
StepΒ 2Β
anΒ exponentialΒ functionΒ is ofΒ theΒ typeΒ Β
π(π₯)Β =Β π
π₯
Β Β whereΒ aΒ is constantΒ andΒ xΒ is variable.Β
by examiningΒ theΒ tableΒ theΒ givenΒ functionΒ is Β
π(π₯)Β =Β 2
π₯+1
Β .Β
HenceΒ optionΒ BΒ is correct.Β
f(x)Β grows by equalΒ factorΒ overΒ equalΒ interval.
Β
QuestionΒ #18Β
Let
Β π
Β Β
=Β Β Β {Β 1,Β 2,Β 3,Β Β Β β¦Β ,2400Β }
Β Β
LetΒ SΒ beΒ theΒ subsetΒ ofΒ theΒ numbers inΒ U thatΒ areΒ multiplesΒ ofΒ 3, andΒ letΒ TΒ beΒ theΒ subsetΒ ofΒ U thatΒ areΒ
multiples ofΒ 7.Β
SinceΒ Β
2400Β Β Γ·Β 3Β Β =Β Β 800
Β ,Β itΒ follows thatΒ Β
π(π)Β =Β π({3Β β Β 1,3Β β Β 2,Β β―Β ,3Β β Β 800})Β =Β 800
Β .Β
(a)Β FindΒ n(T)Β usingΒ aΒ methodΒ similarΒ toΒ theΒ oneΒ thatΒ showedΒ thatΒ Β
π(π)Β =Β 800
Β .Β
(b)Β FindΒ Β
π(πΒ β©Β π)
Β .Β
(c)Β Label theΒ numberΒ ofΒ elementsΒ inΒ eachΒ regionΒ ofΒ aΒ two-loopΒ VennΒ diagramΒ withΒ theΒ universeΒ
π
Β andΒ
subsetsΒ SΒ andΒ T.Β
Questions:FindΒ
π(π
)Β ?Β FindΒ
π(πππ)
Β
Answer:Β
StepΒ 1Β
Solution:Β Β
π(βͺ)Β =Β 2400
Β Β
Β
βͺ=Β Β Β {1,2,3,Β β¦Β .Β .Β ,2400}
Β Β
Β
2400Β Β Γ·Β 3Β Β =Β Β 800
Β Β
Β
π(π )Β =Β π{3Β β Β 1,3Β β Β 2,Β β¦Β .3Β β Β 800}Β =Β 800
Β Β
Β
π(π‘)Β =Β π(7Β β Β 1,7Β β Β 2,7Β β Β 3,Β β¦Β .Β ,7Β β Β 342)Β =Β 342
Β Β
2394Β
StepΒ 2Β
Β
π(π‘)Β =Β 342
Β Β
Β
π Β Β β©
Β Β MultipleΒ ofΒ 7Β andΒ Β
3Β Β =Β Β 21
Β Β
Β
π(π Β β©Β π‘)Β =Β π(21Β β Β 1,21Β β Β 2,Β β¦Β .Β .21Β β Β 114)Β =Β 114
Β Β
2394Β
Β
π(π Β β©Β π‘)Β =Β 114
Β
Β
QuestionΒ #19Β
FindΒ theΒ twenty-eighthΒ positiveΒ multipleΒ ofΒ 4Β andΒ theΒ sumΒ ofΒ theΒ firstΒ twenty-eightΒ positiveΒ multiplesΒ ofΒ 4.Β
Β
π
28
=
Β Β
Β
π
28
=
Β
Β
Answer:Β
StepΒ 1Β
ToΒ calculateΒ theΒ twenty-eighthΒ positiveΒ multipleΒ ofΒ 4Β andΒ theΒ sumΒ ofΒ theΒ firstΒ twenty-eightΒ positiveΒ multiplesΒ
ofΒ 4.Β
StepΒ 2Β
FormulaΒ usedΒ forΒ nthΒ termΒ andΒ sumΒ ofΒ n-termsΒ ofΒ anΒ APΒ series areΒ givenΒ below:Β
Β
π
π
=Β πΒ +Β (πΒ βΒ 1)π
Β Β
Β
π
π
=
π
2
(2πΒ +Β (πΒ βΒ 1)π)
Β Β
Where,Β
Β
πΒ =
Β Β firstΒ termΒ
Β
πΒ =
Β Β commonΒ differenceΒ
Β
πΒ =
Β numberΒ ofΒ termsΒ
StepΒ 3Β Β
Now,Β Β
MultipleΒ ofΒ 4Β is:Β Β
4,Β 8,Β 12,Β 16β¦β¦Β Β
Β
πΒ Β =Β Β 4
Β Β Β
Β
πΒ Β =Β Β 8Β Β βΒ Β 4Β Β =Β Β 12Β Β βΒ Β 8Β Β =Β Β 4
Β Β Β
Β
πΒ Β =Β Β 28Β
Β Β
As theΒ commonΒ differenceΒ is same.Β Therefore,Β theΒ seriesΒ isΒ inΒ AP.Β
StepΒ 4Β
Therefore,Β
Β
π
π
=Β πΒ +Β (πΒ βΒ 1)π
Β Β
Β
π
28
=Β 4Β +Β (28Β βΒ 1)4
Β Β
Β
=Β 112Β
Β
Β
π
π
=
π
2
(2πΒ +Β (πΒ βΒ 1)π)
Β Β
Β
π
28
=
28
2
(2Β β Β 4Β +Β (28Β βΒ 1)4)
Β Β
Β
=Β 1624
Β
Β
QuestionΒ #20Β
Β
πΒ =
Β Β theΒ setΒ ofΒ naturalΒ numbers betweenΒ 10Β andΒ 20Β
Β
π΄Β =Β
Β EvenΒ numbersΒ
Β
π΅Β =
Β Β Multiples ofΒ 3Β whereΒ theΒ numberΒ isΒ less thanΒ 18.Β
Β
πΆΒ =
Β Β compositeΒ numbersΒ
FindΒ Β
(π΄
β²
ππΆ)Β βΒ π΅
Β
Answer:Β
StepΒ 1Β
Β
πΒ =
Β Β theΒ setΒ ofΒ naturalΒ numbers betweenΒ 10Β andΒ 20.Β
Β
πΒ =Β {11,12,13,14,15,16,17,18,19}
Β Β
Β
π΄Β =
Β Β evenΒ numbers Β
βΒ {2,4,6,Β β¦Β Β 18}
Β Β
Β
π΅Β =
Β Β multiples ofΒ threeΒ whereΒ theΒ numberΒ 15Β lessΒ thanΒ 18.Β
Β
π΅Β =Β {3,6,9,12,15}
Β Β
Β
πΆΒ =
Β Β compositeΒ numbersΒ
Β
=Β {4,6,8,9,10,12,14,15,16,18,Β β¦Β }
Β Β
StepΒ 2Β
FindΒ Β
(π΄
β²
ππΆ)Β βΒ π΅
Β Β
Hence,Β Β
π΄
β²
=Β {1,3,5,7,9,11,13,Β β¦Β }
Β Β
Now,Β Β
(π΄
β²
ππΆ)Β =Β {1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,Β β¦Β }
Β Β
andΒ Β
(π΄
β²
ππΆ)Β βΒ π΅Β =Β {1,4,5,7,8,10,11,13,14,16,17,19}
Β Β
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