soutions to practice paper : igher athematics - spta maths · · 2016-05-17soutions to practice...

01

Solutions to Practice Paper C: Higher Mathematics

2 marks

• Rememberthatlimitsonlyexistif themultiplier,inthiscase0·9,liesbetween-1and1

• NocalculaterisavailableduringPaper1soyouhavetobeabletocalculate

− 10 1. .Theeasiestwasistomultiply

‘top’and‘bottom’by10.

HMRN:p24

1.Let the limit be Lthen L = 0·9L -1⇒ L - 0·9L = -1 ⇒ 0·1L = -1

SoL = − ×× = − = −1

0 1

1010

10

110

.

( )( )

ChoiceA

2 marks

• Thegradientformulais:

IfP(x1,y1),Q(x2,y2) then my y

x xPQ =−−

2 1

2 1

.

Youarenotgiventhisformuladuringtheexam.

• − =24

kmultiplybothsidesbyk

so − ×2

1

1

kk =4× k ⇒ -2 = 4k etc....

• Itispossible,toworkthroughthechoices,ineachcasefindingpointsAandBandcalculatingthegradient.Thiswillbemoretimeconsuming

HMRN:p3

2.A(2k,3)andB(k,5)

⇒ = −−

= − = −mk k k kAB3 5

2

2 2

butmkAB So= − =42

4

⇒− = ⇒ = − = −2 42

4

1

2k k

ChoiceB

• The‘differenceof squares’patternis:

A2- B2=(A- B)(A+B)

Inthiscaseyouhavex2- 22

• Forapolynomialequationf(x)= 0thereisacorrespondencebetweenrootsandfactors:-1isaroot↔ x + 1isafactorf(-1)= 0tellsyou-1isaroot.

HMRN:p26

3.x2-4=(x- 2)(x+ 2)soboth(x- 2)and(x+ 2)arefactersSincef (- 1)=0,(x+ 1)isafactorsof (x)=(x+ 1)(x- 2)(x+ 2) ChoiceD

2 marks

• Theprocessis:

u0 u1 u2

multiply by −then add 1

12 multiply by −

then add 1

12

Inthiscase:

4 −1

4×( ) + 112− − 1×( ) + 11

2−

32

HMRN:p23

4.

un+ 1= −1

2 un + 1,uo =4

So u1 = − 1

2 × 4+ 1 = -2 + 1 = -1

andu2 = − 1

2 × (-1) + 1

= 1

21

3

2+ =

Choice C

2 marks

WORKED ANSWERS: EXAM C PAPER 1Practice Paper M Answers Paper 1


• Theinitialstrategyistochangetheequationtotheformat:cos(angle)=number

inthiscasecos x= 22

• Youhavetorecogniseanexactvalue

however 22isnotthenormalform.

• 2

2

2 2

2 2

21

12 2

1

2= = =×

×

is an

alternative.

HMRN:p15,p34

5.

2cos x− 2 = 0 ⇒ 2cos x = 2

⇒ cos x = 2

2

2

2 2

1

2=

×=

So x =π4

π–4

1

1

√2

(1stquadrantonly) ChoiceB

2 marks

• Theinitialstrategyhereistochangetheequationtotheformat: y =mx+c

inthiscasem=− 34

.

• Ingeneralif mab

mba= = −then 1,

theprocessisto‘invert’thefractionandchangethesign.Inthiscasenegativechangestopositive.

HMRN:p4–5

6. 4y = -3x + 2

⇒ y = − +3

4

2

4x

So m m= − ⇒ =3

4

4

31

Choice C

2 marks

7.E(1,-1,-1), F(-1,-1,0), G(-7,-1,3)

EF→

= − =−−

−

−−−

f e

1

1

0

1

1

1

=

−

2

0

1

FG→

= − =−−

−

−−

g f

7

1

3

1

1

0

=

−

6

0

3

So FG EF→

=→

3

E

1 part 3 parts

F G

FdividesEGintheratio1:3 Choice C

2 marks

• If youknowthreepointsarecollinearthenselecttwodifferent‘journeys’betweenthepoints.Inthesolution

EF FGand→ →

wereselected.The‘position

vector’resultegAB→

= −b aisusedtocalculatethecomponentsof these‘journeys’.Collinearityensuresthatone‘journey’isamultipleof theother

‘journey’.InthiscaseFG EF→

=→

3 . Otherselections,inthiscase,wouldgive:

EF EG FG EG GE FE=→ →

=→ →

=→1

434

4, , etc.

Thenusethisinformationtosketchtherelationshipbetweenthepoints(seediagraminsolution)andthereforedeterminetherequiredratio.

HMRN:p44


• Thequestionistryingtoconfuseyoubynotgivingthequadraticin‘standardform’i.e.withnobrackets:x2+2x-8

• Remembertotakeyouranswer,expanditandcheckitworks.Inthiscase:

(x+1)2-9=x2+ 2x +1- 9=x2+2x-8

• bisnot9,whichisacommonmistake.Fortunatelythatisnotoneof thechoices.

HMRN:p13

9. (x+ 4)(x- 2) =x2+2x-8 =(x + 1)(x+ 1)-1-8 =(x+1)2-9Compare:(x+a)2+b sob=-9 ChoiceB

2 marks

2 marks

• Thefunctiongivenisof theformat

f(x)=(g(x))13whereg(x)= 1-x3.

The‘chainrule’isusedinthiscase,so:

f x g x g x

g x x

'( ) ( ( )) '( )

'( )

= ×

= −

13

3

13

2where

• Notethat13

313

3 12 2 2× − = − × × = − ×( )x x x

HMRN:p48–49

10. f x x

f x x x

x

( ) ( )

'( ) ( ) ( )

(

= −

⇒ = − × −

= −

−

1

1

31 3

3

3 2

2

13

23

11 3 23− −x )

ChoiceA

• Analternativemethodwouldbetofirstconsider y =f(x)+ 1.Thisisthegraph y =f(x)shiftedup1unitparalleltothey-axis.Then y =-(f(x)+ 1)isthenewgraph‘flipped’inthex-axis.

• Remembertofollowyourstepsusingeachof thepoints(0,0)and(1,-1)tocheckyourfinalchoiceworksfortheknownpointsonthegraph.

HMRN:p9–102 marks

11.y= -(f(x)+ 1)=-f(x)-1y=f(x)→ y =-f(x)Thisisreflectioninthex-axisy= -f(x)→ y = - f(x)-1followedbyatranslationof 1unitinthenegativedirectionparalleltothey-axis ChoiceB

• The‘syntheticdivisionscheme’isused.

• Takecaretolookfor‘missingterms’.Inthiscasethereisno‘x2term’andthatisrecordedasa0inthetoprow.

• a

b

Polynomial coefficientsThismeansif youdividethe

polynomialbyx-atheremainderisb.

HMRN:p25–26

8.

22 0 −3

2 1 2 1 34

−32

124

The remainder is 3 Choice D

2 marks


• Theconnectionbetweenthecoordinatesof apointPanditspositionvectorpisasfollows:

P and( , , )a b ca

b

c

q =

Themidpointresultforpoints:

P(x1,y1,z1),Q(x2,y2,z2)

Mid point x x y y z z1 2 1 2 1 2

2 2 2

+ + +( ), ,

worksforpositionvectors

HMRN:p44–452 marks

12.

m a b= + =−

+−

1

2

1

2

1

2

0

2

3

1

( )

=−

=

−

1

2

3

5

1

32

52

12

ChoiceB

2 marks

• Inthiscasenoticef ′(x)isaquadratic.So f ′(x) =0isthereforeaquadraticequation.Anysolution(root)of thisequationgivesastationarypoint

• Alternativemethod:finddiscriminantof x2+1 = 0Inthiscasecompareax2 + bx + c = 0givinga = 1,b= 0,c = 1Discriminant= 02 - 4× 1 × 1 = -4Anegativediscriminant⇒noRealroots⇒nostat.pts

HMRN:p20,p27

13.Forstationarypointssetf ′(x) = 0⇒ x2 + 1 = 0 ⇒ x2 = -1andsincex2 ≥0thisequationhasnoRealsolutions.f has no stationarypoints ChoiceA

• Theresultusedis:

xθ° gradient=tanθ

• Alternativelyfrom

2

3a°

thegivendiagram

yougettana°= 32

• Alwaysusetan-1tofindtheangle.

HMRN:p42 marks

14.

m a

a

OA so= ° =

⇒ ° =

−

3

2

3

23

21

tan

tan

Choice D

• The‘lawsof logs’usedare:

mlogb a = logbma

andlogbm -logbn =logb mn

• Infractionswithalgebraicexpressionsyoushouldalwaysfactorise‘top&bottom’expressiontoseeif cancellationcantakeplace.

In this case x - 2canbecancelled.

HMRN:p502 marks

15.log4(x2-4) - 2 log4(x-2)= log4(x2-4) - log4(x-2)2

= log4 xx

2 −−

42 2( )

=log4( )( )( )( )x xx x− +− −

2 22 2

=log4 xx+−

22

Choice C


• Theintegrationresultusedhereiscalleda‘specialintegral’.Theformulausedinthiscaseis:

( ) ( )( )

ax b dx cn ax ba n

n+ = +∫ +

+

+

1

1

wherea = -3,b = 2andn = 13 .Notice

thedivisionby‘a’thecoefficientof x,- 3inthiscase.

• Note: 13

13

33

43

1+ = + =

and 43

4 33

3 41

1

× − = − = −×( )

• Youcould,of course,differentiateeachchoiceusingthe‘chainrule’todeterminethecorrectchoice.The‘inverse’of integrationisdifferentiation.

HMRN:p49

2 marks

16.( )2 3 1

3−∫ x dx

= ( )( )

( )2 33

2 34

43

43

43

−× −

+ = −−

+xc

xc

= − − +14

2 3 43( )x c

ChoiceA

• Rewritinga‘journey’likeAB→interms

of thepositionvectorsaandbof thepointsAandBallowsyoutouseyouralgebraskillstosolveaquestionlikethis.Thebasicresultis:

AB→

= b - a

So,inthiscase,QR→

= - 2PQ→

is translatedintor - q = -2(q - p).Theaimisthentosolvethisequationforrsincebothpandqareknown: r = - q + 2p

• If A(x1,y1,z1) then a =

x

y

z

1

1

1

HMRN:p44–45

2 marks

17.P(- 1,2,5)andQ(- 3,- 1,4)

Q R PQ→ →

= − ⇒ − = − −2 2r q q p( )⇒ r - q = - 2q + 2p ⇒ r = - q + 2pSo

r = −−−

+

−

=

+ × −3

1

4

2

1

2

5

3 2 1( )

11 2 2

4 2 5

+ ×− + ×

⇒ =

r

1

5

6

So R(1,5,6)

ChoiceA

• Thisquestioninvolvesuseof the‘chainrule’.Inthiscasetheresultusedis:

f(x) = bsin(g(x)) ⇒ f ′(x) = b cos(g(x) xg ′(x)withb = - 2andg ′ (x) =3x so g ′(x) =3.

• Yourformulaesheetgives:

f(x) f ′(x)

sin ax a cos ax

HMRN:p48–49

2 marks

18.f(x) = - 2sin3xf ′(x) = -2cos3x ×3= - 6cos3x Choice C


• Tounderstandtherolekplaysinthisquestionhereisadiagramshowingthegraph y =k(x- 1)(x+ 2)fordifferentvaluesof k:

Allthesecurvespassthrough(0,- 2)and(0,1),Theseinterceptscorrespondingtofactors(x+ 2)and(x- 1).

Onlyonecurvepassesthrough−( )1

2 9, andtherewillbeonevalueof kthatgivesthiscurve.Substituting x = - 1

2 and y = 9intheequation y = k(x-1)(x + 2)willdeterminewhatthatvalueis.

HMRN:p29

2 marks

19.y = k(x - 1)(x + 2)

−( )12 9, liesonthecurve

So y = 9when x = - 1

2

⇒ = − −( ) − +( )9 1 212

12k

⇒ = × −( ) × ( )9 32

32

k

⇒ = − ⇒ = − ⇒ = −9 94

36 9 4k k k

ChoiceB

y

x−2 1

2 marks

• Whenlimitsappearontheintegralsignthe‘constantof integration’c is notrequired

• Youareusing:

f x dx x b aa

b

a

b( ) F( ) F F∫ = = −( ) ( )

whereF(x)istheresultof integratingf(x).

• Youshouldknow: a a12 =

HMRN:p49

20.

1

4 10

2 1

2 4 1 0

2

32

12( ) ( )x x

dx+ +

∫ =

= −× +

− −× +

12 4 2 1

12 4 0 1

= − + = − +12 9

12 1

16

12

= − + = =16

36

26

13 Choice D


Strategy• Usingthe‘DoubleAngleFormula’allowsyoutofactorisetheexpression.Theformulasin2A=2sin acos aisgiventoyouintheexam.

Factorise• Thecommonfactorissin x

y

x0 2ππ

Solve

• Notice2cos x- 3 = 0

⇒ 2cos x = 3 ⇒ 2cos x = 32

5 marks

21.

sin 2x - 3 sin x = 0 3

⇒ 2 sin x cos x - 3 sin x = 0

⇒ sin x(2cos x - 3) = 0 3

⇒ sin x =0orcos x = 32

3

Forsin x = 0x=0,π,2π

Forcos x = 32

. x is in 1stor4thquadrants

1stquadrantangleisπ6 .

So

x =6

or6 6 6 6

π π π π π π2 12 11− = − =

3Solutionsare:

x =6 6

0, , , ,π π π π11 2 .

π–62

1

√3

Angles• Forsin xorcos xequaltovalues- 1,0or1youshouldusethe y =sin xor y =cos xgraphtodeterminetheangles.

• Takecarethattheanglesyougive≤ assolutionsareallowed.Inthiscasetheintervalinwhichtheangleslieis0≤ x ≤2π.

• Forquadrantuse:3 showscosinepositive

HMRN:p15,p37

A

T

S

C


Sketch• 1markherewillbeforshowingthecorrectlyshapedgraphwiththemaximumandminimum.

• 1markisforthe‘annotation’i.e.labellingthepointswith(-2,0)and(4,-108)

• Set x =0in y =x3-3x2- 24x-28fory-intercept

HMRN:p212 marks

22. (b)

3

3

y

x0−28 7

(−2, 0)

(4, −108)

Strategy• Evidencethatyouknewtodifferentiatewillgainyouthisstrategywork.

Differentiate• Theruleusedhereis:

f(x)=axn⇒f′(x)=naxn-1

whereaisaconstant

Strategy

• The2ndstrategymarkisforsetting dydx

equaltozero.Theplacesonacurvewherethegradientiszeroarethestationarypoints

Justify• Evidencehastobegivenaboutthe‘nature’of thestationarypoints.Therearethreetypes:

maximum minimum

Stationarypointof inflection

• The‘naturetable’givesthejustificationandwillgainyouthismark.

x-values• Rememberwhensolvingquadraticequationstocheckforcommonfactors,inthiscase3.Thisreducesthesizeof thecoefficientsandmakesthesubsequentfactorisationeasier.

y-values.• Thereisamarkallocatedforthecorrectcalculationof thetwoy-valuesforthestationarypoints

• Remembertousethe‘original’formula

y =...andnotthe dydx...gradient

formulaforthiscalculation

Communication• Statementsmustbemadestating‘maximum’or‘minimum’.

HMRN:p20–21

7 marks

22. (a)y = x3 - 3x2 - 24x - 28 3

⇒ dydx

= 3x2 - 6x - 24 3

Forstationarypointssetdydx = 0 3

⇒3x2 - 6x -24= 0⇒3(x2 - 2x - 8)= 0⇒3(x +2)(x - 4)= 0⇒ x + 2 = 0or x - 4= 0⇒ x = -2or x = 4. 3

3when x = - 2 y =(-2)3-3× (-2)2 - 24× (-2) -28= - 8- 12 +48- 28= 0so(- 2,0)isamaximumstationarypointwhen x =4y =43 - 3×42 - 24×4- 28

= 64 - 48 - 96 - 28 = - 108 3so(4,-108)isaminimum

stationarypoint 3

x:dydx = 3(x + 2)(x - 4):

Shapeof graph:

nature:

++ −

−2 4

max min


Simplify

• Notice : 1

5

21

5

1

5

1 1

5 5

1

5

= × =

×

×=

HMRN:p15,p35–36

Expansion• Theformula: sin(A±B)= sin acosB± cos asinBisgiventoyouinyourexam.

Simplify• Thevaluescos90°=0andsin90°=1areknownfromthegraphs.

–remembernocalculatorsinPaper1:

sin90° = 1

cos90° = 0

6 marks

23. (b)sin θ° = sin (2y-90)°= sin 2y°cos 90° - cos 2y° sin 90° 3= sin 2y° × 0 - cos 2y° × 1 3= -cos 2y° = - (2 cos 2y° - 1) 3= -2cos 2y° + 1 = 1-2cos 2y°

1

2

y°

√5

now cos y°= 1

5 3

So sin θ°=1 - 2 × 1

5

2

3

= − × = − =1 21

51

2

5

3

5 3

Strategy• Useof the‘Doubleangle’formula

Calculation• Theexactvalueof cos y°isobtainedusingPythagoras’Theoremin∆ APDandthenusingSOHCAHTOA.

Value

• Substitutionof 1

5forcos y°inthe

expression

4 marks

Log Law• The‘law’usedislogbm-logbn = logb

mn24. log log

log

.

2 2

2

2 2

22

22 2

4

2

x

x

x

x

− =

⇒ =

⇒ = ( ) =

⇒ =

3 3

3

3

Exponential form• Rewritelogba = c as a = bc

Start Solution

• 22

2 2 2( ) = × =

Finish Solutions

• x xx

22

22 2 2 4= ⇒ × = × ⇒ =

HMRN:p50

1 mark

Angle• “Findqintermsofy.”meansyouareaimingfor:q=(anexpressioninvolving y only)Soqcannotappearonthe‘right-handside’.

• Youareusing:theanglesuminatriangleis180º

23. (a)From the diagram

q + 180 -2y = 90⇒ q = 90 - 180 + 2y⇒ q = 2y - 90 3

180° − 2y°

B

C

P

θ°2y°


Strategy• Inquestionswhereyouaregivenf ′(x)

(ordydx )andaskedtofindf(x)(ory)

thenyouneedtouseintegration:

f x f x

f x

( ) ( )

( )

differentiation

integrati

→ ′oon← ′f x( )

Preparation• Multiplyingoutbracketsisessentialbeforeyouintegrate

Integrate

• Youareusing x dxxn

cnn

∫ =+

++1

1

Substitution• The‘constantof integration’ciscrucial

inthisquestion.f(2)= 13 means

when x = 2theformulagives13 . c can

thereforebefound.

Calculation• Nocalculator—sopracticeyourfractionwork!

HMRN:p31

5 marks

25.f ′(x) = x2(x - 1) = x3 - x2 3

⇒ = −

= − +

=

⇒ −

∫f x x x dx

x xc

f

( )

( )

3 2

4 3

4 3

4 3

21

32

4

2

3

now

++ =

⇒ − + =

⇒ = + − = − = −

=

c

c

c

so f x x

1

3

48

3

1

31

3

8

34 3 4 1

1

4( ) 44 31

31− −x

3 3

3

3

1


Strategy• Yourfirststepistoexpandksin(x-a)°

• Onyourformulaesheetis:

sin(A± B) = sin acosB±cos asinB

Thisistheexpansionyoushoulduse

• Noticethatthereisanimpliedstepintheworking:

ksin(x- a)°=k(sin x°cos a°- cos x°sin a°)

=ksin x°cos a°-kcos x°sin a°

• If thisexpansiondoesnotappearinyourworkingyouwillnotbeawardedthismark.

Coefficients• Themethodis:

k sin x° cos a°

cos x°1sin x°3

k cos x° sin a°

−

−

sokcos a°= 3andksin a°= 1

Angle• Acommonmistakeistodivideinthewrongorder.Theresultyouareusingis sin

costan

aa

a°

°°=

Inthiscasesinceisonthe‘top’of thefractionbutcomesfromthe‘bottom’equationsothefractionis 1

3 not 3

1.

Amplitude• Youshouldtrytounderstandthemethodforcalculatingk.Thiswillhelpyoudealwithmoreunusualquestionsthatsometimesarise-theserequireunderstanding,nottheblindapplicationof aformula.

HMRN:p53–54

1. (a)3sin x°- cos x°= ksin(x - a)°⇒ 3sin x°- cos x°= k sin x°cos a°-k cos x°sin a° 3nowequatethecoefficientsof sin x°andcos x°

k a

k a

cos

sin

°

°

==

3

1

sincebothsin a° andcos a°arepositive,a° is in the 1stquadrant.

3

Divide:

k a

k aa

sin

costan

°

°°= ⇒ =1

3

1

3

so a° −⋅

°== ⋅tan.1 1

318 4 3

(to1decimalplace)

Square and add:(k cos a°)2 + (k sin a°)2 = 32+12

⇒ k2cos 2a°+ k2sin 2a°=9+ 1⇒ k2(cos2a°+ sin 2a°)= 10⇒ k2 × 1 = 10 ⇒ k2 =10

⇒ k k= >10 0( ) 3

So 3

10 18 4

sin cos

sin( )

x x

x

° °

°

−

= − ⋅

(correctto1decimalplace)

4 marks

worked answers: exam C PaPer 2Practice Paper M Answers Paper 2


1. (b)3sin x° - cos x° = 1⇒ 10 sin(x -18·4...)°= 1 3

⇒sin(x - 18·4...)°= 1

10

⇒ x°- 18·4...°= sin -1 110

=18·4...° 3so x =18·43...+18·43...

=36·86...

⇒ x ..= ⋅36 9 (todecimalplace)

(0≤ x ≤ 90) 3

3 marks

Strategy• Usingyourresultfrompart(a)youcanrewritethegivenequationinaformthatiseasiertosolve

Solve for (x-a)°• Youraimistomovetoanequationof theformsin(angle)=number

• Sincetheintervalof allowablevaluesof x is0≤ x ≤ 90onlythe1stquadrantvalueisconsideredinthisparticularexample

Solve for x• x-18·4= 18·4⇒ x = 18·4+ 18·4

HMRN:p54

Strategy• Theformulagiventoyouintheexamis:“ a b a b⋅ = cosθ , whereθ is the anglebetweenaandb”soyoumustknowtorearrangethisintotheform

usedinthisquestion: cosθ = ⋅a b

a b

Asisthecaseforallformulae:LEARNTHEM!

Dot product• Thereisacheckyoucandowiththe‘dotproduct’

If a·b>0theanisacute(0°<θ°<90°)

If a·b<0thentheangleisobtuse(90°<θ°<180°)

Magnitudes• Youwillbeawarded1markforeachcorrectanswer.

Angle• Manymistakesaremadeinthiscalculation.Use:

INV cos ( ) )1 0 3 8 1 1÷ × asthekeyingsequencefollowedby

EXE (trueformostcalculators)

HMRN:p46

2.

Use cos θ =v w

v w

.

( )( )

when andv w= =

−

−

2

3

5 3

1

1 3

v w. =−

. −

= − × + ×

2

3

5

1

1

3

2 1 3 (( )− + × =1 5 3 10 3

| |v = ( 2)2− + + = + +

=

3 5 4 9 25

38

2 2

3

| |w = + − + = + +

=

1 1 3 1 1 9

11

2 2 2( )

3

so cos =10

= cos 1 10

θ

θ

38 11

38 11⇒

−

3

⇒ θ = 60·71...=.. 60·7 (to1dec.place)

w

v

A

C

B

θ

5 marks


3.x2 - 2x + c2 + 2 = 0 3Discriminant = (-2)2- 4 × 1 × (c2 + 2) 3= 4 - 4(c2 + 2) = 4 - 4c2 - 8= - 4c2 -4 = -4(c2 + 1) 3Since c2 + 1 > 0 then - 4(c2 + 1) < 0So for all values of c the Discriminant < 0 and the equation has no Real roots 3

4 marks

Strategy• Thepositive/zero/negativenatureof thediscriminantof thisquadraticequationhastobedetermined.

Substitution• Comparingax2+bx +c´=0

with1x2-2x+(c2+ 2)=0

givesa= 1,b= -2andc´=c2+2(differentc’s!)

Sob2- 4ac´=(-2)2- 4× 1× (c2+2)

Simplify• Becarefulwithnegativeseg-4(c2+ 2)=-4c2-8

Proof• -4×(c2+1)isnegative×positive=negative c2+ 1isalwayspositivesincec2isalwayspositiveorzero.

HMRN:p27

4. (a)

y x x

dydx

x x

= −

⇒ = −

1

32

4

3 2

2

3

3

Thetangenthasgradient-4

so dydx

= −4 3

⇒ x2 - 4x = -4⇒ x2- 4x +4= 0 3⇒ (x - 2)(x - 2) = 0⇒ x =2whichistherequired

x-coordinate 3

5 marks

Strategy• Knowingtodifferentiatewillearnyouthismark.

Differentiate

• Noticethecoefficients:313

1× = and2×(-2) = -4

Strategy• This2ndstrategymarkisforsettingthegradientequalto-4

Solve• 2marksareavailable:forstartingtheprocessandthencompletingit.Youshouldrecogniseaquadraticequationandwriteitin‘standardform’i.e.x2-4x+4=0.

HMRN:p20


4. (b)When x = 2

y = × − × = − = −1

32 2 2

8

38

16

33 2 3

Apointonthetangentis

216

3,−

andthegradient= -4Sotheequationis:

y x− −

= − −

16

34 2( )

⇒ y +16

3 = -4x +8

⇒ 3y +16= -12x +24⇒ 3y + 12x = 8 3

2 marks

Calculation• Tofindtheequationof thetangentyouwillneedtoknowthecoordinatesof apointonthattangent.Allyouknowsofaristhex-coordinate(x= 2)andsocalculationof they-coordinateisessential.

Equation• Hereyouareusingy - b=m(x- a)wherem= -4andthepoint(a,b)is

2 163

,−( )• Donotusedecimalapproximationsincoordinatework-youwilllosemarksif youdoeg-5.3shouldnotbeusedfor − 16

3.

HMRN:p20

5. (a)C Cot e t= − 4

In this case Ct =3.5whent = 3

3⇒ = ⇒

⇒ = =

−35 35

35 7 409

34

34

34

. .

. .

CC

C

oo

o

ee

e … 3

Theconcentrationjustafteradministrationwas7.4mg/ml 3(to1decimalplace)

3 marks

Strategy• Inaquestionlikethisyoushouldknowthatyouwillhavetopickoutvaluesforthelettersintheformulaanddoasubstitution.Sometimesithelpstolabeltheformula:

Ct = Co e− t

4

concentrationafter t hours

concentrationat the start

the time elapsed(t hours)

Change of Subject.• Ctisthesubjectof thegivenformula.YouareaskedtofindCo-theconcentrationinitially.ThismeansyouaimforCo=(anumber)

Calculation• Onthecalculatoruse.

3 5i × ÷ex ( )3 4 EXE

HMRN:p50–51


5. (b)Required to find t so that

Ct =1

2Co

⇒ =

⇒ =

⇒ = −

⇒ = − =

−

−

1

21

2

4

4

4

4

12

12

C Co ot

t

e

e

t

t

e

e

log

log 22 772. …

3

3

3

Thisis2hoursand0.772...× 60min. Ittakes2hours46minutes 3(tothenearestminute)

4 marks

Interpretation• Youarebeingaskedtocalculatetfor

Ct,theconcentrationafterthours,toequalhalf of theinitialconcentrationi.e. 1

2Co

1st step to solving• Bothsidesof theequationcanbedividedbyCo.Thiscreatesanexponentialequationwithonlythevariablet.

Calculation

• The ln keycalculates‘loge’

HMRN:p50–51

Log Statement• Youusethisconversion: c=ba↔logbc=a

Inthiscase a b e ct= − = =4

12

, and

6. (a)x2 + y2 - 4x - 6y + 8 = 0Centre: C1(2, 3) 3For C1(2,3)and A(1,5) 3

m

m

CA =−−

=−

= −

⇒ =

5 3

1 2

2

12

1

21

3

PointonthetangentisA(1,5)

andthegradient = 1

2sotheequationis:

y x− = −51

21( )

⇒2y - 10 = x - 1⇒ 2y - x =9 3

4 marks

Centre• Thisusestheresult:

circle:x2+y2+2gx+2fy+c=0

Centre:(-g ,-f )

Theprocessinvolveshalvingandchangingsignsof thecoefficientsof xandytogetthecoordinatesof thecentreof thecircle.

Gradient of radius• Thegradientformulais:

P(x1,y1),Q(x2,y2)⇒ =−−

my y

x xPQ2 1

2 1

Equation• Usey- b=m(x- a).Inthiscaseyouusem = 1

2with(a,b)beingthepoint

A(1,5)

HMRN:p3–4,p39

Strategy• Thetangentisperpendiculartotheradiustothepointof contacti.e.C1P

• Perpendiculargradientsareobtained

byusingm mab

ba

= ⇒ = −1 . In this

case -2isthoughtof as − 21

… …


Rearrangment• Change2y-x=9tox=2y-9readyforsubstitution

6. (b)For intersection of line and circle solve:

2 9

2 2 18 02 2

y x

x y x y

− =+ + + − =

3

Substitutex = 2y -9incircleequation: 3(2y - 9)2 + y2 +2(2y - 9)+ 2y - 18= 0⇒ 4y2 - 36y+81+ y2 +4y - 18 + 2y -18= 0⇒ 5y2 - 30y +45= 0 3⇒ 5(y2 - 6y +9)= 0⇒ 5(y -3)(y - 3)= 0⇒ y = 3 3andsincethereisonlyone solutionthelineisatangent tothecircle 3

5 marks

Strategy• Thestrategymarkhereisawardedforsubstitutionof thelineequationintothecircleequation.

‘Standard form’• Youshouldrecogniseaquadraticequationandthereforewriteitinthestandardorder,namely:

5y2-30y+45=0

Solve• Itisalwayseasiertofactorisequadraticexpressionsif anycommonfactorisremovedfirst.Inthiscase5isthecommonfactor

Justify• Howdoyouprovealineistangenttoacircle?Youfindthepointsof intersection.If thelineisatangenttherewillbeonly1point.Youwriteaclearstatementtothiseffecttogainthis‘communication’work.

HMRN:p40

6. (c)when y = 3 x = 2 × 3 - 9 = -3 3so the point of contact is B(-3,3)For A(1,5) and B(-3,3)

AB = − − + −

= + = =

( ( )) ( )1 3 5 3

4 2 20 2 5

2 2

2 2 3

2 marks

Other coordinate• Havingdeterminedthevalueof the

y-coordinateinpart(b)younowhavetosubstitutethisback(usethelineequation)tofindthex-coordinate

Distance• Youusethedistanceformula:P(x1,y1),Q(x2,y2)

PQ = − + −( ) ( )x x y y1 22 2

1 2

HMRN:p7


7. 3

3

3

yy = f ′(x)

x0 8–3

−2

3 marks

Stationary Points• Anystationarypointonthegraphy=

f(x)hasazerovalueforthegradient.Soony= f′(x),thegraphthatshowsthegradientvalues,anx-axisintercept(zerovalue)indicatesastationarypointontheoriginalgraph.Inthiscaseyoursketchshouldcrossthex-axisat(-2,0)and( 8

3,0)

Correct relationship• Between-2and 8

3theoriginalgraph

goesdownhill

⇒gradientisnegative⇒ y = f ′(x) graphisbelowthex-axis

• Lessthan-2andgreaterthan 83

originalgraphgoesuphill⇒gradientpositive⇒ y = f ′(x)graphisabovethex-axis

Shape• Differentiatingacubicproducesaquadratic⇒yourgraphshouldshowaparabola.

HMRN:p19

8. (a)Volume = x × x × h = x2h 3

so x h hx x

2 12

12

2 262

62 125

2= ⇒ = =

Basearea= x2

Sidearea= xh

= × =xx x

125

2

125

22 3

Totalarea=4×sidearea+Basearea

⇒ = × + = +A( )x x x x x41252

2502 2

3

3 marks

Strategy• Volume=length×breadth×height

Strategy

• Findinghintermsof x hx

=( )1252 2

andthensubstitutingintheareaexpressionsisthekeytosolvingthisquestion.

Proof• Makesureallstepsareshown.Theanswerisgivensotheexaminerneedstoseeeachstepof yourreasoning.


8. (b)A(x) = 250x-1 + x2 3

⇒ ′ = − +

= − +

−A ( )x x x

xx

250 2

2502

2

2 3

ForstationaryvaluessetA'(x) = 0 3

⇒−

+ = ×250

2 02

2

xx x( )

⇒ -250+ 2x3 = 0 ⇒ 2x3=250⇒ x3 = 125⇒ x =5 3 x:

+−

5

A'( )xx

x= +250

22 :

3So x = 5givesaminimumvalueforthearea.

5 marks

Preparation

• Youshouldknowthattheterm 250x

cannotbedirectlydifferentiated.Changingtheexpressionto250x -1fitsittothefollowingrule:

f(x) = axn ⇒ f ′(x) = naxn-1

in this case a =250andn = -1.

Differentiation• Itisusefulforsubsequentworktochangenegativeindicestopositive.Inthiscase − = −−250 2 250

2x

x

Strategy• ‘A′(x)=0’hastobestatedtogainthismark.

Solve• Youshouldmultiplybothsidesby

x2.Theaimistoridtheequationof fractions

• Cubingapositivenumbergivesapositive,butcubinganegativenumbergivesanegative,sox3=125hasonlyonepositivesolution.Thisisnotlikesquaring:x2=25givesx=5orx=-5.

Justify• 1markisallocatedforthe‘naturetable’.

HMRN:p20–22

Shape of graph: nature: min


Strategy• Thestrategyistofindtheintersectionpointsof liney= 5andparabolay=x2-6x+10asthex-valuesof thesepointswillsubsequentlybecrucialinsettingupintegralstofindareas.

9.First find the points of intersection of y = 5 with the curve:Solve: y

y x x

== − +

5

6 102

x x

x x

x x

x x

2

2

6 10 5

6 5 0

1 5 0

1 5

− + =⇒ − + =⇒ − − =⇒ = =

( )( )

or

3

3

Diagram: y y = 5

x0 1 5

3

shadedareainthisdiagramis

givenby: 5 6 10

5 6 10

6 5

21

5

21

5

2

− − +

= − + −

= − + −

∫∫

( )x x dx

x x dx

x x dxx

x x x

1

5

32

1

5

33 5

∫= − + −

3 3

3

= − + × − ×

− − + × − ×

= −

53

3 5 5 5 13

3 1 5 1

1

32

32

2253

75 2513

3 5

521243

1563

1243

323

2

+ − + − +

= − = − = cm 3

SoAreaof plate = × −5 6323

( tan )area of rec gle

= − = − =

=

30323

903

323

583

1913

2cm 3

8 marks

Solve• Alwaysrearrangeaquadraticequationinto‘standardform’,inthiscasex2-6x+5=0sothatthefactorisationallowsthisprocess:(factor1)(factor2)=0⇒factor1= 0orfactor2= 0

Strategy• Didyouknowtointegratetofindthisarea?

Limits• Lefttorightonthediagram(1and5)

Bottomtotopontheintegral(1and5)

Integration• Youshouldalways‘simplify’firstbeforeyouintegrate.Inthiscasedon’tintegrate5-(x2-6x+10),simplifyfirstto-x2+6x-5andthenintegrate.

Evaluate• Evenwithacalculatoravailablethesecalculationsareverydifficulttogetcorrect.Youshouldtakegreatcareanddouble-checkyourworkingalways.

Area• Finallymakesureyouanswerthequestion. 32

32cm isnottherequired

answer!Yourstrategywastosubtracttheparabolicareafromtherectangle.

HMRN:p32–33

Strategy• Thisstrategymarkisforhowyouwillsplitareasup.Inthesolutiongiventheapproachis:

(RectangleArea)-(ParabolicArea).

Analternativeapproachwouldhavebeentoaddtogether3areastocalculatetheplateareaasshowninthisdiagram. areaA+areaB+areaC

areaA

areaC

area B

soutions to practice paper : igher athematics - spta maths · · 2016-05-17soutions to practice...

Documents