Soutions to Practice Paper : igher athematics - SPTA Maths · · 2016-05-17Soutions to Practice...
Transcript of Soutions to Practice Paper : igher athematics - SPTA Maths · · 2016-05-17Soutions to Practice...
Page 101
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
Page 102
Solutions to Practice Paper C: Higher Mathematics
• 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
Page 103
Solutions to Practice Paper C: Higher Mathematics
• 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
Page 104
Solutions to Practice Paper C: Higher Mathematics
• 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
Page 105
Solutions to Practice Paper C: Higher Mathematics
• 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
Page 106
Solutions to Practice Paper C: Higher Mathematics
• 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
Page 107
Solutions to Practice Paper C: Higher Mathematics
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
Page 108
Solutions to Practice Paper C: Higher Mathematics
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
Page 109
Solutions to Practice Paper C: Higher Mathematics
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°
Page 110
Solutions to Practice Paper C: Higher Mathematics
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
Page 111
Solutions to Practice Paper C: Higher Mathematics
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
Page 112
Solutions to Practice Paper C: Higher Mathematics
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
Page 113
Solutions to Practice Paper C: Higher Mathematics
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
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Solutions to Practice Paper C: Higher Mathematics
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
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Solutions to Practice Paper C: Higher Mathematics
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
… …
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Solutions to Practice Paper C: Higher Mathematics
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
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Solutions to Practice Paper C: Higher Mathematics
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.
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Solutions to Practice Paper C: Higher Mathematics
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
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Solutions to Practice Paper C: Higher Mathematics
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