2
Instructions
Individual, exam-style questions The questions contained in this booklet match the style of questions that are typically asked
in exams. This booklet is not however, a practice exam. Elevate’s research with top students
identified that top students do more practice questions than anyone else. They begin the
process of testing their knowledge early in the year.
Therefore, we have provided exam-format questions that are sorted by topic so that you can
answer them as you learn the information, rather than waiting until the very end of the year to
complete exams.
Comments, questions? Let us know if you need any further advice by visiting www.elevateeducation.com. You can
comment on any of our material, or head to the FAQ section and ask us a question. Also, you
can find us on social media so you can stay up to date on any brand new tips we release
throughout the year.
Other information Every effort has been made to ensure the accuracy of the information expressed in this
booklet, but no warranty or fitness is implied. If you’d like to provide any feedback on this
booklet, let us know at [email protected]. Finally, no part of this publication may
be reproduced, stored in a retrieval system, or transmitted by any means without prior written
consent of the publisher.
3
Trigonometry
a) Simplify
sin 180° − 𝑎 𝑐𝑜𝑠68°sin 22°. 𝑐𝑜𝑠(360° − 𝑎)
b) Evaluatewithoutacalculator
tan 300°. sin 120°. cos 90° − 𝑎sin(180° − 𝑎)
c) Sin310°
d) Cos140°
e) Sin200°
f) Sin251°
g) : ;<= >>°
?@; A:
h) ;<= CD°EF .?@;GHC
?@; G>D°EF ;<=:C°
i) ForwhichvaluesofBisthefollowingequationundefined?j) SolvefroA:2cos(3A+10°) = −1𝑓𝑜𝑟𝐴∈[-360°; 360]
A=36,7° + 𝑘120°𝑜𝑟𝐴 = 76,7° + 𝑘120°A=36.7°, 76.7°, 196.7°, −43.3, −83.3°, 276,7°, 316,7°, −163,3°, −203,3°, −323,3°, −283,3°
4
Trigonometry Solutions
(a) 𝑠𝑖𝑛𝑎𝑐𝑜𝑠𝑎𝑡𝑎𝑛𝑎
(b) tan 360° − 60° sin 180° − 𝟔𝟎 𝑠𝑖𝑛𝑎
𝑠𝑖𝑛𝑎 (−tan60°)(𝑠𝑖𝑛60°) (-√3)(√G
A)
−32
(c) Sin(360°-310°)-sin50°
(d) Son(180°-140°)-cos40°
(e) j
Sin(180°-200°)-sin20° (f)Sin(180°-251°)-sin71°(g):(;<=>>°)
A:°
:?@;A:°?@;A:°
4
5
(h)𝑠𝑖𝑛𝑥. 𝑐𝑜𝑠319°𝑐𝑜𝑠𝑥. 𝑠𝑖𝑛49° 𝑠𝑖𝑛𝑥. 𝑐𝑜𝑠319°𝑐𝑜𝑠𝑥. 𝑐𝑜𝑠319°𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥tanx
(i)
I;<=F?@;F
=;<=F?@;F
Solution1+sinB=0sinB=-1B=-90°+k360°(B=270° + 𝑘360°ORcosB=0B=90° + 𝑘360°OrB=270° + 𝑘360°(j)Solution3A+10° = 120° + 𝑘360°𝑜𝑟3𝐴 + 10° = 240° + 𝑘360°3A+10° =±120°, ±240°A=36,7°, 76,7°, −43,3°, −83,3°
6
Circles Determinetheequationofthecirclecentretheoriginand
a) Radius3√2unitsb) PassingthroughthepointQ(-4;-2).
Solutiona) Substituter=3 2into
X2+y2=r2∴X2+y2=(3 2)2
∴X2+y2=18b) SubstituteQ(-4;-2)
X2+y2=r2(-4)2+(-2)2=r2
r2=20∴X2+y2=20
Twopoints(a;2)and(b;2)(a<b)lieonthecircleX2+y2=16.Determinethedistancebetweenthesetwopoints.
SolutionDrawaroughsketch.
LetP(x;y)=P(a;2)andQ(x;y)=Q(b;2)TheequatonofthelinethroughPandQisy=2.
Substitutey=2intoX2+y2=16X2+22=16X2=16-4X2=12
X=±√12X=±2√3
FromthesketchP(x;y)=P(-2√3;2)andQ(x;y)=Q(2√3;2).
ThepointsareP(-2√3;2)andQ(2√3;2)
TherequireddistancePQ=2(2√3)=4√3(orusethedistanceformula
7
Determinetheequationofthecirclewithcentreattheoriginand:
a) Radius2√7
X2+y2=r2
∴X2+y2=(2√7)2
∴X2+y2=28
b) Passingthroughthepoint(2;-1)
X2+y2=r2
∴22+(-1)2=r2
∴4+1=r2
Determinetheradiusofeachofthefollowingcircle:
a) ax2+ay2=b(a>0;b>0)
a(x2+y2)=b
X2+y2=^_
r2=^_
∴r√^_
Determinetheradiusofthecircle,withcentreattheorigin,whichpassesthroughthegivenpoint.Answersinsimplestsurdformwhereapplicable.
a) (2;1)
X2+y2=r2
∴22+(-1)2=r2
∴4+1=r2
∴ 5 = 𝑟-2
Whichoftheequationsdonotrepresentcircles?
a) X2+y2=9
X2+y2=r2
∴X2+y2=(9)2
8
∴X2+y2=81(validcircle)
b) X2+y2+9=0
X2+y2=r2
∴X2+y2=(-9)2
∴X2+y2=81(validcircle)
c) X2+y2
InvalidcircleWorkoutthetotalarea
SolutionA1=½absinc =½(4)(4)sin46
=57A2=½absinc
=½(9)(9)sin6636.9
Totalarea=42.6
4cm
9cm
46’
66’
A1A2
9
Quadratic Formula
𝑥 =−𝑏 ± 𝑏A − 4𝑎𝑐
2𝑎 Solve5x2+6x+1=0
SolutionA=5,b=6,c=1
𝑥 =−6 ± 36A − 4(5)(1)
2(5)
𝑥 =−6 ± 36 − 20
10
𝑥 =−6 ± 16
10
𝑥 =−6 ± 410
X=-0.2or-16x2+5x-6
Solution:Acis6x(-6)=-36,andbis5
Factorsofac=-36:1,2,3,4,6,9,12,18,36-4x9=-36and-4+9=5
6x2–4x+9x-62x(3x-2)+3(3x-2)
(2x+3)(3x-2)Check:(2x+3)(3x-2)=6x2-4x+9x-6=6x2+5x-6(yes)
Whataretherootsof:6x2+5x-6
SolutionSubstitutea=6,b=5andc=-6intotheformula:
𝑥 =−𝑏 ± 𝑏A − 4𝑎𝑐
2𝑎
𝑥 =−5 ± 5A − 4(6)(−6)
2(6)
=−5 ± 25 + 144
12
=−5 ± 169
12
=−5 ± 1312
𝑥 = (EcdHG)HA
= eHA=A
G
𝑥 = (EcEHG)HA
=HeHA=EG
G
10
Word SumsJaceentersaracewherehehastocycleandrun.Hecyclesadistanceof25km,andthenrunsfor20km.Hisaveragerunningspeedishalfhisaveragecyclingspeed.Joecompletestheraceinlessthan2½hours,whatcanwesayabouthisaveragespeed?
Formulas:Distance/time=speed
• Distance=20km• Averagespeed=skm/h
SoTime=Distance/Averagespeed=20/shours
Jacecompletestheraceinlessthan2½hours
• Thetotaltime<2½• Ac
A;+20/s<2½
Startwith: AcA;+20/s<2½
Multiplyby2s: 25+40<5s
Simplify: 65<5s
Dividebothsidesby5:13<s
Sohisaveragespeedrunningisgreaterthan13km/handhisaveragespeedcyclingisgreaterthan26km/h
11
Functions Determinetheequationofthefunctionintheformofy=ax+p+q.
Solution1=a1+p-34=a1.ap
-1=a0+p–32=a0.apap=2
4=a1.2∴a=2
2p=2∴p=1Y=2x+1–3
given: 𝑥=2x-3Determine
a) (1)
𝑥 =2(1)-31=-1
b) (𝑥 + 2)
(𝑥 + 2)=2(x+2)-3 (𝑥 + 2)=2x+4-3
(𝑥 + 2)=2x+1ParabolaDeterminetheturningpointFormula:y=a(x+p)2+qY=-2(x+1)2+8
Turingpointsare:(-1;8)
Y=-3A(1;1)B(0;-1)
12
Patterns Findthemissingterminthefollowingsequence:8,______,16,______,24,28,32
Solution:
Tofindthepattern,lookcloselyat24,28and32.Eachterminthenumbersequenceisformedbyadding4totheprecedingnumber.So,themissingtermsare8+4=12and16+4=20.Checkthatthepatterniscorrectforthewhole
sequencefrom8to32.Whatisthevalueofninthefollowingnumbersequence?16,21,n,31,36
Solution:Wefindthatthenumberpatternofthesequenceis“add5”tothepreceding
number.So,n=21+5=26