Maths Lit Practice Questions

Senior

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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.

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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°

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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°

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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

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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

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∴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

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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

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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

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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

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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)

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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