Transcript
  • SPECIALIST MATHEMATICSWritten examination 1

    Friday 10 November 2017 Reading time: 9.00 am to 9.15 am (15 minutes) Writing time: 9.15 am to 10.15 am (1 hour)

    QUESTION AND ANSWER BOOK

    Structure of bookNumber of questions

    Number of questions to be answered

    Number of marks

    10 10 40

    • Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpenersandrulers.

    • StudentsareNOTpermittedtobringintotheexaminationroom:anytechnology(calculatorsorsoftware),notesofanykind,blanksheetsofpaperand/orcorrectionfluid/tape.

    Materials supplied• Questionandanswerbookof11pages• Formulasheet• Workingspaceisprovidedthroughoutthebook.

    Instructions• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.• AllwrittenresponsesmustbeinEnglish.

    At the end of the examination• Youmaykeeptheformulasheet.

    Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

    ©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2017

    SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2017

    STUDENT NUMBER

    Letter

  • 2017SPECMATHEXAM1 2

    THIS PAGE IS BLANK

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    InstructionsAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegms–2,whereg=9.8

    Question 1 (3marks)Findtheequationofthetangenttothecurvegivenby3xy2+2y = xatthepoint(1,–1).

    Question 2 (4marks)

    Find 11 21

    3

    x xdx

    +( )∫ , expressingyouranswerintheformlog ,eab

    whereaandbarepositiveintegers.

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    Question 3 (3marks)Letz3 + az2 + 6z + a=0,z ∈ C,whereaisarealconstant.

    Giventhatz = 1 – iisasolutiontotheequation,findallothersolutions.

    Question 4 (3marks)Thevolumeofsoftdrinkdispensedbyamachineintobottlesvariesnormallywithameanof298mLand astandarddeviationof3mL.Thesoftdrinkissoldinpacksoffourbottles.

    Findtheapproximateprobabilitythatthemeanvolumeofsoftdrinkperbottleinarandomlyselected four-bottlepackislessthan295mL.Giveyouranswercorrecttothreedecimalplaces.

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    Question 5 (4marks)Relativetoafixedorigin,thepointsB,CandDaredefinedrespectivelybythepositionvectors

    b i j k c i j k and d i j,= − + = − + = −2 2 2, a whereaisarealconstant.

    GiventhatthemagnitudeofangleBCDisπ3

    ,finda.

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    Question 6 (3marks)

    Let fx

    x( ) .( )

    =1

    arcsin

    Find f ′(x)andstatethelargestsetofvaluesofxforwhich f ′(x)isdefined.

    Question 7 (4marks)

    Thepositionvectorofaparticlemovingalongacurveattimetisgivenby

    r i j( ) cos ( ) sin ( ) , .t t t t= + ≤ ≤3 3 04π

    Findthelengthofthepaththattheparticletravelsalongthecurvefromt=0tot = π4

    .

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    CONTINUES OVER PAGE

  • 2017SPECMATHEXAM1 8

    Question 8 –continued

    Question 8 (4marks)

    Aslopefieldrepresentingthedifferentialequationdydx

    xy

    =−+1 2

    isshownbelow.

    y

    x1O 2–2 –1

    2

    –1

    –2

    1

    a. Sketchthesolutioncurveofthedifferentialequationcorrespondingtothecondition y(–1)=1ontheslopefieldaboveand,hence,estimatethepositivevalueofxwheny=0. Giveyouranswercorrecttoonedecimalplace. 2marks

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    b. Solvethedifferentialequation dydx

    xy

    =−+1 2

    withtheconditiony(–1)=1.Expressyouranswer

    intheformay3 + by + cx2 + d =0,wherea,b,canddareintegers. 2marks

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    Question 9 (5marks)Aparticleofmass2kgwithinitialvelocity3 2

    i j+ ms–1experiencesaconstantforcefor 10seconds.Theparticle’svelocityattheendofthe10-secondperiodis 43 18

    i j− ms–1.

    a. Findthemagnitudeoftheconstantforceinnewtons. 2marks

    b. Findthedisplacementoftheparticlefromitsinitialpositionafter10seconds. 3marks

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    END OF QUESTION AND ANSWER BOOK

    Question 10 (7marks)

    a. Showthat ddx

    x xa

    xa

    x

    a xarccos arccos ,

    =

    −2 2wherea>0. 1mark

    b. Statethemaximaldomainandtherangeof f x x( ) arccos .=

    2

    2marks

    c. Findthevolumeofthesolidofrevolutiongeneratedwhentheregionboundedbythegraphofy = f(x),andthelinesx=–2andy=0,isrotatedaboutthex-axis. 4marks

  • SPECIALIST MATHEMATICS

    Written examination 1

    FORMULA SHEET

    Instructions

    This formula sheet is provided for your reference.A question and answer book is provided with this formula sheet.

    Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

    Victorian Certificate of Education 2017

    © VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2017

  • SPECMATH EXAM 2

    Specialist Mathematics formulas

    Mensuration

    area of a trapezium 12 a b h+( )

    curved surface area of a cylinder 2π rh

    volume of a cylinder π r2h

    volume of a cone 13π r2h

    volume of a pyramid 13 Ah

    volume of a sphere 43π r3

    area of a triangle 12 bc Asin ( )

    sine ruleaA

    bB

    cCsin ( ) sin ( ) sin ( )

    = =

    cosine rule c2 = a2 + b2 – 2ab cos (C )

    Circular functions

    cos2 (x) + sin2 (x) = 1

    1 + tan2 (x) = sec2 (x) cot2 (x) + 1 = cosec2 (x)

    sin (x + y) = sin (x) cos (y) + cos (x) sin (y) sin (x – y) = sin (x) cos (y) – cos (x) sin (y)

    cos (x + y) = cos (x) cos (y) – sin (x) sin (y) cos (x – y) = cos (x) cos (y) + sin (x) sin (y)

    tan ( ) tan ( ) tan ( )tan ( ) tan ( )

    x y x yx y

    + =+

    −1tan ( ) tan ( ) tan ( )

    tan ( ) tan ( )x y x y

    x y− =

    −+1

    cos (2x) = cos2 (x) – sin2 (x) = 2 cos2 (x) – 1 = 1 – 2 sin2 (x)

    sin (2x) = 2 sin (x) cos (x) tan ( )tan ( )tan ( )

    2 21 2

    x xx

    =−

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    Circular functions – continued

    Function sin–1 or arcsin cos–1 or arccos tan–1 or arctan

    Domain [–1, 1] [–1, 1] R

    Range −

    π π2 2

    , [0, �] −

    π π2 2

    ,

    Algebra (complex numbers)

    z x iy r i r= + = +( ) =cos( ) sin ( ) ( )θ θ θcis

    z x y r= + =2 2 –π < Arg(z) ≤ π

    z1z2 = r1r2 cis (θ1 + θ2)zz

    rr

    1

    2

    1

    21 2= −( )cis θ θ

    zn = rn cis (nθ) (de Moivre’s theorem)

    Probability and statistics

    for random variables X and YE(aX + b) = aE(X) + bE(aX + bY ) = aE(X ) + bE(Y )var(aX + b) = a2var(X )

    for independent random variables X and Y var(aX + bY ) = a2var(X ) + b2var(Y )

    approximate confidence interval for μ x zsnx z s

    n− +

    ,

    distribution of sample mean Xmean E X( ) = µvariance var X

    n( ) =σ 2

  • SPECMATH EXAM 4

    END OF FORMULA SHEET

    Calculus

    ddx

    x nxn n( ) = −1 x dx n x c nn n=

    ++ ≠ −+∫ 1 1 1

    1 ,

    ddxe aeax ax( ) = e dx a e c

    ax ax= +∫ 1

    ddx

    xxe

    log ( )( ) = 1 1xdx x ce= +∫ log

    ddx

    ax a axsin ( ) cos( )( ) = sin ( ) cos( )ax dxa

    ax c= − +∫ 1

    ddx

    ax a axcos( ) sin ( )( ) = − cos( ) sin ( )ax dxa

    ax c= +∫ 1

    ddx

    ax a axtan ( ) sec ( )( ) = 2 sec ( ) tan ( )2 1ax dxa

    ax c= +∫ddx

    xx

    sin−( ) =−

    12

    1

    1( ) 1 0

    2 21

    a xdx xa c a−=

    + >

    −∫ sin ,ddx

    xx

    cos−( ) = −−

    12

    1

    1( ) −

    −=

    + >

    −∫ 1 02 2 1a x dxxa c acos ,

    ddx

    xx

    tan−( ) =+

    12

    11

    ( ) aa x

    dx xa c2 21

    +=

    +

    −∫ tan( )

    ( )( ) ,ax b dx

    a nax b c nn n+ =

    ++ + ≠ −+∫ 1 1 1

    1

    ( ) logax b dxa

    ax b ce+ = + +−∫ 1 1

    product rule ddxuv u dv

    dxv dudx

    ( ) = +

    quotient rule ddx

    uv

    v dudx

    u dvdx

    v

    =

    2

    chain rule dydx

    dydududx

    =

    Euler’s method If dydx

    f x= ( ), x0 = a and y0 = b, then xn + 1 = xn + h and yn + 1 = yn + h f (xn)

    acceleration a d xdt

    dvdt

    v dvdx

    ddx

    v= = = =

    2

    221

    2

    arc length 12 2 2

    1

    2

    1

    2

    + ′( ) ′( ) + ′( )∫ ∫f x dx x t y t dtxx

    t

    t( ) ( ) ( )or

    Vectors in two and three dimensions

    r = i + j + kx y z

    r = + + =x y z r2 2 2

    � � � � �ir r i j k= = + +ddt

    dxdt

    dydt

    dzdt

    r r1 2. cos( )= = + +r r x x y y z z1 2 1 2 1 2 1 2θ

    Mechanics

    momentum

    p v= m

    equation of motion

    R a= m

    2017 Specialist Mathematics 1InstructionsFormula sheet


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