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SPECIALIST MATHEMATICSWritten examination 1
Friday 6 November 2015 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
9 9 40
• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpenersandrulers.
• Studentsarenotpermittedtobringintotheexaminationroom:notesofanykind,acalculatorofanytype,blanksheetsofpaperand/orcorrectionfluid/tape.
Materials supplied• Questionandanswerbookof10pageswithadetachablesheetofmiscellaneousformulasinthe
centrefold.• Workingspaceisprovidedthroughoutthebook.
Instructions• Detachtheformulasheetfromthecentreofthisbookduringreadingtime.• Writeyourstudent numberinthespaceprovidedaboveonthispage.
• AllwrittenresponsesmustbeinEnglish.
Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.
©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2015
SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2015
STUDENT NUMBER
Letter
3 2015SPECMATHEXAM1
TURN OVER
InstructionsAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.
Question 1 (3marks)
ConsidertherhombusOABCshownbelow,whereOA a→=�i andOC
→= + +� � �i j k ,andaisapositivereal
constant.
C
O
B
A
a. Finda. 1mark
b. ShowthatthediagonalsoftherhombusOABCareperpendicular. 2marks
2015SPECMATHEXAM1 4
Question 2 (4marks)A20kgparcelsitsonthefloorofalift.
a. Theliftisacceleratingupwardsat1.2ms–2.
Findthereactionforceoftheliftfloorontheparcelinnewtons. 2marks
b. Findtheaccelerationoftheliftdownwardsinms–2sothatthereactionoftheliftfloorontheparcelis166N. 2marks
Question 3 (4marks)Thevelocityofaparticleattimetsecondsisgivenbyr i j k( ) ( )t t t= − + −4 3 2 5 ,wherecomponentsaremeasuredinmetrespersecond.
Findthedistanceoftheparticlefromtheorigininmetreswhent=2,giventhat� � �r i k( )0 2= − .
5 2015SPECMATHEXAM1
TURN OVER
Question 4 (4marks)
a. Findallsolutionsof z i z C3 8= ∈, incartesianform. 3marks
b. Findallsolutionsof( ) ,z i i z C− = ∈2 83 incartesianform. 1mark
Question 5 (3marks)Findthevolumegeneratedwhentheregionboundedbythegraphofy=2x2–3,theliney=5andthey-axisisrotatedaboutthey-axis.
2015SPECMATHEXAM1 6
Question 6 (4marks)Theaccelerationams–2ofabodymovinginastraightlineintermsofthevelocityvms–1isgivenby a = 4v2.
Giventhatv = ewhenx=1,where xisthedisplacementofthebodyinmetres,findthevelocityofthebodywhenx=2.
7 2015SPECMATHEXAM1
TURN OVER
Question 7 (5marks)a. Solvesin( ) sin( ), [ , ]2 0 2x x x= ∈ π . 3marks
b. Find x x x x: ) ), , ,cosec( cosec(2 02 2
< ∈
∪
π ππ . 2marks
2015SPECMATHEXAM1 8
Question 8–continued
Question 8 (7marks)
a. Showthat tan( ) log sec2 12
2x dx x ce∫ = ( ) + . 2marks
Thegraphof f x x( ) ( )=12arctan isshownbelow.
O
y
x
b. i. Writedowntheequationsoftheasymptotes. 1mark
ii. Ontheaxesabove,sketchthegraphof f –1,labellinganyasymptoteswiththeirequations. 1mark
9 2015SPECMATHEXAM1
TURN OVER
c. Find f 3( ). 1mark
d. Findtheareaenclosedbythegraphof f,thex-axisandtheline x = 3 . 2marks
2015SPECMATHEXAM1 10
END OF QUESTION AND ANSWER BOOK
Question 9 (6marks)
Considerthecurverepresentedby x xy y2 232
9− + = .
a. Findthegradientofthecurveatanypoint(x,y). 2marks
b. Findtheequationofthetangenttothecurveatthepoint(3,0)andfindtheequation ofthetangenttothecurveatthepoint 0 6,( ) .
Writeeachequationintheformy = ax + b. 2marks
c. Findtheacuteanglebetweenthetangenttothecurveatthepoint(3,0)andthetangenttothecurveatthepoint 0 6,( ) .
Giveyouranswerintheformkπ,wherekisarealconstant. 2marks
SPECIALIST MATHEMATICS
Written examinations 1 and 2
FORMULA SHEET
Instructions
Detach this formula sheet during reading time.
This formula sheet is provided for your reference.
© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2015
SPECMATH 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: 43 π r
3
area of a triangle: 12 bc Asin
sine rule: aA
bB
cCsin sin sin
= =
cosine rule: c2 = a2 + b2 – 2ab cos C
Coordinate geometry
ellipse: x ha
y kb
−( )+
−( )=
2
2
2
2 1 hyperbola: x ha
y kb
−( )−
−( )=
2
2
2
2 1
Circular (trigonometric) functionscos2(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
+ =+
−1 tan( ) tan( ) tan( )tan( ) tan( )
x y x yx 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 2x x
x=
−
function sin–1 cos–1 tan–1
domain [–1, 1] [–1, 1] R
range −
π π2 2, [0, �] −
π π2 2,
3 SPECMATH
Algebra (complex numbers)z = x + yi = r(cos θ + i sin θ) = r cis θ
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)
Calculusddx
x nxn n( ) = −1
x dx
nx c nn n=
++ ≠ −+∫ 1
111 ,
ddxe aeax ax( ) =
e dx
ae cax ax= +∫ 1
ddx
xxelog ( )( ) = 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 dx
aax c= +∫
ddx
xx
sin−( ) =−
12
1
1( )
1 02 2
1
a xdx x
a c a−
=
+ >−∫ sin ,
ddx
xx
cos−( ) = −
−
12
1
1( )
−
−=
+ >−∫ 1 0
2 21
a xdx x
a c acos ,
ddx
xx
tan−( ) =+
12
11
( )
aa x
dx xa c2 2
1
+=
+
−∫ tan
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
constant (uniform) acceleration: v = u + at s = ut +12
at2 v2 = u2 + 2as s = 12
(u + v)t
TURN OVER