2014 specialist maths exam 2

8/19/2019 2014 Specialist Maths Exam 2

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Student Name:………………………………………

SPECIALIST MATHEMATICS

TRIAL EXAMINATION 2

2014

Reading Time: 15 minutesWriting time: 2 hours

Instructions to students

This exam consists of Section 1 and Section 2.Section 1 consists of 22 multiple-choice uestions and should !e ans"ered on the detacha!leans"er sheet on page 2# of this exam. This section of the paper is "orth 22 mar$s.Section 2 consists of 5 extended-ans"er uestions% all of "hich should !e ans"ered in thespaces pro&ided. Section 2 !egins on page 1' of this exam. This section of the paper is "orth5( mar$s.There is a total of (' mar$s a&aila!le.Where more than one mar$ is allocated to a uestion% appropriate "or$ing must !e sho"n.Where an exact &alue is re uired to a uestion a decimal approximation "ill not !e accepted.)nless other"ise stated% diagrams in this exam are not dra"n to scale.The acceleration due to gra&it* should !e ta$en to ha&e magnitude 2m+s g "here (.#= g Students ma* !ring one !ound reference into the exam.

Students ma* !ring into the exam one appro&ed , S calculator or , S soft"are and% ifdesired% one scientific calculator. ,alculator memor* does not need to !e cleared.ormula sheets can !e found on pages 2/-2( of this exam.

This paper has !een prepared independentl* of the 0ictorian ,urriculum and ssessmentuthorit* to pro&ide additional exam preparation for students. lthough references ha&e !een

reproduced "ith permission of the 0ictorian ,urriculum and ssessment uthorit*% the pu!lication is in no "a* connected "ith or endorsed !* the 0ictorian ,urriculum and

ssessment uthorit*. THE HEFFERNAN GRO P 2014

This Trial xam is licensed on a non transfera!le !asis to the purchasing school. t ma* !e

copied !* the school "hich has purchased it. This license does not permit distri!ution orcop*ing of this Trial xam !* an* other part*.

P ! O ! " o # 1 1 $ 0S u r r e % H i & & s N o r t ' ( I C ) 1 2 *

P ' o n e 0 ) + $ ) , - 0 2 1F . # 0 ) + $ ) , - 0 2 -

i n / o t ' e ' e / / e r n . n r o u ! c o 3 ! . u4 4 4 ! t ' e ' e / / e r n . n r o u ! c o 3 ! . u

T H E

G R O PH E F F E R N A N


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2

SECTION 1

5uestion 1

The follo"ing e uations each represent a h*per!ola.Which h*per!ola has the maximum and minimum points of its t"o !ranches located on the

y-axis3

A!

x 2 − 4 y 2 = 4"!

y 2 − 4( x −1) 2 =1C!

4( x −1) 2 − y 2 =16!

y 2 −( x −1) 2 = 4E!

( y −1) 2 − 4 x 2 = 4

5uestion 2

The graph of y = 1ax 2 +bx +c

has% as its as*mptotes% the x and y axes and the line x = 6 .

The y-coordinate of the maximum turning point on the graph is −1

9.

The &alues of a%b and c are

A!

a = 0, b = 6, c =1

3"!

a = 1, b = − 6, c = 0

C!

a=

0, b=

6, c=

0

6!

a = − 1, b = 6, c =1

3E!

a = 6, b = − 1, c = 3

5uestion )

The graph of y = f ( x ) has a period of π and f is not defined at4

5ator

4

π π == x x .

The rule for f could !e

A!

−=

42cosec67

π x x f

"!

f ( x ) = sec 2 x − π 2

C! ( )π −= x x f 2sec676! ( )π −= x x f cosec67

E!

+=

2cosec67

π x x f

8 T9 9 RN N R;)< 2'14 Specialist Maths Trial Exam 2


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5uestion 4

The implied range of the function "ith rule f ( x ) =2arctan xa

−

b is

A! 6%7 bb −−− π π "! 62%27 bb −−−C! 62%27 baba −−−

6!

−

ba

ba 2

%2

E!

−−− bb

2%

2

π π

5uestion -

The graph of y = ax + π

2 intersects "ith the graph of y =arccos( x ) exactl* three times if

A! 12

−= r r z θ

.The expression

1

( z ) 2 is e ual to

A! 67cis θ −

"! 627cis2 θ −−r

C! 627cis1 θ −−r

6! 627cis2 θ −r

E! 627cis1 θ −r

5uestion *

n the complex plane% a straight line passes through the origin. This line could !e defined !*the set of points C z z ∈% % such that

A!4

67-rg π = z

"!

z z =1C!

z + z =16!

z − 1 + i = z + 1 − i

E! 16m76Re7 =+ z z



4/34

4

5uestion $

The solutions to the e uation z 3 = 3i ha&e principal arguments of

A!

− π 6

,π

6 and

π

2"!

π

6,5 π

6 and

3 π

2

C!

− 7 π 6

,π

2 and

π

6

6!

− π 2

,0 andπ

2

E!

− π 2

,π

6 and

5 π

6

5uestion +

The roots of the e uation z 7 +az 3 −1 =0, where a ∈ R % are displa*ed on an rgand diagram.That diagram could !e


O O

O

O

O

E !

C ! 6 !

A ! " !

6R e 7 z

6R e 7 z

6R e 7 z

6R e 7 z

6R e 7 z

6m 7 z

6m 7 z

6m 7 z

6m 7 z

6m 7 z


5/34


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/

5uestion 10

∫ dx x 67sin 2 is e ual to

A!

1

2( x

−sin(2 x ))

+c

"!

13

sin 3 ( x ) +c

C!

x − 12

sin(2 x ) +c

6!6cos75

67sin 5

x

x

E!

1

2 x − 1

2sin(2 x )

+c

5uestion 11

)sing a suita!le su!stitution% ∫ −−2

'

2617 dx x x is e ui&alent to

A! ∫

−

2

'

25

21

duuu

"! ∫ −

−

1

1

25

21

duuu

C! ∫

−

2

'

21

25

duuu

6! ∫

−

1

'

25

21

22 duuu

E! ∫

−

2

'

25

21

2 duuu



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(

5uestion 14

=et u~ = i~−2 j~ −k ~ and v~ =2 i~−3 j~ + 3 k ~ .

The scalar resolute of u~ in the direction of v~ is

A!

2 − 3"!

2 + 3C!

8 − 34

6! 62741

@@@ k ji −−

E! /241

@@@

−+ k jt i

5uestion 1- ABC is a parallelogram. The position &ectors of A% B%C and are gi&en respecti&el* !*

@@@@@@@@@@@@ and5%4%2 ji!d k ! jick m jbk a +=++=+== .

The &alues of m and ! are

A!

m = 0 and n = 2"!

m = − 1 and n = − 1C!

m = 1 and n = 26!

m = 3 and n = 1E!

m = 4 and n = − 1

5uestion 1,

The diagram !elo" sho"s a particle "hich is in e uili!rium "hilst !eing acted on !* threeforces of magnitude 11% / and " ne"tons.

The magnitude of force "# in ne"tons% is gi&en !* the expression

A! 62 +11 2 −132cos(105 o )

"!

6sin(105 o )

×sin(75 o )

C!

6 2 +11 2sin(105 o )

6! 62 +11 2 −132cos(75 o )

E!

11sin(75 o ) ×sin(105

o

)


" N1 1 N

/ N

°> 5


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#

5uestion 1*

The position &ectors of particles A and B at time t %t ≥ 0 are gi&en respecti&el* !*a~

=( t 2 +1) i~+( t +2) j

~ and b

~=( t − t 2 ) i

~+ t 2 j

~ "herei~

and j~ are unit &ectors in the east

and north directions respecti&el*.


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1'

5uestion 20

crane lifts an enclosed crate containing a load of 24'$g. The acceleration of the crate is5m+s2 &erticall* up"ards.The reaction of the !ase of the crate on the load% expressed in ne"tons% is

A!

−240 g −1200"!

−240 g +1200C! 240 g − 56! 240 g −1200E! 240 g +1200

5uestion 21

particle of mass 5$g is mo&ing in a straight line "ith a &elocit* of /m+s "hen it is acted on !* a constant force of magnitude " ne"tons% acting in the same direction.

fter tra&elling a further 5' metres% the &elocit* of the particle is 12m+s.The &alue of " is

A! 4./"! 5.4C! >.26! #.'E! 2 .2

5uestion 22

particle mo&es in a straight line "ith respect to a fixed point. The &elocit*-time graph of the particle is sho"n !elo".

nitiall* the particle is 2 units from the fixed point.ts possi!le displacement from the fixed point 1' seconds later could !e

A! A .5"! A 1.5C! A 1.'

6! 1.'E! 1.5


1 2 4 5 / > ( # 1 ' 1 1 t

%

O- 1

- 2-

- 4

1

2

4


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11

SECTION 2

5uestion 1 711 mar$s6

,onsider the function f "ith rule xy − x 3 = 2 .

.! Sho" thatdy

dx=2 x − 2

x 2.

1 mar$

7! S$etch the graph of f on the set of axes !elo". ndicate clearl* an* axes interceptsand stationar* points as "ell as the as*mptotes of the graph.

mar$s

c! indd 2 ydx 2

.

1 mar$


x

y


12/34


13/34

1

=et the gradient of the graph of f !e m.

d! f there are three points on the graph of f "here the gradient is m% find the possi!le&alues of m.

mar$s



14/34

14

The region enclosed !* the graph of xy − x 3 = 2 % the x-axis and the lines x = 1 and x = 2 %is rotated a!out the x-axis to form a solid of re&olution.

e! i! Write do"n a definite integral that gi&es the &olume of this solid ofre&olution.

2 mar$s

ii! ind the &olume of this solid of re&olution.

1 mar$



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15

5uestion 2 712 mar$s6

=et C z i z ∈+= 11 "here51 .

.! ind 67-rg 1 z .

1 mar$

7!

z1 is a root of the e uation z 6 = 64 where z ∈ C .

i! Sho" that z 1 is also a root.

1 mar$

ii! ind all the other roots of the e uation.

1 mar$

The line &% "hich lies in the complex plane% has the e uation z − 1 − 2 i = z − 1 .

c! Sho" that the ,artesian e uation of & is y = 1 .2 mar$s



16/34

1/



17/34

1>

d! ind% in ,artesian form% the point of intersection !et"een & and the line67-rg67-rg 1 z z = .

2 mar$s

;n the rgand diagram !elo"% B%2:C C z z z ∈=

is sho"n.

e! ;n the diagram a!o&e% s$etch B%16m7:Cand%/

67-rg: C z z z C z z z ∈=∈= π .

2 mar$s


' z (R e1

- 1

- 1- 2

- 2

-

-

1

2

2

m ' z (

B%2:C C z z z ∈=


18/34

1(

/! ;n the rgand diagram in part e!% shade the region defined !*

{ } }C z z zC z z z zC z z z ∈≥∩∈≤∩∈≤ %16m7:C6%7-rg67-rg:B%2:C 1 .1 mar$



19/34

1#

! ind the area of the region shaded in part / .2 mar$s



20/34

2'

5uestion ) 712 mar$s6

The position &ector of point A is a~ =−2 i~+2 j~ % for point B it is b~ =5 i~−5 j~ % for pointC it is

c~ = 4 i~−6 j~ and for point it is d ~ = −4 i~−2 j~ relati&e to the origin O.

.! )se a &ector method to sho" that → AB passes through the origin.2 mar$s

7! Sho" that →→ C A and are perpendicular.2 mar$s



21/34

21

c! Resol&e → A into t"o &ector components% one parallel to→

AC and one

perpendicular to→

AC .

mar$s

d! ind the cosine of ∠CAD % the angle !et"een →→ A AC and % using a &ector

method.

2 mar$s



22/34

22


23/34


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24



25/34

25

d! 9ence &erif* that y = f ( x ) is a solution to the differential e uation

sin( y ) dydx

= cos 2 ( y )

2 mar$s

e! Sol&e the differential e uation sin( y ) dydx = cos

2 ( y ) .

2 mar$s



26/34


27/34

2>

5uestion - 71 mar$s6

sno"!oarder mo&es do"n a long% straight s$i slope "hich has a constant gradient and can !e regarded as a plane. 9e tra&erses the slope from side to side as &ie"ed from a pea$ near!*and sho"n in the diagram !elo".

The position &ector of the sno"!oarder relati&e to a chairlift station near!* is gi&en !*

@@ 5

12sin2'567 jt i

t t r +

+=

π

"here t = 0 corresponds to the time% in seconds% that the sno"!oarder started his run.

The j~ component of the position &ector is a unit &ector in the direction straight do"n theslope.

The i~ component of the position &ector is a unit &ector perpendicular to the j~ componentand directed to"ards the left of the slope as *ou loo$ do"n it.Foth components are measured in metres.

.! ind the speed of the sno"!oarder 12 seconds after he starts his run. xpress *ourans"er in metres+second correct to 2 decimal places.

2 mar$s


p a t h o f s n o " ! o a r d e r

p a t h s t r a i g h td o " n

t h e s l o p e

@i

@ j


28/34

2(

7! ind the time7s6 "hen the magnitude of the acceleration of the sno"!oarder is amaximum.

mar$s



29/34

2#

=ater in the da*% the sno"!oarder learns ho" to ma$e Gumps. 9e !egins !* practicing on astraight launching pad "hich is inclined at an angle of ' ° to the horiEontal sno" on "hich itrests. The sno"!oarder is to !e to"ed in a straight line along the sno" and up the launching

pad.

;n his first attempt% the sno"!oarder releases the to" Gust as he reaches the launching pad. tthis point his speed is >.2m+s. 9e tra&els a further 5 metres up the launching pad% deceleratinguniforml* !efore momentaril* coming to rest.

c! Sho" that the coefficient of friction !et"een the sno"!oarderDs !oard and the surface

of the launching pad is 10.368 − g3 g

.

mar$s


°'

t o "

l a u n c h i n g

p a d


30/34

'

d! ind the acceleration of the sno"!oarder !ac$ do"n the launching pad after hemomentaril* comes to rest.

2 mar$s

;n a later attempt% the sno"!oarder tra&els up the entire length of the launching pad at aspeed of 1' m+s. t the top of the launching pad% he releases the to" and is air!orne until helands on the sno" !elo". Whilst air!orne% the sno"!oarder is su!Gect onl* to gra&itationalforce "ith air resistance !eing negligi!le. 9e lands 1 .>( metres horiEontall* from "here he

!ecame air!orne.

e! ind the length of the launching pad. xpress *our ans"er in metres correct to onedecimal place.

mar$s



31/34

1

S eci.&ist M.t'e3.tics For3u&.sMensur.tion

area of a trapeEium: hba 6721 +

cur&ed surface area of a c*linder: rhπ 2

&olume of a c*linder: hr 2π

&olume of a cone: hr 21

π

&olume of a p*ramid: Ah1

&olume of a sphere:4

r π

area of a triangle: Abc sin21

sine rule:C

c

B

b

A

a

sinsinsin==

cosine rule: C abbac cos2222 −+=

Coordin.te eo3etr%

ellipse: 16767

2

2

2

2

=−+−b

k y

a

h xh*per!ola: 1

67672

2

2

2

=−−−b

k y

a

h x

Circu&.r 8tri ono3etric9 /unctions167sin67cos 22 =+ x x67sec67tan1 22 x x =+ 67cosec167cot 22 x x =+

6sin76cos76cos76sin76sin7 y x y x y x +=+ 6sin76cos76cos76sin76sin7 y x y x y x −=−6sin76sin76cos76cos76cos7 y x y x y x −=+ 6sin76sin76cos76cos76cos7 y x y x y x +=−

6tan76tan716tan76tan7

6tan7 y x y x

y x−

+=+6tan76tan716tan76tan7

6tan7 y x y x

y x+

−=−

67sin21167cos267sin67cos62cos7 2222 x x x x x −=−=−=

6cos76sin7262sin7 x x x =67tan1

6tan7262tan7

2 x

x x

−=

function 1sin − 1cos − 1tan −

domain

range

H1%1I− H1%1I− *

−2

%2

π π H%'I π

−2

%2

π π

A& e7r. 8Co3 &e# nu37ers9θ θ θ cis6sin7cos r ir yi x z =+=+=

r y x z =+= 22 π π ≤


32/34

2

C.&cu&us

( ) 1−= !! !x xdxd ∫ −≠++= + 1%1

1 1 !c x!

dx x !!

( )axax aee

dx

d

=ce

adxe axax

+=∫ 1

( ) x

xdxd

e1

67log = c xdx x e

+=∫ log1( ) 6cos76sin7 axaax

dxd = cax

adxax +−=∫ 6cos716sin7

( ) 6sin76cos7 axaaxdxd −= cax

adxax +=∫ 6sin7167cos

( ) 67sec6tan7 2 axaaxdxd = ∫ += caxadxax 6tan7

167sec 2

( )2

1

1

167sin

x x

dx

d

−=− '%sin1 1

22>+

=−

−

∫ ac

a

xdx

xa

( )2

1

1

167cos

x x

dxd

−−=− '%cos1 1

22>+

=

−− −∫ aca

xdx

xa

( )2

1

1

167tan

x x

dxd

+=− c

a

xdx

xa

a +

=

+−∫ 122 tan

product rule:dxdu

%dxd%

uu%dxd +=67

uotient rule:2%

dxd%

udxdu

%%u

dxd −=

chain rule:dxdu

dudy

dxdy =

ulerDs method: %and6%7f '' b ya x x f dxdy ===

67andthen 11 !!!!! xhf y yh x x +=+= ++

acceleration:

==== 22

2

21

%dxd

dxd%

%dt d%

dt

xd a

constant 7uniform6 acceleration: at u% += 221

at ut s += asu% 222 += t %u s 6721

+=



33/34

(ectors in t o .nd t'ree di3ensions

@@@@k z j yi xr ++=

r z y xr =++= 222@ 21212121@2@1cos. z z y y x xr r r r ++== θ

@@@

@

@k

dt dz

jdt dy

idt dx

dt

r d r ++==

Mec'.nics

momentum: @@ %m p =e uation of motion: @@ am * =friction: + " µ ≤



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4

S P E C I A L I S T M A T H E M A T I C S

T R I A L E X A M I N A T I O N 2

I N S T R C T I O N Si l l i n t h e l e t t e r t h a t c o r r e s p o n d s to * o u r c h o i c e . x a m p l e :

T h e a n s " e r s e l e c t e d i s F . ; n l * o n e a n s " e r s h o u l d ! e s e l e c t e d .

1 .

2 .

.

4 .

5 .

/ .> .

( .

# .

1 ' .

1 2 .

1 .

1 4 .

1 5 .

1 / .

1 > .1 ( .

1 # .

2 ' .

2 2 .

2 1 .

1 1 .

A

A

A

A

A

"

"

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C

C

C

C

C

CC

C

C

C

6

6

6

6

6

66

6

6

6

E

E

E

E

E

EE

E

E

E

A C 6 E

AA "" CC 66 EE

A

A

A

AA

A

A

A

A

A

AA

A

A

A

"

"

"

"

"

""

"

"

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C

C

C

C

C

CC

C

C

C

6

6

6

6

6

66

6

6

6

E

E

E

E

E

EE

E

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E

S T , E + T + A M E - $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

J ) = T < = - , 9 ; , N S W R S 9 T

2014 specialist maths exam 2

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