Exam 2 - Section 2 - Soln 2012

8/16/2019 Exam 2 - Section 2 - Soln 2012

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PERTH MODERN SCHOOL

Trial WACE Examination, 2012

Quetion!An"er #oo$let

MATHEMAT%CSSPEC%AL%ST &C!&D

Se'tion T"o(Cal'ulator)aume*

Student Number: In figures

In words ______________________________________

Your name ______________________________________

Teacher ______________________________________

Time allo"e* +or ti e'tionReading time before commencing work: ten minutesWorking time for this section: one hundred minutes

Material re-uire*!re'ommen*e* +or ti e'tionTo be provided by the supervisor

This Question/Answer ook!et"ormu!a Sheet #retained from Section $ne%

To be provided by the candidateStandard items: &ens' &enci!s' &enci! shar&ener' eraser' correction f!uid/ta&e' ru!er' high!ighters

S&ecia! items: drawing instruments' tem&!ates' notes on two unfo!ded sheets of A( &a&er'and u& to three ca!cu!ators satisf)ing the conditions set b) the *urricu!um*ounci! for this e+amination,

%m.ortant note to 'an*i*ate

No other items ma) be used in this section of the e+amination, It is /our res&onsibi!it) to ensurethat )ou do not ha-e an) unauthorised notes or other items of a non.&ersona! nature in thee+amination room, If )ou ha-e an) unauthorised materia! with )ou' hand it to the su&er-isore+ore reading an) further,


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CALCLATOR)ASSMED & MATHEMAT%CS SPEC%AL%ST &C!&D

Stru'ture o+ ti .a.er

SectionNumber of uestionsa-ai!ab!e

Number of uestions tobe answered

Working time#minutes%

0arksa-ai!ab!e

1ercentageof e+am

Section $ne:*a!cu!ator.free 2 2 34 34 55

Section Two:*a!cu!ator.assumed

65 65 644 644 72

Total 634 644

%ntru'tion to 'an*i*ate

6, The ru!es for the conduct of Western Austra!ian e+terna! e+aminations are detai!ed in theYear 12 Information Handbook 2012 , Sitting this e+amination im&!ies that )ou agree toabide b) these ru!es,

8, Write )our answers in the s&aces &ro-ided in this Question/Answer ook!et, S&are &agesare inc!uded at the end of this book!et, The) can be used for &!anning )our res&onsesand/or as additiona! s&ace if reuired to continue an answer,• 1!anning: If )ou use the s&are &ages for &!anning' indicate this c!ear!) at the to& of the

&age,• *ontinuing an answer: If )ou need to use the s&ace to continue an answer' indicate in

the origina! answer s&ace where the answer is continued' i,e, gi-e the &age number,"i!! in the number of the uestion#s% that )ou are continuing to answer at the to& of the&age,

5, So" all /our "or$in 'learl/, Your working shou!d be in sufficient detai! to a!!ow )ouranswers to be checked readi!) and for marks to be awarded for reasoning, Incorrectanswers gi-en without su&&orting reasoning cannot be a!!ocated an) marks, "or an)uestion or &art uestion worth more than two marks' -a!id working or 9ustification isreuired to recei-e fu!! marks, If )ou re&eat an answer to an) uestion' ensure that )oucance! the answer )ou do not wish to ha-e marked,

(, It is recommended that )ou *o not ue .en'il' e+ce&t in diagrams,

See next .ae


4/21

MATHEMAT%CS SPEC%AL%ST &C!&D 3 CALCLATOR)ASSMED

Se'tion T"o( Cal'ulator)aume* 4100 Mar$5

This section has tirteen 41&5 uestions, Answer all uestions, Write )our answers in the s&aces&ro-ided,

Working time for this section is 644 minutes,

Quetion 6 47 mar$5

In two residentia! suburbs' A and ' from 6;( to 6' the median house &rice' M do!!ars'

increased at a rate gi-en b)dM

kM dt

= ' where t is the time' in )ears and k is a constant s&ecific

to each suburb,

"or suburb A' the median &rice at the start of 6;( was


5/21

CALCLATOR)ASSMED 8 MATHEMAT%CS SPEC%AL%ST &C!&D

Quetion 9 48 mar$5

The fo!!owing es!ie matri+' L ' a&&!ies to a &o&u!ation of beet!es in which the fema!e beet!es inthe &o&u!ation !i-e for a ma+imum of 5 )ears and on!) &ro&agate in their third )ear of !ife,

6883

4 4 3

4 44 4

L

=

#a% What is the &robabi!it) that a newborn fema!e beet!e wi!! sur-i-e to the 5rd )ear of its !ife=#6 mark%

#b% Initia!!) there are 344 fema!es in each age grou&, >ow man) fema!es wi!! there bea!together after 8 )ears= #8 marks%

#c% *omment on the !ong.term &o&u!ation of fema!e beet!es &redicted b) this mode!, #8 marks%

See next .ae

6 8 6

8 3 3× =

8

344 6444

344 6834

344 644

L

=

Tota! @ 8534 beet!es

The tota! number of fema!e beet!es starts at 6344 initia!!)'then near!) doub!es after 6 )ear to 834' fa!!s back to 8534after 8 )ears and then returns to 6344 after 5 )ears' with344 in each age grou& as at the start of the c)c!e,

This c)c!e then re&eats end!ess!)' according to the mode!,


6/21


Quetion 10 4: mar$5

#a% A triang!e with -ertices at ( , )6 6 A ' ( , )5 6 B and ( , )5 (C is ref!ected in the x .a+is and then

rotated 4° antic!ockwise about the origin,

#i% "ind the matri+ T that wi!! combine these two transformations in the order gi-en,#5 marks%

#ii% "ind the coordinates of C after transformation b) T , #6 mark%

#b% Another transformation matri+ is gi-en b).

. .

4 7 4

6 8 4 7 R

− = − −

,

Betermine the area of triang!e ABC after it has been transformed b) T andthen b) R , #5 marks%

See next .ae

4 6 6 4 4 6

6 4 4 6 6 4T

− = × = −

'( , )

4 6 5 (

6 4 ( 5

( 5C

=

$rigina! area of triang!e A* @ 5 s units,

Beterminant of T is .6' so no change in area,

Beterminant of R is 4,57' so fina! area . .4 57 5 6 4;= × = s units,


7/21

CALCLATOR)ASSMED : MATHEMAT%CS SPEC%AL%ST &C!&D

Quetion 11 4: mar$5

A function is defined as ( ) 8 5 8 f x x x= + + − ,

#a% C+&ress ( ) f x without the use of abso!ute -a!ue bars, #5 marks%

or f(x) = (x + 2)2

+ (3 – 2x)2

#b% Sketch the gra&h of ( ) f x #8 marks%

x.64 .3 3 64

y

3

64

#c% So!-e ( ) 3 f x x≤ + , #8 marks%

See next .ae

( ) ( )

( ) ( ) ( ) .

( ) ( ) .

( ) .

.

8 5 8 8

8 5 8 8 6 3

8 5 8 6 3

6 5 8

3 8 6 3

5 6 6 3

x x x

f x x x x

x x x

x x

f x x x

x x

− + + − < −= + + − − ≤ ≤ + − − >

− < −= − − ≤ ≤ − >

3 3 4

3 5 6 5

So!ution:

4 5

x x x

x x x

x

+ = − ⇒ =+ = − ⇒ =

≤ ≤


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Quetion 12 47 mar$5

"ind the e+act area bounded b) the x .a+is' the y .a+is' the function .( ) 4 838 x f x e= and thetangent to ( ) f x when ; x = ,

x3 64

y

64

84

See next .ae

( ) ( )

.

( )

'( )

'( )

( )

( )

8

4 83

8

88

88

; ; 88

4 (8 8

8

; 8

8

;8

Tangent:

8 ;8

88

Area:

88

; ; (

( ;

x

f e

e f x

e f

e y e x

e y x e

e f x dx x e dx

e e

e

=

=

=

− = −

= −

− −

= − −

= −

∫ ∫

units8


9/21


Quetion 1& 47 mar$5

A !ight is &ositioned at the to& of a -ertica! &ost 68 m high, A sma!! ba!! is dro&&ed from the sameheight as the !ight but at a &oint ( m awa),

If the distance tra-e!!ed b) the ba!! t seconds after re!ease is gi-en b) . 8( t ' how fast is the

shadow of the ba!! mo-ing a!ong the horiDonta! ground ha!f a second after the ba!! is dro&&ed=

68

(

(,:t8

+

See next .ae

"inddx

dt when .4 3t = ,

Esing simi!ar triang!es'

.

.

( )

.

.

.

8

8

5

68 (

(

(;

(

8 (;

(

4 3

27;4637 2

(

t

x

xt

dx

dt t

t

dx

dt

=

=

−=

== − ≈ −

>ence s&eed of shadow is 637,2 m/s,


10/21


Quetion 13 46 mar$5

#a% Ese &roof b) contradiction to &ro-e that 8 is irrationa!, #( marks%

#b% Ese a -ector method to &ro-e that the diagona!s of the rhombus $1QR are &er&endicu!ar,

#( marks%

$

1 Q

R

See next .ae

et andOP RQ OR PQ= = = =a ,

Then andOQ RP = + = −a a ,

So dot &roduct of diagona!s is

( ) ( )

8 8

4 since

OQ RP • = + • −

= −

= =

a a

a

a

>ence the diagona!s are &er&endicu!ar,

Assume that 8 is rationa! and can be written as the sim&!ified fractiona

b

'

where a and b are integers with no common factors,

.

( )

8 8

8 8 8 8

8 8

8 8 is an e-en integer,

If is e-en then it can be written as 8

>ence 8 8 ( 8

>ence 8 is an e-en integer,

ut if and are both e-en integers then the) ha-e 8 as a

aa b a

b

a a n

n b n b

b n b

a b

= ⇒ = ⇒

=

= ⇒ =

= ⇒

common factor'

which contradicts the assum&tion that the) ha-e no common factors,

Thus the assum&tion is incorrect' and 8 must be irrationa!,

as $1QR is a rhombus


11/21


Quetion 18 412 mar$5

A*BC"F> is a rectangu!ar &rism with

8 5OA = + +i ; $

( 8 2OB = + +i ; $

2 ; OD = − + +i ; $

65 82 7OE = + −i ; $

#a% "ind a -ector euation for the &!ane C"F> in the form c• =r n , #5 marks%

#b% "ind a -ector euation for the !ine &assing through A and C, #8 marks%

See next .ae

A

*B

C "

F>

65 6 68

82 8 83

7 5

68 65

83 82 ;;3

7

68

83 ;;3

AE

c OE

= = − = − −

= • = • =

− −

• = −

n

n

r

uuur

uuur

( )8 5 68 83

OA AE λ

λ

= += + + + + −

r

i ; $ i ; $


12/21


#c% The &oint 1 !ies on the !ine through AC so that the siDe of HPD∠ is 4°, "ind the shortest&ossib!e distance from A to 1,

#2 marks%

See next .ae

65 ; 3

82 7 55

7 7 4

3 68 (

55 83 56

4 5

6 68 2 68 ;

8 83 ; 83 7

5 7

OH OE EH

HP

DP

HP DP

λ

λ

λ

λ λ

λ λ

λ λ

− = + = + = −

− = − = − − + + − +

= + − = −

− − −

•

uuur uuur uuur

uuur

uuur

uuur uur

( ) ( ) ( ) ( ) ( ) ( )

. , .

.

.

.

.

8

4

68 ( 68 ; 83 56 83 7 5 7 4

;34 ;34 657 4

4 8 4 ;

68 8 (

4 8 83 3

6 ;

5( 3 ;5 units

AP

AP

λ λ λ λ λ λ

λ λ

λ λ

=

− + + − − + − + − − =

− + == =

= × =

= ≈

u

uuur

uuur

6 ? 68

8 ? 83

5 G :

G

3

55

4

@

68 G (

83 G 7

G : ? 5

..


13/21

CALCLATOR)ASSMED 1& MATHEMAT%CS SPEC%AL%ST &C!&D

Quetion 17 49 mar$5

et sinw cos θ θ = + and sin ! cos φ φ = + ,

#a% Ese Cu!erHs formu!a to e+&ress the &roduct w! in e+&onentia! form, #6 mark%

#b% Ese w and ! to show that sin( ) sin cos cos sinθ φ θ φ θ φ + = + , #( marks%

#c% >ence show thatcos sin

cos

855 sin

( (8 8d c

π θ θ π θ θ θ

+ + = + + ÷ ÷ ÷

∫ , #( marks%

See next .ae

( ) w! e e eθ φ θ φ += =

( ) ( )

( ) ( ) ( )

( )

( ) cos sin cos sin

cos sin cos cos sin sin cos sin sin cosCuating imaginar) &arts gi-es

sin cos sin sin cos

as reuired,

e

θ φ θ θ φ φ

θ φ θ φ θ φ θ φ θ φ θ φ

θ φ θ φ θ φ

+ = + +

+ + + = − + +

+ = +

sin cos

cos cos sin sin cos

cos sin

sin

8

8

5

6

( ( 8

5( ( (

5( (

(

d

d

c

π π

π π π θ θ θ θ

π π θ θ θ

π θ

= =

+ + ÷ ÷

= + + ÷ ÷

= + + ÷

∫

∫


14/21


Quetion 1: 46 mar$5

A &artic!e mo-es a!ong the x .a+is' with dis&!acement x cm from the origin' after t seconds'

gi-en b) cos5

t x a

π = ÷

' where a is a &ositi-e constant, After 6 second' the &artic!e is 68 cm from

the origin,

#a% "ind the -a!ue of a , #6 mark%

#b% Show that the motion of the &artic!e is sim&!e harmonic, #8 marks%

#c% "ind the s&eed of the &artic!e as it &asses through the origin, #8 marks%

#d% "ind the distance tra-e!!ed b) the &artic!e during the first minute of its motion, #5 marks%

See next .ae

cos68 8(5

a aπ

= ⇒ =

cos

( ) sin

( ) cos

8

8

8(5

8(5 5

8(5 5

S>05

x

x

x

x x

π

π π

π π

π

= ÷

= − ÷

= − ÷ ÷

= − ⇒ ÷

&

&&

&&

0a+imum s&eed when &assing through origin' so

sin

( ) sin

65

8(5 5

; s&eed ; cm/s

x

x

π

π π

π π

= ÷

= − ÷

= − ⇒ =

&

&

1eriod is5

87 seconds

π

π = ' so in 74 seconds wi!!

mo-e through e+act!) 64 c)c!es,

Am&!itude is 8( cm' so in one com&!ete c)c!emo-es 8( ( 7× = cm,

>ence wi!! tra-e! 64 7 74× = cm,


15/21


Quetion 16 49 mar$5

A com&!e+ ineua!it) is gi-en b) 5 5 5 5 ! + − ≤ ,

#a% Sketch the region in the com&!e+ &!ane defined b) this ineua!it), #5 marks%

Re(!).7 .5 5 7

"#(!)

.:

.7

.5

5

75 5

#b% "ind the minimum and ma+imum -a!ues of ! , #5 marks%

#c% "ind the minimum and ma+imum -a!ues of arg ! , #5 marks%

See next .ae

Radius of circ!e is 5,

Bistance from #4' 4% to circ!e centre is ( )8 85 5 5 7+ = ,

0in ! is 7 5 5− =

0a+ ! is 7 5 :+ =

0inimum -a!ue is8

π ,

0a+imum -a!ue is3

88 7 7

π π π + × = ,

a

5

7sin a @ 5/7

a @π

7


16/21


Quetion 19 410 mar$5

The -e!ocit) of a bod) mo-ing in a straight !ine is gi-en b) 5 (dx

xdt

= + ' where x is the

dis&!acement' in metres' from a fi+ed reference &oint at time t seconds, When 6t = ' 8 x = ,

#a% "ind an e+&ression for x in terms of t , #3 marks%

#b% What is the e+act -e!ocit) of the bod) when

#i% 5 x = = #6 mark%

#ii% 5t = = #8 marks%

#c% What is the acce!eration of the bod) when 6t = = #8 marks%

See next .ae

ln( )

,

(

(

( (

( (

6

5 (

65 (

(

5 (

6 8 66

5 ( 66

66 5(

t

t

t

dx dt x

x t c

x ae

t x a e

x e e

e x

−

−

−

=+

+ = +

+ =

= = ⇒ =

+ =

−=

∫ ∫

( )5 ( 5 63 m/s$ = + =

( )( 5 ( ;

;;

66 5 66

( (

665 ( 5 66 m/s

(

e e x

e$ e

− −= =

= + × = +

( ( ))

8

5 (

(

( 5 ( 8

(( m/s

$ x

d$ dx

dt dt

= +

=

= × +

=

66e

; G 5

(

66e;


17/21

CALCLATOR)ASSMED 1: MATHEMAT%CS SPEC%AL%ST &C!&D

Quetion 20 4: mar$5

et ( )3 5 2

3 5 63

n n n P n = + + ,

#a% C-a!uate ( )6 P and ( )( P , #6 mark%

#b% 1ro-e b) induction that ( ) P n is a!wa)s an integer' when n is a &ositi-e integer, #7 marks%

See next .ae

( )

( )

6 6

( 88;

P

P

=

=

( ) ( )

( ) ,

( ) ( ) ( )( )

(

3 5

3 5

6 6 is an integer when 6

2 Assume that is an integer where is a &ositi-e integer' so that3 5 63

k?6 is the ne+t consecuti-e integer after k,

6 6 2 66

3 5 63

6

P P n n

k k k P k k "

k k k P k

P k

= ⇒ =

+ + =

+ + ++ = + +

+ )

( )

( )

( )

( ) ( )

3 ( 5 8 5 8

3 5( 8

3 5( 5 8

( 5 8

3 64 64 3 6 5 5 6 2 2

3 5 63

2 526 5 63 5 63

26 8 5 8 6

3 5 63

6 8 5 8 6

>ence' if is an integer' then 6 is a!so an integer' as bo

k k k k k k k k k

k k k P k k k

k k k P k k k k k

P k " k k k k

P k P k

+ + + + + + + + += + +

+ = + + + + +

+ = + + + + + + +

+ = + + + + +

+

( ) ( )

th and are integers,

Since 6 is an integer' then 8 must be an integer' and so on for a!! &ositi-e integer ,

" k

P P n


18/21


A**itional "or$in .a'e

Question number: _________

En* o+ -uetion


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2012 Tem.late

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Exam 2 - Section 2 - Soln 2012