Exam 2 - Section 2 Solutions - 2011[1]

8/16/2019 Exam 2 - Section 2 Solutions - 2011[1]

1/16

SCHOOL

Trial WACE Examination, 2011Question/Answer Booklet

ATHE AT!CSS"EC!AL!ST #C/#$

Se%tion Two&Cal%ulator'assume(

Student Number: In figures

In words ______________________________________

Your name ______________________________________

Time allowe( )or t*is se%tionReading time before commencing work: ten minutesWorking time for this section: one hundred minutes

aterials re+uire(/re%ommen(e( )or t*is se%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(ates

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-isor

.e)ore reading an) further,

SOL T!O S


2/16

ATHE AT!CS S"EC!AL!ST #C/#$ 2 CALC LATO 'ASS E$

Stru%ture o) t*is a er

SectionNumber of

uestionsa-ai!ab!e

Number of uestions to

be answered

Working time#minutes%

0arksa-ai!ab!e

1ercentageof e+am

Section $ne:*a!cu!ator.free 2 2 34 (4 55

Section Two:*a!cu!ator.assumed 65 65 644 74 28

Total 694 644

!nstru%tions to %an(i(ates

6, The ru!es for the conduct of Western Austra!ian e+terna! e+aminations are detai!ed in theYear 12 Information Handbook 2011 , Sitting this e+amination im&!ies that )ou agree toabide b) these ru!es,

9, 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 re uired 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, S*ow all -our workin %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 ustification isre uired 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 use en%il ' e+ce&t in diagrams,

See next a e


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CALC LATO 'ASS E$ # ATHE AT!CS S"EC!AL!ST #C/#$

Se%tion Two& Cal%ulator'assume( 340 arks5

This section has t*irteen 31#5 uestions, Answer all uestions, Write )our answers in the s&aces&ro-ided,

Working time for this section is 644 minutes,

Question 6 37 marks5

The tem&erature' I °*' of a !i uid in an insu!ated f!ask at an) time t seconds can be described

b) the differentia! e uation .4 445dI

I dt

= − ,

#a% ;ow !ong wi!! it take for the !i uid in the f!ask to fa!! b) 64


4/16



#a% "ind the distance between the &oints with &o!ar coordinates ,9

35

π ÷

and ,3

69 .2

π ÷

'

where distances are in centimetres and ang!es in radians, #9 marks%

#b% The gra&hs of θ α = ' r b= and r nθ = are shown be!ow together with the &oints A and which ha-e &o!ar coordinates of ( , )6 9 and ( , )(b , "ind the -a!ues of , , b nα and the &o!ar coordinates of &oint *, #5 marks%

x

y

A

*

See next a e

9 9 9

9 3 59 9 At right ang!es,5 2 9 9

3 6965 cm,

d

d

π π π π π π − + = − = ⇒ ÷ = +

=

.

.

. , ,

@sing A' 9 and 6 9 4 3

@sing ' 4 3 ( 9

9 ' 9 ' n 4 39

n n

b

b C

α

π α π

= = × ⇒ =

= × =

= = = ÷


5/16

CALC LATO 'ASS E$ 9 ATHE AT!CS S"EC!AL!ST #C/#$

Question : 34 marks5

The &oint A has &osition -ector 5 9 (− +i ; k ,

#a% "ind the -a!ue of a if the -ectors OA and 5 5a + −i ; k are &er&endicu!ar, #6 mark%

#b% "ind the si e of the ang!e between OA and the z .a+is' to the nearest degree, #9 marks%

#c% "ind the -a!ue of b if the &oint #8' b ' 9% !ies in the &!ane containing the &oint #.6' 9' 3% andwith norma! -ector OA , #9 marks%

#d% "ind the -a!ue of c if the &oint #63' .6(' c % !ies on the straight !ine through A and the&oint #.6' 9' 3%, #5 marks%

See next a e

59 5 4

( 5

5 672

a

a

a

− • = − =

=

Ang!e between OA and the !ine =r k

( ) ( ) ( )

( ) ( ) ( )

1!ane 5 9 (5 6 9 9 ( 3 65 65

5 8 9 ( 9 657

x y z k

k

b

b

− + =− − + = ⇒ =

− + ==

( )

?irection of !ine gi-en b)

5 6 (

9 9 (( 3 6

5 ( 639 ( 6(

( 6

5 ( 63 5( 5 66

c

c

λ

λ λ

−

− − = − −

− + − = − −

+ = ⇒ == + −=


6/16



When an ob ect is at a distance u cm from a !ens of foca! !ength f cm' an image is formed at adistance of v cm from the !ens,

The -ariab!es are re!ated b) the formu!a6 6 6 f u v

= + ,

An ob ect is mo-ing with a constant s&eed of 9 cm/s towards a !ens of foca! !ength 94 cm,

At the instant when the image is 54 cm from the !ens' in what direction and with what s&eed is itmo-ing=

See next a e

( )

( )

9

9

Bi-en 9 find when 54

6 6 624

94 54

6 6 6 9494 94

(44

94

(44 924 94

6 cm/s awa) from the !ens # since C-e%

9

du dvv

dt dt

uu

uv

u v u

dvdu u

dv dv dudt du dt

= − =

= + ⇒ =

= + ⇒ =−

−=−

= ×

−= × −−

=


7/16



#a% The gra&h of ( ) y f x= is shown be!ow,

x

y

Sketch the gra&hs of #9 marks%

x

y y = f (| x|)#ii%

x

y y = | f ( x)|#i%

#b% The e uation (ax b x+ = − has so!utions .4 9 x = − and 5 x = − , "ind the -a!ues of a and b , #5 marks%

See next a e

. . .So!utions when 4 9 ( 9 #using 4 9%

and when 5 8 #using 5%

a b x

a b x

− + = = −

− + = = −


8/16



1ro-e b) deduction that6 sin 9 cos 9

tan6 sin 9 cos 9

θ θ θ

θ θ + − =+ + ,

See next a e

( )

( )

( )( )

9

9

9

9

6 9 sin cos 6 9sin

6 9 sin cos 9cos 6

9 sin 9 sin cos

9 cos 9 sin cossin sin coscos sin costan

LHS θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ

+ − −= + + −

+=+

+= +=


9/16

CALC LATO 'ASS E$ : ATHE AT!CS S"EC!AL!ST #C/#$

Question 1# 37 marks5

Two com&!e+ numbers are gi-en b) 5u i= and 5 5 59

iv

−= ,

#a% D+&ress 5u v in the form ( )cos sinr iθ θ + where π θ π − ≤ ≤ and 4r ≥ , #9 marks%

#b% "ind a!! so!utions for z in the form ire θ ' gi-en that ( 5 z u v= , #9 marks%

#c% Show that the sum of a!! the so!utions from &art #b% is 4, #9 marks%

See next a e

95 5

32 2

5 and 5 are e/ua! in magnitude and o&&osite in direction and sum to 4,

5 and 5 are e/ua! in magnitude and o&&osite in direction and sum to 4,

;ence sum of a!! roots is 4,

e e

e e

π π

π π

−

−


10/16



At a schoo! with 647 boarders' boarders can either eat breakfast or not, The canteen managerestimates that of those boarders who eat breakfast one morning' 3< of them wi!! not eatbreakfast the ne+t morning and of those boarders who do not eat breakfast one morning' 33< ofthem eat breakfast the fo!!owing morning,

#a% If 33 boarders eat breakfast on 0onda)' how man) boarders shou!d the canteen managere+&ect to eat breakfast on Wednesda)= #5 marks%

#b% In the !ong term' what &ro&ortion of boarders can be e+&ected to eat breakfast= #9 marks%

See next a e

. .

. .

.

.9

4 >3 4 33 33

4 43 4 (3 35

>6 >262 4(

D+&ect >9 students for breakfast

T P

T P

= =

=

.. .

4 33 664 33 4 43 69

or

>> >> 66 as increases

> >> > 69nT P n

=+

→ ⇒ = +


11/16



The gra&hs of the function ( ) log ( )9 e f x x k = + ' where k is a constant' and its in-erse ( )6 f x− 'intersect where 9 x = − and at one other &oint,

#a% "ind the e+act -a!ue of k , #9 marks%

#b% Sketch the gra&hs of ( ) f x and its in-erse ( )6 f x− on the a+es be!ow' gi-ing e uations ofan) as)m&totes and showing the coordinates of a!! &oints of intersection and a+es.interce&ts correct to 9 decima! &!aces, #3 marks%

x.( (

y

.(

(

x = -2.37

y = -2.37

#4' 6,89%

#4' .6,58%

#.6,58' 4% #6,89' 4%

#5,32' 5,32%

#.9' .9%

See next a e

( )

( )

ln( )

and in-erse intersect a!ongSo!-e 9 9

9 9 96

9

f x y x

f

k

k e

=− = −

− = − +

= +


12/16



#a% The matri+ e uation AX B= cou!d be used to so!-e the fo!!owing s)stem of e uations,

9 5 39 ( 6

3

a b c

b c a

c b

+ = −− − =

= +Write down suitab!e matrices for , and A X B, #?$ N$T S$EFD Y$@R DQ@ATI$NS%

#9 marks%

#b% If 5 PQ P I = + where6 44 6

I =

and8 >3 7

Q− = −

' find matri+ P , #5 marks%

See next a e

9 5 6( 6 9

4 6 6

363

A

a

X b

c

B

− = − − = − =

( )

( ) 6

55

5

66 >3 (

PQ P I P Q I I

P Q I

P

−

− =− =

= −−

= −


13/16

CALC LATO 'ASS E$ 1# ATHE AT!CS S"EC!AL!ST #C/#$


The dis&!acement ( ) x t metres' of a sma!! &artic!e undergoing sim&!e harmonic motion is gi-en b)( ) cos sin x t A t B t ω ω = + ' where , and A B ω are &ositi-e constants,

#a% Show that ''( ) ( )9 4 x t x t ω + = , #9 marks%

The bod) &asses through the origin #where ( ) 4 x t = % fi-e times &er second' ( ) .4 6 3 x = m and'( ) .4 8 3 x = ms .6 ,

#b% "ind the e+act -a!ues of the constants , and A B ω , #5 marks%

#c% What is the am&!itude of motion' correct to the nearest mi!!imetre= #9 marks%

See next a e

'( ) ( )

''( ) ( )

( )

''( ) ( )

9

9

9

sin cos

cos sin

4

x t A t B t

x t A t B t

x t

x t x t

ω ω ω

ω ω ω

ω

ω

= − += − += −

+ =

.

.

.

. ( . )

1assing origin 3 times &er sec means fre uenc) is 9,3 c)c!es/sec,

9 9 3 3

6 3 cos 4 sin 46 3

8 3 3 6 3 sin 4 cos 45

9

A B

A

B

B

ω π π

π

π

= × == +

=

= − +

=

. .

.

99 56 3 6 38(62

9

6 38( m to nearest mm,

π + = ÷

≈


14/16



D-er) odd integer I can be written as 64 I n c= + ' where 6' 5' 3' 8' >c = ,

#a% Show how the integers 958' 5 and .53 can be written this wa), #6 mark%

#b% ) considering the fi-e different cases for c ' or otherwise' &ro-e that the s uare of e-er)odd integer ends in 6' 3 or >, #8 marks%

See next a e

( )

958 64 95 85 64 4 5

53 64 ( 3

= × += × +− = × − +

( )

( )

9 9 9

9 9 9

9 9

664 6 644 94 6 64 64 9 6

This number must end in 6' as for an) number in form 64 ' is the units digit,

5

64 5 644 24 > 64 64 2 >' which ends in >

3

64 3 644 644

c I n I n n n n

p q q

c

I n I n n n n

c

I n I n n

== + ⇒ = + + = + ++

== + ⇒ = + + = + +

== + ⇒ = + ( )

( )

( )

9

9 9 9

9 9 9

93 64 64 64 9 3' which ends in 3

8

64 8 644 6(4 (> 64 64 6( ( >' which ends in >

>

64 > 644 674 76 64 64 67 7 6' which ends in 6

;ence in a!! &ossib!e cases' the s uare e

n n

c

I n I n n n n

c

I n I n n n n

+ = + + +

== + ⇒ = + + = + + +

== + ⇒ = + + = + + +

nds in 6' 3 or >,


15/16


Question 1: 3: marks5

Re!ati-e to itse!f' an anti.ba!!istic missi!e #A 0% !aunch site detects a ba!!istic missi!e at66 67 64− +i ; k km headed at constant -e!ocit) for a target at 53 6(+ +i ; k km, The ba!!isticmissi!e is e+&ected to hit the target in 34 seconds,

#a% ;ow c!ose does the ba!!istic missi!e come to the A 0 !aunch site= #3 marks%

#b% The !aunch site &!ans to fire an A 0 to hit the ba!!istic missi!e, The hit is timed to take&!ace at the instant the ba!!istic missi!e comes within 7km of the target, Assuming the A 0instant!) achie-es a constant -e!ocit) of 6634 ms .6 as it is !aunched' how !ong from thetime of detection shou!d the defence site fire it= #( marks%

En( o) +uestions

.

( ) .

.

)

Eet missi!e 0 be at A and target T at initia!!), Then

53 66 9( 9( 4 (76

6( 67 59 km and 59 4 2( km/s34

6 64 > > 4 67

*!osest when # 4

66

AB !

OA t ! !

− = − − = = = − − − −

+ × • =

+

uuur

uuur

.

. .

. . .

. .

. .

. . .

. .66 >38

4 (7 4 (767 4 2( 4 2( 4 when 66 >38 seconds,

64 4 67 4 67

66 4 (7 62 85>67 4 2( 64 5(8 and 96 672 km

64 4 67 8 7(7t

t

t t

t

t

O" t O"

t =

− + • = = − −

+ = − + = − = −

uuuur uuuur

.

.

. .

.

55(6 km, *o!!ide at *' 7 km from and 55 km from A after 34 (4 9(( seconds,

(6

66 9( 54 56855

67 59 8 832 and 56 (63 km,(6

64 > 9 832

Time for A 0 to h

AB

OC OC

= × ≈

= − + = = −

uuur

uuur uuur

. . .

. . .

it 0 56 (63 6 63 98 568 seconds,

0ust !aunch after (4 9(( 98 568 69 >5 seconds,

= ÷ =

− =


16/16

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Exam 2 - Section 2 Solutions - 2011[1]

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