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S Y D N E Y B O Y S H I G H S C H O O LM O O R E P A R K , S U R R Y H I L L S

2004

TRIAL HIGHER SCHOOL

CERTIFICATE EXAMINATION

Mathematics Extension 2

Sample SolutionsSection Marker

A Mr Hespe

B Mr Kourtesis

C Mr Parker

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Section A

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Section B

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Section C

Q6

(a) With R v , to make the algebra easier take R mkv=

(i) ( )dv

m mg mkvdt

= +

( )

( ) ( )

1

1

1 1( )

1ln

0,

1 1ln ln

1

kt

kt kt

dv dt k g kvdt dv g kv k g kv

t g kv ck

t v u

g kvc g ku t

k k g ku

g kve

g ku

g kv g ku e v g ku e gk

= + = = + +

= + +

= =

+= + =

+

+ =

+

+ = + = +

( )

( )

2

2 2

1

1

0, 0

kt

kt

x g ku e g dt k

g kue gt c

k k

t x

g kuc

k

= +

+ = +

= =

+ =

( )

2

2

1

1

kt

kt

g ku g ku x e gt

k k k

g ku gt e

k k

+ + = +

+=

QED

+

mg

mkv

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(ii) The two particles meet when1 2

x x=

[NB You are allowed to assume the formula for 2x !]

ie ( ) ( )2 21 1kt kt g ku gt g gt

e h ek k k k

+ = +

( ) ( ) ( )2 21 1 1kt kt kt g u gt g gt

e e h ek k k k k

+ = +

( )1

1 1

ln ln

1ln

kt

kt kt

ue h

k

hk hk u hk e eu u u

u hk ukt kt

u u hk

ut

k u hk

=

= = =

= =

=

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Q7

(i)

(ii) Since ||| BWF VUF

3

3

VU UF h x

BW FW ba

axh

b

= =

=

(iii) Since ||| BWF VUF then3

VF VU h

BF BW a= =

BV BF VF =

||| BLM RFS then BV LM BF VF LM

BF RS BF a

= =

( )

1 13

3

1 13

VF LM h LM

BF a aa

ax LM LM x b xb

a a b ba

a b xLM

b

= =

= = =

=

QED

(iv) Clearly whenx = 0 then PQ = a and whenx = b then PQ = 2a, so given

the linear relationship ofPQ in terms ofx then

( )2

0a a a

PQ a x PQ x ab b

= = +

Alternative solution

2a

a

BW

By Pythagoras Theorem

4a2

= BW2

+ a2

BW2

= 3a2

BW = 3a

a1

a

x

PQ = a + 2a1

a1

a 2=

x

b 2a

1=

a

bx

PQ =a

bx + a

2a

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(v) Area of slice is area of trapezium KLMNand rectangle KNQP

( )

2

1 3

2

3

a b xax aKLMN x a

b b b

a xb

= + +

=

21

a xKNQP a x a a

b b

= + = +

So cross sectional area is given by

( )

( )

22

22

2

2

31

3

1 3

1 3

a x xa

b b

a x x ba

b b

a x b

b

ax b

b

+ +

+ = +

+ + =

= + +

So the cross sectional volume is ( )2

1 3a

x b xb

+ +

So the volume, V, is given by ( )2

0

1 3

ba

x b dxb

+ +

( )

( )

( )

( )

2

0

2 2

0

2 22

2

1 3

1 32

1 32

3 32

b

b

a

V x b dxb

a xbx

b

a bb

b

a b

= + +

= + +

= + +

= +

[NB This is not a solid formed by rotation, so shouldntappear in the

answer!]

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Q8

(a) Method 1 Method 2

( )

2 2

2

1 1 4 1 1 4

( ) ( ) 4

( )

2

( )

( )

0

a b t a b a bb a b a a b ab

ab a b

a ab b

ab a b

a b

ab a b

+ = +

+

+ + + =

+

+=

+

=

+

1 1 4a b t

+

( )2

0 2a b a b ab +

1 1 1 2

2a b a bab ab

+ +

Also1 1 2

a b ab+

So1 1 2 4 4

a b a b t ab+ =

+

Method 3 (reductio ad absurdum)

Assume1 1 4

a b t+ <

( ) ( )

( ) ( )

2

2 2

4

4

4 0

a b

ab t

a b ab t a b

a b ab a b

+