Leaving Certificate Higher Level - educateplus · PDF fileHigher Level, Educate.ie Sample Paper Solutions 1 ... When you multiply complex numbers in polar form you add their arguments

1Higher Level, Educate.ie Sample Paper Solutions

Sample Paper 1 Educate.ie Paper 1 Solutions ....................................................................... 2

Educate.ie Paper 2 Solutions ..................................................................... 16

Sample Paper 2 Educate.ie Paper 1 Solutions ..................................................................... 26








Sample Paper 6 Educate.ie Paper 1 Solutions ................................................................... 111

Educate.ie Paper 2 Solutions ................................................................... 122

Sample Paper 7 Educate.ie Paper 1 Solutions ................................................................... 133

Educate.ie Paper 2 Solutions ................................................................... 143

ContentS

educate.ie MAtHeMAtICS SoLUtIonS

Leaving Certificate Higher Level

Educate.ie’s Leaving Certificate Higher Level Sample Paper Mathematics Solutions are also available on www.educateplus.ie.

2 Mathematics Leaving Certificate

LC HigHer LeveL SoLutionS SampLe paper 1 (© Educate.ie)

Question 2 (25 marks)

( )( )3 2 3 2

3 6 6 2 1

− +

= + − − =

3 23 2

3 23 2

3 23 2

5 2 61

5 2 6

+−

=+−

×++

=+

= +

( )( )

( )( )

[theme: All about irrationals, root 2 and root 3]

z i

iz

z

12 1212

1212

1

26 6

2 126

126

= +

= +

cos sin

cos sin

π π

π π

22

12

12

12 12

2 2 2

2 1 0 2

= +{ }

= +{ }=

cos sin

( )

π πi

iz

Question 1 (e)

Question 2 (a)Draw a circle with centre (0, 0) through the point (1, 1).The radius of this circle is r = − + − =( ) ( ) .1 0 1 0 22 2

Question 2 (b) (i)

Question 2 (b) (ii) Question 2 (b) (iii)

zw i i i= +

+

= +2

6 62 2

323

4 56

5cos sin cos sin cos sinπ π π π π π66

4 56

150

= = =zw zw, arg π o

Question 1 (d)When you multiply complex numbers in polar form you add their arguments.

(De Moivre’s Theorem)

2

0�2 �1

y

21

1

P

�2

�1

2

x2 is the distance of (0, 0) to P on this scale.

Or (use the difference of 2 squares)

( )( )

( ) ( )

3 2 3 2

3 23 2 1

2 2

− +

= −= − =

( )

( ) ( )

3 2

3 2 3 2 2

3 2 6 2

5 2 6

2

2 2

+

= + +

= + +

= +

[ (cos sin )] (cos sin )r i r n i nn nq q q q+ = +



z r z= =, arg θ

z r i= +(cos sin )θ θz i

z z

= +

= = =

26 6

26

30

cos sin

, arg

π π

π o

Im

Re

1

2

z

4

0�1�2�3�4 1 2 3 4

�1

�2

�3

�4

3

kz

30o

izzw

30o

Question 1 (a)

kz i kz kz= +

⇒ = = =4

6 64

630cos sin , argπ π π o

w iz= = +

+

= +

+

12 2

26 6

22 6

cos sin cos sin

cos

π π π π

π π ii

i

w w

sin

cos sin

, arg

π π

π π

π

2 6

2 23

23

2 23

1

+

= +

= = = 220o

Question 1 (b)

Question 1 (c)

Multiplying by i rotates a complex number by 90o anticlockwise.

sample paper 1: paper 1

Write i as a complex number in polar form: i i i= + = +

0 1 1

2 2cos sinπ π

[When you multiply complex numbers in polar form you add their arguments.]

Higher Level, Educate.ie Sample 1, Paper 1 3



( )( )3 2 3 2

3 6 6 2 1

− +

= + − − =

3 23 2

3 23 2

3 23 2

5 2 61

5 2 6

+−

=+−

×++

=+

= +

( )( )

( )( )

[theme: All about irrationals, root 2 and root 3]

z i

iz

z

12 1212

1212

1

26 6

2 126

126

= +

= +

cos sin

cos sin

π π

π π

22

12

12

12 12

2 2 2

2 1 0 2

= +{ }

= +{ }=

cos sin

( )

π πi

iz

Question 1 (e)

Question 2 (a)Draw a circle with centre (0, 0) through the point (1, 1).The radius of this circle is r = − + − =( ) ( ) .1 0 1 0 22 2

Question 2 (b) (i)

Question 2 (b) (ii) Question 2 (b) (iii)

zw i i i= +

+

= +2

6 62 2

323

4 56

5cos sin cos sin cos sinπ π π π π π66

4 56

150

= = =zw zw, arg π o

Question 1 (d)When you multiply complex numbers in polar form you add their arguments.

(De Moivre’s Theorem)

2

0�2 �1

y

21

1

P

�2

�1

2

x2 is the distance of (0, 0) to P on this scale.

Or (use the difference of 2 squares)

( )( )

( ) ( )

3 2 3 2

3 23 2 1

2 2

− +

= −= − =

( )

( ) ( )

3 2

3 2 3 2 2

3 2 6 2

5 2 6

2

2 2

+

= + +

= + +

= +

[ (cos sin )] (cos sin )r i r n i nn nq q q q+ = +



z r z= =, arg θ

z r i= +(cos sin )θ θz i

z z

= +

= = =

26 6

26

30

cos sin

, arg

π π

π o

Im

Re

1

2

z

4

0�1�2�3�4 1 2 3 4

�1

�2

�3

�4

3

kz

30o

izzw

30o

Question 1 (a)

kz i kz kz= +

⇒ = = =4

6 64

630cos sin , argπ π π o

w iz= = +

+

= +

+

12 2

26 6

22 6

cos sin cos sin

cos

π π π π

π π ii

i

w w

sin

cos sin

, arg

π π

π π

π

2 6

2 23

23

2 23

1

+

= +

= = = 220o

Question 1 (b)

Question 1 (c)

Multiplying by i rotates a complex number by 90o anticlockwise.


Write i as a complex number in polar form: i i i= + = +

0 1 1

2 2cos sinπ π

[When you multiply complex numbers in polar form you add their arguments.]

Sample 1

Paper 1




lim lim( )

lim( )n n

nn

n

nn

nn→∞ →∞ →∞+

=+

=+

=2

3 423

23

234 4

Question 3 (a) (i)

Question 3 (a) (ii) Question 3 (a) (iii)

Question 3 (b) (i)

steps for proof by induction

1. Prove result is true for some starting value of n∈�.2. Assume result is true for n = k.3. Prove result is true for n = (k +1).

[You can use the technique shown for finding the limit or you can find the limit by inspection.]

Terms of a geometric sequence: a ar ar ar arn, , , ,............,2 3 1−

1. Prove true for n = 1: S a rr

a1

111

=−−

=( )

[Therefore, true for n = 1.]

2. Assume true for n = k: Assume S a ar ar a rrk

kk

= + + + =−−

−............. ( )1 11

3. Prove true for n = k + 1: Prove S a ar ar ar a rrk

k kk

+−

+

= + + + + =−−1

111

1............. ( )

Proof:

( ............. ) ( )

(

a ar ar ar a rr

ar

a r

k kk

k

k

+ + + + =−−

+

=−

−1 11

1 )) ( )

( )

+ −−

=− + −

−

=−−

+

+

ar rr

a ar ar arr

a rr

k k k k

k

11 1

11

1

1

lim ,n

nr r→∞

= <0 1 lim ,n

nr r→∞

= ∞ >1

Therefore, assuming true for n = k means it is true for n = k + 1. So true for n = 1 and true for n = k means it is true for n = k + 1. This implies it is true for all n∈�.

lim ( ) limn

n

n

n

→∞ →∞= = ∞

13

2 13

2lim limn

n

n

n

→∞ →∞

=

= × =

2 13

2 13

2 0 2

lim ,n

nr r→∞

= <0 1lim lim ( ) lim( ) ( limn n n

n

n

n

n

nS a rr

ar

r ar

r→∞ →∞ →∞ →∞

=−−

=−

− =−

−11 1

11

1 ))

( )=−

− =−

ar

ar1

1 01

Question 3 (b) (ii)


log log ( )

log ( )

( ) ( )

2 2

2

2 4 4 2

2

4 4

4 4

4 2 2 2 44 4

12

x x

x x

x xx x

+ + =

+ =

+ = = = =

+ − ==

=− ± − −

=− ± +

=− ±

=− ±

= − ±

>

0

4 4 4 1 42 1

4 16 162

4 322

4 4 22

2 2 2

0

2

x

x

( ) ( )( )( )

⇒⇒ = −x 2 2 2

Question 2 (c) (i)

x b b aca

=− ± −2 4

2

x xx x

2

2

2 12 1 0

− >

− − >

x x

x

2

2

2 1 0

2 2 4 1 12 1

2 4 42

2 82

2 2 22

1 2

− − =

=− − ± − − −

=± +

=±

=±

= ±

( ) ( ) ( )( )( )

x x x< − > + ∈1 2 1 2, , �

Question 2 (c) (ii)

[Solve the equality using the quadratic formula.]

SoLutionS:

Carry out the region test, or whatever test you use, to locate the regions that satisfy the inequality:

1 2+1 2−

ª 2 41.≈ −0 41.

−1 0 3

x x2

2

2 1 03 2 3 12 0

− − >

− −= >( ) ( )

( )True

x x2

2

2 1 00 2 0 1

1 0

− − >

− −= − >( ) ( )

( )False

x x2

2

2 1 01 2 1 12 0

− − >

− − − −= >( ) ( )

( )True




lim lim( )

lim( )n n

nn

n

nn

nn→∞ →∞ →∞+

=+

=+

=2

3 423

23

234 4

Question 3 (a) (i)

Question 3 (a) (ii) Question 3 (a) (iii)

Question 3 (b) (i)



[You can use the technique shown for finding the limit or you can find the limit by inspection.]

Terms of a geometric sequence: a ar ar ar arn, , , ,............,2 3 1−

1. Prove true for n = 1: S a rr

a1

111

=−−

=( )


2. Assume true for n = k: Assume S a ar ar a rrk

kk

= + + + =−−

−............. ( )1 11

3. Prove true for n = k + 1: Prove S a ar ar ar a rrk

k kk

+−

+

= + + + + =−−1

111

1............. ( )

Proof:

( ............. ) ( )

(

a ar ar ar a rr

ar

a r

k kk

k

k

+ + + + =−−

+

=−

−1 11

1 )) ( )

( )

+ −−

=− + −

−

=−−

+

+

ar rr

a ar ar arr

a rr

k k k k

k

11 1

11

1

1

lim ,n

nr r→∞

= <0 1 lim ,n

nr r→∞

= ∞ >1


lim ( ) limn

n

n

n

→∞ →∞= = ∞

13

2 13

2lim limn

n

n

n

→∞ →∞

=

= × =

2 13

2 13

2 0 2

lim ,n

nr r→∞

= <0 1lim lim ( ) lim( ) ( limn n n

n

n

n

n

nS a rr

ar

r ar

r→∞ →∞ →∞ →∞

=−−

=−

− =−

−11 1

11

1 ))

( )=−

− =−

ar

ar1

1 01

Question 3 (b) (ii)


log log ( )

log ( )

( ) ( )

2 2

2

2 4 4 2

2

4 4

4 4

4 2 2 2 44 4

12

x x

x x

x xx x

+ + =

+ =

+ = = = =

+ − ==

=− ± − −

=− ± +

=− ±

=− ±

= − ±

>

0

4 4 4 1 42 1

4 16 162

4 322

4 4 22

2 2 2

0

2

x

x

( ) ( )( )( )

⇒⇒ = −x 2 2 2

Question 2 (c) (i)

x b b aca

=− ± −2 4

2

x xx x

2

2

2 12 1 0

− >

− − >

x x

x

2

2

2 1 0

2 2 4 1 12 1

2 4 42

2 82

2 2 22

1 2

− − =

=− − ± − − −

=± +

=±

=±

= ±

( ) ( ) ( )( )( )

x x x< − > + ∈1 2 1 2, , �

Question 2 (c) (ii)

[Solve the equality using the quadratic formula.]

SoLutionS:


1 2+1 2−

ª 2 41.≈ −0 41.

−1 0 3

x x2

2

2 1 03 2 3 12 0

− − >

− −= >( ) ( )

( )True

x x2

2

2 1 00 2 0 1

1 0

− − >

− −= − >( ) ( )

( )False

x x2

2

2 1 01 2 1 12 0

− − >

− − − −= >( ) ( )

( )True

Sample 1

Paper 1



Question 5 (25 marks)Question 5 (a)

A

x

yB

′ = ⇒ + − =∴ = −f x x xx( ) ( )( )

,0 6 2 1 0

2 1

f x x xf A

( ) ( ) ( )( ) ( ) ( ( )) ( , )

= + −

− = − + − − = ⇒ −

2 5 22 2 2 5 2 2 0 2 0

2

2 is the locaal minimum.is the locf B( ) ( ) ( ( )) ( )( ) ( , )1 1 2 5 2 1 9 3 27 1 272= + − = = ⇒ aal maximum.

′′ = ⇒ − − =∴ = −

= + −

− = − + −

f x xx

f x x xf

( )

( ) ( ) ( )( ) ( ) (

0 12 6 0

2 5 22 5

12

2

12

12

2 22

5 16

12

32

2

94

272

12

272

( ))

( ) ( )( )( )

( , )

−

= +

= =

∴ −C is point of infleection.

C

x

yB

A

Question 5 (b)

f x x xf x x x x

x x

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

( )[ (

= + −

′ = + − + − += + −

2 5 22 2 5 2 2 2

2 2

2

2 1

++ + −= + − − + −= + − += + −

′′

2 5 22 2 2 5 22 2 3 36 2 1

) ( )]( )[ ]( )[ ]( )( )

xx x xx xx x

ff x x xx xxx

( ) {( )( ) ( )( )}{ }{ }

( )

= + − + −= − − + −= − −= − +

6 2 1 1 16 2 16 2 1

6 2 1

to find turning pointS (tp):

Put dydx

f x= ′ =( ) 0 and solve for x.

d ydx

d ydx

2

2

2

2

0

0

< ⇒

> ⇒

TP

TP

Local Maximum

Local Minimum

′′ − = − − = > ⇒′′ = − = − > ⇒

ff

( ) ( )( ) ( )

2 6 3 18 01 6 3 18 0

Local MinimumLocal Maxximum

to find point of infLeCtion:

Put d ydx

f x2

2 0= ′′ =( ) and solve for x.


Question 3 (c)

a r= =

231000

1100

,


kx x bx x x kxkx x bx x x kxk

3 2

3 2 2

6 6 1 3 26 6 2 3 2

− + − = + − +

− + − = − − +

( )( )( )( )( )

xx x bx kx x kx x kxkx x bx kx k x

3 2 3 2 2

3 2 3

6 6 2 2 4 3 66 6 2 2

− + − = + − − − −

− + − = + −( ) 22 4 3 6+ − − −( )k x

− = −=

∴ =

6 2 22 8

4

kkk

b kb= − −= − − = −

4 34 3 4 16( )

( )( )( ), ,

x x xx+ − + =

∴ = − −1 3 4 2 0

1 312

xx

xx

x x

x x xx x x

+>

++ > +

+ − + >+ − + >

12

11 2 1

1 2 1 01 2 1 0

2 2

2

( ) ( )

( ) ( )( )[ ( )](xx x xx x+ − − >+ − − >

1 2 2 01 2 0)[ ]

( )[ ]

( )( ),

x xx+ − − =

∴ = − −1 2 0

2 1

Question 4 (a)

Question 4 (b)

[A cubic equation equals a linear factor by a linear factor by a linear factor.The coefficients of x in each linear factor must multiply together to give k and the constants in each linear factor must multiply together to give −6. Find k and b by lining up the coefficients.]

[Multiply across by (x + 1)2. This is a positive value and so will not reverse the inequality.]

Solve the equality:

SoLutionS:


−20

−1−3

( )( )( )( ) ( )x x+ − − >− >

1 2 02 1 0 False

− 32

( )( )( )( ) ( )x x+ − − >− >1 2 0

1 2 0 False( )( )( )( ) ( )x x+ − − >− − >

1 2 001

212 True

− < < − ∈2 1x x, �

0 523 0 5232323 12

231000

23100 000

. . ....... ..........⋅ ⋅

= = + + +

= +−

= +

= +

=

12 11212

23990

259495

231000

1100

231000

99100




A

x

yB

′ = ⇒ + − =∴ = −f x x xx( ) ( )( )

,0 6 2 1 0

2 1

f x x xf A

( ) ( ) ( )( ) ( ) ( ( )) ( , )

= + −

− = − + − − = ⇒ −

2 5 22 2 2 5 2 2 0 2 0

2

2 is the locaal minimum.is the locf B( ) ( ) ( ( )) ( )( ) ( , )1 1 2 5 2 1 9 3 27 1 272= + − = = ⇒ aal maximum.

′′ = ⇒ − − =∴ = −

= + −

− = − + −

f x xx

f x x xf

( )

( ) ( ) ( )( ) ( ) (

0 12 6 0

2 5 22 5

12

2

12

12

2 22

5 16

12

32

2

94

272

12

272

( ))

( ) ( )( )( )

( , )

−

= +

= =

∴ −C is point of infleection.

C

x

yB

A

Question 5 (b)

f x x xf x x x x

x x

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

( )[ (

= + −

′ = + − + − += + −

2 5 22 2 5 2 2 2

2 2

2

2 1

++ + −= + − − + −= + − += + −

′′

2 5 22 2 2 5 22 2 3 36 2 1

) ( )]( )[ ]( )[ ]( )( )

xx x xx xx x

ff x x xx xxx

( ) {( )( ) ( )( )}{ }{ }

( )

= + − + −= − − + −= − −= − +

6 2 1 1 16 2 16 2 1

6 2 1

to find turning pointS (tp):

Put dydx

f x= ′ =( ) 0 and solve for x.

d ydx

d ydx

2

2

2

2

0

0

< ⇒

> ⇒

TP

TP

Local Maximum

Local Minimum

′′ − = − − = > ⇒′′ = − = − > ⇒

ff

( ) ( )( ) ( )

2 6 3 18 01 6 3 18 0

Local MinimumLocal Maxximum

to find point of infLeCtion:

Put d ydx

f x2

2 0= ′′ =( ) and solve for x.


Question 3 (c)

a r= =

231000

1100

,


kx x bx x x kxkx x bx x x kxk

3 2

3 2 2

6 6 1 3 26 6 2 3 2

− + − = + − +

− + − = − − +

( )( )( )( )( )

xx x bx kx x kx x kxkx x bx kx k x

3 2 3 2 2

3 2 3

6 6 2 2 4 3 66 6 2 2

− + − = + − − − −

− + − = + −( ) 22 4 3 6+ − − −( )k x

− = −=

∴ =

6 2 22 8

4

kkk

b kb= − −= − − = −

4 34 3 4 16( )

( )( )( ), ,

x x xx+ − + =

∴ = − −1 3 4 2 0

1 312

xx

xx

x x

x x xx x x

+>

++ > +

+ − + >+ − + >

12

11 2 1

1 2 1 01 2 1 0

2 2

2

( ) ( )

( ) ( )( )[ ( )](xx x xx x+ − − >+ − − >

1 2 2 01 2 0)[ ]

( )[ ]

( )( ),

x xx+ − − =

∴ = − −1 2 0

2 1

Question 4 (a)

Question 4 (b)

[A cubic equation equals a linear factor by a linear factor by a linear factor.The coefficients of x in each linear factor must multiply together to give k and the constants in each linear factor must multiply together to give −6. Find k and b by lining up the coefficients.]

[Multiply across by (x + 1)2. This is a positive value and so will not reverse the inequality.]

Solve the equality:

SoLutionS:


−20

−1−3

( )( )( )( ) ( )x x+ − − >− >

1 2 02 1 0 False

− 32

( )( )( )( ) ( )x x+ − − >− >1 2 0

1 2 0 False( )( )( )( ) ( )x x+ − − >− − >

1 2 001

212 True

− < < − ∈2 1x x, �

0 523 0 5232323 12

231000

23100 000

. . ....... ..........⋅ ⋅

= = + + +

= +−

= +

= +

=

12 11212

23990

259495

231000

1100

231000

99100

Sample 1

Paper 1



′ = − +

= ′ = − +

= − + +∫ ∫

f x x x

f x f x dx x x dx

x x x c

f

( )

( ) ( ) ( )

( )

9 4 5

9 4 5

3 2 5

1

2

2

3 2

==

= − + + =− + + =

∴ = −

∴ = − +

11 3 1 2 1 5 1 1

3 2 5 15

3 2 5

3 2

3 2

f cc

cf x x x x

( ) ( ) ( ) ( )

( ) −−5

Question 6 (b)


Q aet Q ae at Q a

a ae

e

bt

b

b

=

= = == =

∴ =

=

05730

0

12

12

5730

12

5730

::

( )

(

years

))

ln( ) ( )ln

ln

12

1

57302 5730

25730

=

− =

∴ = −

−

bb

b yr

a is the initial amount of Carbon-14

Qa

e= = =−

ln

. . %2

57302000

0 785 78 5

Question 7 (a)

Question 7 (b)

Question 7 (c) (i)

[At t = 0 years the amount of Carbon-14 present is a. After a half-life of 5730 years, the amount of Carbon-14 present will fall by a half of a.]

[Q divided by a (the initial amount) will give you the fraction of Carbon-14 present at any time.]

Q ae dQdt

abe

dQdtdQdt

abe

bt bt

t

t

= ⇒ =

==

=

−

4000

1000

25ln7730

4000

25730

1000

25730

4000

−

−

−=

abe

e

eln

ln

lnn

ln

.25730

1000

25730

30000 696

−

= =e


m x yy xy xy

= = −

− = +

− = +

− =

272 1 1

12

272

272

272

12

272

272

274

4 54

, ( , ) ( , )( )

554 2754 4 81 0

xx y

+− + =

Question 5 (c) Question 5 (d)

A x x dx

x x x dx

x x

= + −

= + + −

= − − +

−

−

∫

∫

(( ) ( ))

(( )( ))

(

2 5 2

4 4 5 2

2 3 1

2

2

0

2

2

0

3 2 22 20

24

33

122

20

6 2

2

0

4 3 2

2

0

12

4 3 2

x dx

x x x x

x x x

+

= − − + +

= − − + +

−

−

∫ )

00

0 2 2 6 2 20 28 8 24 40

16

2

0

12

4 3 2

x = − − − − − + − + −

= + − − +=

−

{ ( ) ( ) ( ) ( )}

Question 6 (25 marks)Question 6 (a) (i)

sin

( cos )

( sin )

2

12

12

12

1 2

2

x dx

x dx

x x c

∫∫= −

= − +

3

4 1

2

0

2

32

20

2

32

2 2 0

32

2 0

32

2

e dx

e

e e

e e

x

x

ln

ln

ln

ln

[ ]

( )

( )( )

∫=

= −

= −

= −

= 992

x x x x3 21 1 1+ = + − +( )( )

xx

dx

x x xx

dx

x x dx

x x x c

3

2

2

13

3 12

2

111 1

1

1

++

=+ − +

+

= − +

= − + +

∫

∫

∫

( )( )( )

( )

2

2

2 22

22

2 2

0

1

0

1

0

1

1 0

12

12

12

12

x

x

x

dx

dx

∫

∫=

=×

= −

=

ln

ln{ }( ) ( )

222

2 1ln

( )−

Question 6 (a) (ii)

Question 6 (a) (iii)

Question 6 (a) (iv)

Question 6 (a) (v)

xxdx

xxdx

x x dx

x x

+

= +

= +

= +

=

∫

∫

∫ −

1

1

2

1

9

1

9

1

9

23 1

9

12

12

32

12

( )

[ ]

{223

23

23

23

9 2 9 1 2 127 2 3 1 2 1

1

32

12

32

12( ) ( ) } { ( ) ( ) }

( ) ( ) ( ) ( )+ − +

= + − −

= 88 6 223

643

+ − −

=



′ = − +

= ′ = − +

= − + +∫ ∫

f x x x

f x f x dx x x dx

x x x c

f

( )

( ) ( ) ( )

( )

9 4 5

9 4 5

3 2 5

1

2

2

3 2

==

= − + + =− + + =

∴ = −

∴ = − +

11 3 1 2 1 5 1 1

3 2 5 15

3 2 5

3 2

3 2

f cc

cf x x x x

( ) ( ) ( ) ( )

( ) −−5

Question 6 (b)


Q aet Q ae at Q a

a ae

e

bt

b

b

=

= = == =

∴ =

=

05730

0

12

12

5730

12

5730

::

( )

(

years

))

ln( ) ( )ln

ln

12

1

57302 5730

25730

=

− =

∴ = −

−

bb

b yr

a is the initial amount of Carbon-14

Qa

e= = =−

ln

. . %2

57302000

0 785 78 5

Question 7 (a)

Question 7 (b)

Question 7 (c) (i)

[At t = 0 years the amount of Carbon-14 present is a. After a half-life of 5730 years, the amount of Carbon-14 present will fall by a half of a.]

[Q divided by a (the initial amount) will give you the fraction of Carbon-14 present at any time.]

Q ae dQdt

abe

dQdtdQdt

abe

bt bt

t

t

= ⇒ =

==

=

−

4000

1000

25ln7730

4000

25730

1000

25730

4000

−

−

−=

abe

e

eln

ln

lnn

ln

.25730

1000

25730

30000 696

−

= =e


m x yy xy xy

= = −

− = +

− = +

− =

272 1 1

12

272

272

272

12

272

272

274

4 54

, ( , ) ( , )( )

554 2754 4 81 0

xx y

+− + =


A x x dx

x x x dx

x x

= + −

= + + −

= − − +

−

−

∫

∫

(( ) ( ))

(( )( ))

(

2 5 2

4 4 5 2

2 3 1

2

2

0

2

2

0

3 2 22 20

24

33

122

20

6 2

2

0

4 3 2

2

0

12

4 3 2

x dx

x x x x

x x x

+

= − − + +

= − − + +

−

−

∫ )

00

0 2 2 6 2 20 28 8 24 40

16

2

0

12

4 3 2

x = − − − − − + − + −

= + − − +=

−

{ ( ) ( ) ( ) ( )}


sin

( cos )

( sin )

2

12

12

12

1 2

2

x dx

x dx

x x c

∫∫= −

= − +

3

4 1

2

0

2

32

20

2

32

2 2 0

32

2 0

32

2

e dx

e

e e

e e

x

x

ln

ln

ln

ln

[ ]

( )

( )( )

∫=

= −

= −

= −

= 992

x x x x3 21 1 1+ = + − +( )( )

xx

dx

x x xx

dx

x x dx

x x x c

3

2

2

13

3 12

2

111 1

1

1

++

=+ − +

+

= − +

= − + +

∫

∫

∫

( )( )( )

( )

2

2

2 22

22

2 2

0

1

0

1

0

1

1 0

12

12

12

12

x

x

x

dx

dx

∫

∫=

=×

= −

=

ln

ln{ }( ) ( )

222

2 1ln

( )−

Question 6 (a) (ii)


Question 6 (a) (iv)

Question 6 (a) (v)

xxdx

xxdx

x x dx

x x

+

= +

= +

= +

=

∫

∫

∫ −

1

1

2

1

9

1

9

1

9

23 1

9

12

12

32

12

( )

[ ]

{223

23

23

23

9 2 9 1 2 127 2 3 1 2 1

1

32

12

32

12( ) ( ) } { ( ) ( ) }

( ) ( ) ( ) ( )+ − +

= + − −

= 88 6 223

643

+ − −

=

Sample 1

Paper 1



Q aeQ Q aeQ Q ae

Q aeQ ae

bt

bt

bt

bt

bt

=

− =

+ =

=

=

0 050 05

0 951 050 95

1

2

1

2

.

.... QQQ

ee

e

b t t

t t

bt

btb t t

1 050 951 05

1

2

1 2

1 2

1 2

.

ln ..

( )

(

( )= =

∴

= −

−

−

))ln .

.ln .

.ln

=

=

−

=

0 951 05

0 951 05

25730

827b

yearrs

Question 7 (f)Q aeQ aeQ a eQ bt a y mx c

bt

bt

bt

=

=

= += + → = +

ln ln( )ln ln ln( )ln ln ( )b is the slope of the graph and ln a is the y-intercept.

b

b

=−−

= − ×

= − = − ×

− −

−

1 08 0 842500 4500

1 2 10

25730

1 21 10

4 1

4

. . .

ln .

year

yeear−1

Question 7 (e)

ln . ..a a e= ⇒ = =1 38 3 9751 38 mg


x pp x

p xx

x

= −= −

= − ≥∴ ≥≤

20 000 100100 20 000

200 0 01 020 000

20 000

.

C xdCdx

= +

= =

120 000 40

40 constant

It costs 40 to produce one more phone at any level of production.

R x pp xR x x x x

= ×= −

∴ = − = −

200 0 01200 0 01 200 0 01 2

.( . ) .

R x xx x

x x

= − ≥

− ≥− ≥

200 0 01 020 000 0

20 000 0

2

2

.

( )

Question 8 (a) Question 8 (b) (i)

Question 8 (b) (ii) Question 8 (c) (i)

Question 8 (c) (ii) Question 8 (d)

Solve the equality:

SoLution: 0 20 000£ £x

R x xdRdx

x

dRdx

x

xx

= −

= −

= ⇒ − =

=∴ =

200 0 01

200 0 02

0 200 0 02 0

200 0 021

2.

.

.

.00 000200 10 000 0 01 10 000 1 000 0002RMax. = − =( ) . ( )

x xx

( ),

20 000 00 20 000

− =∴ =


Qa

e

t

t

t= =

= −

∴ = −

−

0 93

0 93 25730

5730

25730.

ln( . ) ln

ln

ln

(( . )ln

0 932

600= years

Question 7 (c) (ii)

t (years) 1000 2000 3000 4000 5000 6000

Q (mg) 3.54 3.14 2.78 2.47 2.18 1.94

ln Q 1.26 1.14 1.02 0.90 0.78 0.66

Question 7 (d) (i)

1000 2000 3000 4000 5000 6000

t (years)

0

0.40

0.20

0.60

1.20

ln Q

0.80

1.00

1.40

(4500, 0.84)

(2500, 1.08)

Question 7 (d) (ii)



Q aeQ Q aeQ Q ae

Q aeQ ae

bt

bt

bt

bt

bt

=

− =

+ =

=

=

0 050 05

0 951 050 95

1

2

1

2

.

.... QQQ

ee

e

b t t

t t

bt

btb t t

1 050 951 05

1

2

1 2

1 2

1 2

.

ln ..

( )

(

( )= =

∴

= −

−

−

))ln .

.ln .

.ln

=

=

−

=

0 951 05

0 951 05

25730

827b

yearrs

Question 7 (f)Q aeQ aeQ a eQ bt a y mx c

bt

bt

bt

=

=

= += + → = +

ln ln( )ln ln ln( )ln ln ( )b is the slope of the graph and ln a is the y-intercept.

b

b

=−−

= − ×

= − = − ×

− −

−

1 08 0 842500 4500

1 2 10

25730

1 21 10

4 1

4

. . .

ln .

year

yeear−1

Question 7 (e)

ln . ..a a e= ⇒ = =1 38 3 9751 38 mg


x pp x

p xx

x

= −= −

= − ≥∴ ≥≤

20 000 100100 20 000

200 0 01 020 000

20 000

.

C xdCdx

= +

= =

120 000 40

40 constant

It costs 40 to produce one more phone at any level of production.

R x pp xR x x x x

= ×= −

∴ = − = −

200 0 01200 0 01 200 0 01 2

.( . ) .

R x xx x

x x

= − ≥

− ≥− ≥

200 0 01 020 000 0

20 000 0

2

2

.

( )

Question 8 (a) Question 8 (b) (i)

Question 8 (b) (ii) Question 8 (c) (i)

Question 8 (c) (ii) Question 8 (d)

Solve the equality:

SoLution: 0 20 000£ £x

R x xdRdx

x

dRdx

x

xx

= −

= −

= ⇒ − =

=∴ =

200 0 01

200 0 02

0 200 0 02 0

200 0 021

2.

.

.

.00 000200 10 000 0 01 10 000 1 000 0002RMax. = − =( ) . ( )

x xx

( ),

20 000 00 20 000

− =∴ =


Qa

e

t

t

t= =

= −

∴ = −

−

0 93

0 93 25730

5730

25730.

ln( . ) ln

ln

ln

(( . )ln

0 932

600= years

Question 7 (c) (ii)

t (years) 1000 2000 3000 4000 5000 6000

Q (mg) 3.54 3.14 2.78 2.47 2.18 1.94

ln Q 1.26 1.14 1.02 0.90 0.78 0.66

Question 7 (d) (i)

1000 2000 3000 4000 5000 6000

t (years)

0

0.40

0.20

0.60

1.20

ln Q

0.80

1.00

1.40

(4500, 0.84)

(2500, 1.08)

Question 7 (d) (ii)

Sample 1

Paper 1



Question 8 (g)

x b b aca

=− ± −2 4

2

Breakeven points from graph: (780, 145 000), (15 200, 720 000)

CaLCuLuS:

grapH: The maximum profit is the greatest distance between the R and C graphs.PMax. = 960 000 − 440 000 = 520 000

xp xp

== −

∴ = − =

8 000200 0 01

200 0 01 8000 120

,.

. ( )

Question 8 (h) (i)

Question 8 (h) (ii)

R Cx x x

x x

x

=

− = +

− + =

=− − ±

200 0 01 120 000 4016 000 12 000 000 0

16 000

2

2

.

( ) (−− −

∴ =

16 000 4 12 000 0002

789 15 211

2) ( )

,x

P R C x x xx x

dPdx

= − = − − −

= − + −

= −

200 0 01 120 000 400 01 160 120 000

0 02

2

2

..

. xx

dPdx

x

xx

P

+

= ⇒ − + =

=∴ =

= −

160

0 0 02 160 0

160 0 028000

0 01 8000

.

.

. ( )Max.22 160 8000 120 000 520 000+ − =( )


x R () C ()

0 0 120 000

4000 640 000 280 000

8000 960 000 440 000

12 000 960 000 600 000

16 000 640 000 760 000

20 000 0 920 000

Question 8 (e) (i)

20000180001200010000

(15200, 720000)

60004000

600 000

1600014000 x0

500 000

800 000

700 000

200 000

100 000

400 000

300 000

1 000 000

900 000

C

R

(euro)

(euro)

2000

Profit Loss

8000

(780, 145000)

Maximum

Profit

960 000

440 000

Question 8 (e) (ii)

Question 8 (f) (See graph above)



Question 8 (g)

x b b aca

=− ± −2 4

2

Breakeven points from graph: (780, 145 000), (15 200, 720 000)

CaLCuLuS:

grapH: The maximum profit is the greatest distance between the R and C graphs.PMax. = 960 000 − 440 000 = 520 000

xp xp

== −

∴ = − =

8 000200 0 01

200 0 01 8000 120

,.

. ( )

Question 8 (h) (i)

Question 8 (h) (ii)

R Cx x x

x x

x

=

− = +

− + =

=− − ±

200 0 01 120 000 4016 000 12 000 000 0

16 000

2

2

.

( ) (−− −

∴ =

16 000 4 12 000 0002

789 15 211

2) ( )

,x

P R C x x xx x

dPdx

= − = − − −

= − + −

= −

200 0 01 120 000 400 01 160 120 000

0 02

2

2

..

. xx

dPdx

x

xx

P

+

= ⇒ − + =

=∴ =

= −

160

0 0 02 160 0

160 0 028000

0 01 8000

.

.

. ( )Max.22 160 8000 120 000 520 000+ − =( )


x R () C ()

0 0 120 000

4000 640 000 280 000

8000 960 000 440 000

12 000 960 000 600 000

16 000 640 000 760 000

20 000 0 920 000

Question 8 (e) (i)

20000180001200010000

(15200, 720000)

60004000

600 000

1600014000 x0

500 000

800 000

700 000

200 000

100 000

400 000

300 000

1 000 000

900 000

C

R

(euro)

(euro)

2000

Profit Loss

8000

(780, 145000)

Maximum

Profit

960 000

440 000

Question 8 (e) (ii)

Question 8 (f) (See graph above)

Sample 1

Paper 1



t (days) 0p6

p3

p2

23p 5

6p

p 1000 1000e 10001000

e1000 1000e

p 1000 2700 1000 370 1000 2700

2000

3

3000

4000

�

Period

1000

0

6

�

2

�

3

2�

6

5�

t (days)

p

Range

Question 9 (f) (i)

Period days= =23

2 1p .

Question 9 (f) (ii)

Range =

1000 1000e

e,

This represents the population in a cycle from its maximum to minimum values.

This is the time for the population to return to its original value at any point in the cycle.



dpdt

mp mt

dpp

m mt dt

p m mtm

c

p mt ct p N

=

=

= +

= += =

Ú Ú

cos

cos

ln sin

ln sin, :0 lln sin ln

ln sin ln

ln sin

sin

N c c Np mt NpN

mt

p Ne mt

= + fi == +

ÊËÁ

ˆ¯̃

=

=

0

p Nep N N Ne

tt

t

t

== =

== = =-

sin

sin:sin ln

sin (ln ) .

3

3

13

1

2 23 2

2 0 255 days 3368 minutes

dpdt

kp

dpp

k dt

p kt ct p N N c c Np kt N

=

=

= += = = + fi == +

Ú Úln

, : ln lnln lnl

At 0 0

nn ln

ln

p N ktpN

kt

pN

e

p Ne

kt

kt

- =

ÊËÁ

ˆ¯̃

=

=

\ =

p Nep N N Neet

t

t

t

t

== fi ==

=

= = =

3

3

3

2 22

2 32

30 231 333

lnln . days minutes

cos sinmxdx mx cmÚ = +1

Question 9 (a)

[Separate the variables.]

Question 9 (b)

Question 9 (c)

Question 9 (d)

Question 9 (e)

Question 9 (f)

p et p e e

t p e

t=

= = = =

= =

10000 1000 1000 1000

61000

3

3 0 0

3

sin

sin ( )

sin

:

:p (( )

sin ( ):

:

16

13

1000 2700

31000 1000 1000

21

3 0

p

pp

p

= ≈

= = = =

= =

e

t p e e

t p 0000 1000 1000 370

23

1000 100

3 1

3

12

23

e ee

t p e

sin ( )

sin ( ):

p

pp

= = ≈

= = =

−

00 1000

56

1000 1000 2700

0

3 56

e

t p e e

=

= = = ≈p p: sin ( )



t (days) 0p6

p3

p2

23p 5

6p

p 1000 1000e 10001000

e1000 1000e

p 1000 2700 1000 370 1000 2700

2000

3

3000

4000

�

Period

1000

0

6

�

2

�

3

2�

6

5�

t (days)

p

Range

Question 9 (f) (i)

Period days= =23

2 1p .

Question 9 (f) (ii)

Range =

1000 1000e

e,

This represents the population in a cycle from its maximum to minimum values.

This is the time for the population to return to its original value at any point in the cycle.



dpdt

mp mt

dpp

m mt dt

p m mtm

c

p mt ct p N

=

=

= +

= += =

Ú Ú

cos

cos

ln sin

ln sin, :0 lln sin ln

ln sin ln

ln sin

sin

N c c Np mt NpN

mt

p Ne mt

= + fi == +

ÊËÁ

ˆ¯̃

=

=

0

p Nep N N Ne

tt

t

t

== =

== = =-

sin

sin:sin ln

sin (ln ) .

3

3

13

1

2 23 2

2 0 255 days 3368 minutes

dpdt

kp

dpp

k dt

p kt ct p N N c c Np kt N

=

=

= += = = + fi == +

Ú Úln

, : ln lnln lnl

At 0 0

nn ln

ln

p N ktpN

kt

pN

e

p Ne

kt

kt

- =

ÊËÁ

ˆ¯̃

=

=

\ =

p Nep N N Neet

t

t

t

t

== fi ==

=

= = =

3

3

3

2 22

2 32

30 231 333

lnln . days minutes

cos sinmxdx mx cmÚ = +1

Question 9 (a)

[Separate the variables.]

Question 9 (b)

Question 9 (c)

Question 9 (d)

Question 9 (e)

Question 9 (f)

p et p e e

t p e

t=

= = = =

= =

10000 1000 1000 1000

61000

3

3 0 0

3

sin

sin ( )

sin

:

:p (( )

sin ( ):

:

16

13

1000 2700

31000 1000 1000

21

3 0

p

pp

p

= ≈

= = = =

= =

e

t p e e

t p 0000 1000 1000 370

23

1000 100

3 1

3

12

23

e ee

t p e

sin ( )

sin ( ):

p

pp

= = ≈

= = =

−

00 1000

56

1000 1000 2700

0

3 56

e

t p e e

=

= = = ≈p p: sin ( )

Sample 1

Paper 1



s x yx y

r

s x y y

12 2

2 2

22 2

9 09

0 0 3

6 8 0

:

( , ),

:(

+ − =

+ ==

+ − + =−

Centre

Centre gg f

r g f c

, ) ( , )− =

= + − = + − = =

0 3

0 9 8 1 12 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

h k r h k r

h k r h

− + − = + ⇒ + − = +

− + − = + ⇒

0 3 1 3 1

0 0 3

2 2 2 2 2

2 2 2 ++ = +

− − = + − +

− + − − = + + +

k r

k k r r

k k k k r r

2 2

2 2 2 2

3

3 1 3

3 3 1 3

( )

( ) ( ) ( )

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

r rk rk rr k

+ − −− − = + −

− + = − −+ =

1 32 3 3 2 4 26 9 4 8

4 17 6


Question 2 (a)

Question 2 (b)

a b a b a b2 2− = − +( )( )

a b a b a b2 2− = − +( )( )

(0, 0)

r1

r

3

(0, 3)( , )h k

s

s1

s2

h k r

r k r k

h k k

h k k

2 2 2

2 22

2 2

3

4 17 6 6 174

6 174

3

6

+ = +

+ = ⇒ =−

+ =−

+

+ =

( )

−− +

=

−

=−

−

= − −

17 124

6 54

6 516

16 6 5 16

2 2

22

2

2 2

k

h k k

h k

( )

( ) kkh k k k kh k kh k

2

2

2

2

16 6 5 4 6 5 416 10 5 2 5

16 5 2 1

= − + − −

= − −

∴ = −

( )( )( )( )

( )(22 5k − )


A(5, 10)

B( 3, 4)�

C( 1, 2)� �

A B

m

( , ), ( , )

, ( , )

5 10 3 45 3

210 4

21 7

4 103 5

−

=− +

=

=−

− −=

Mid-point

−−−

= ⇒ = −

− = − −

− = − −− = − +

⊥68

34

43

7 13 7 4 13 21 4 44

43

m

k y xy xy xx

: ( ) ( )( ) ( )

++ − =3 25 0y

A C

m

( , ), ( , )

, ( , )

5 10 1 25 1

210 2

22 4

2 101

− −

=− −

=

=− −− −

Mid-point

55126

2 12

4 22 4 1 22 8 2

12

=−−

= ⇒ = −

− = − −

− = − −− = − ++

⊥m

l y xy xy xx

: ( ) ( )( ) ( )

22 10 0y − =

4 3 25 02 10 0 4

4 3 25 04 8 40 0

5 15 0 3

x yx y

x yx y

y y

x

+ − =+ − = ×−

+ − =− − + =

− + = ⇒ =

( )

++ − =+ − =

∴ =

2 3 10 06 10 0

4

( )xx

Circumcentre D(4, 3)r AD= = − + − = + = =( ) ( )5 4 10 3 1 49 50 5 22 2

Equation of circumcircle: ( ) ( ) ( )( ) ( )x yx y− + − =

− + − =

4 3 504 3 50

2 2 2

2 2



Question 1 (a)

[Solve the equations simultaneously for k and l to find their point of intersection D.]

Question 1 (b)

17Higher Level, Educate.ie Sample 1, Paper 2


s x yx y

r

s x y y

12 2

2 2

22 2

9 09

0 0 3

6 8 0

:

( , ),

:(

+ − =

+ ==

+ − + =−

Centre

Centre gg f

r g f c

, ) ( , )− =

= + − = + − = =

0 3

0 9 8 1 12 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

h k r h k r

h k r h

− + − = + ⇒ + − = +

− + − = + ⇒

0 3 1 3 1

0 0 3

2 2 2 2 2

2 2 2 ++ = +

− − = + − +

− + − − = + + +

k r

k k r r

k k k k r r

2 2

2 2 2 2

3

3 1 3

3 3 1 3

( )

( ) ( ) ( )

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

r rk rk rr k

+ − −− − = + −

− + = − −+ =

1 32 3 3 2 4 26 9 4 8

4 17 6


Question 2 (a)

Question 2 (b)

a b a b a b2 2− = − +( )( )

a b a b a b2 2− = − +( )( )

(0, 0)

r1

r

3

(0, 3)( , )h k

s

s1

s2

h k r

r k r k

h k k

h k k

2 2 2

2 22

2 2

3

4 17 6 6 174

6 174

3

6

+ = +

+ = ⇒ =−

+ =−

+

+ =

( )

−− +

=

−

=−

−

= − −

17 124

6 54

6 516

16 6 5 16

2 2

22

2

2 2

k

h k k

h k

( )

( ) kkh k k k kh k kh k

2

2

2

2

16 6 5 4 6 5 416 10 5 2 5

16 5 2 1

= − + − −

= − −

∴ = −

( )( )( )( )

( )(22 5k − )


A(5, 10)

B( 3, 4)�

C( 1, 2)� �

A B

m

( , ), ( , )

, ( , )

5 10 3 45 3

210 4

21 7

4 103 5

−

=− +

=

=−

− −=

Mid-point

−−−

= ⇒ = −

− = − −

− = − −− = − +

⊥68

34

43

7 13 7 4 13 21 4 44

43

m

k y xy xy xx

: ( ) ( )( ) ( )

++ − =3 25 0y

A C

m

( , ), ( , )

, ( , )

5 10 1 25 1

210 2

22 4

2 101

− −

=− −

=

=− −− −

Mid-point

55126

2 12

4 22 4 1 22 8 2

12

=−−

= ⇒ = −

− = − −

− = − −− = − ++

⊥m

l y xy xy xx

: ( ) ( )( ) ( )

22 10 0y − =

4 3 25 02 10 0 4

4 3 25 04 8 40 0

5 15 0 3

x yx y

x yx y

y y

x

+ − =+ − = ×−

+ − =− − + =

− + = ⇒ =

( )

++ − =+ − =

∴ =

2 3 10 06 10 0

4

( )xx

Circumcentre D(4, 3)r AD= = − + − = + = =( ) ( )5 4 10 3 1 49 50 5 22 2

Equation of circumcircle: ( ) ( ) ( )( ) ( )x yx y− + − =

− + − =

4 3 504 3 50

2 2 2

2 2



Question 1 (a)

[Solve the equations simultaneously for k and l to find their point of intersection D.]

Question 1 (b)

Sample 1

Paper 2



z x

xx

= − =−

− = −∴ =

0 68 104

2 72 107 28

.

.. ( )Lower Quartile

Interquartile range = UQ − LQ = 12.72 − 7.28 = 5.44

146 µ = 10

68%

Standard deviation = 4Median = 10 x z

x z

x z

= =−

=

= =−

=

= =−

= −

10 10 104

0

14 14 104

1

6 6 104

1

:

:

:

The standard normal distribution has a mean of 0 and standard deviation of 1 and has the same shape and area distribution as the original normal distribution.

P z Z

z x

xx

( ) .

.

.. ( )

< =

∴ = =−

= −∴ =

0 75

0 68 104

2 72 1012 72 Upper Quartile


empiriCaL ruLe: In any normal distribution with mean x and standard deviation s.1. 68% of the data falls within 1s of the mean x.2. 95% of the data falls within 2s of the mean x.3. 99.7% of the data falls within 3s of the mean x.

Question 4 (a) (i) Question 4 (a) (ii)

z x=

− µσ

Question 4 (b)

Question 4 (c) (i) Question 4 (c) (ii)

Question 4 (c) (iii)

UQ = Z

75%

zz Z=LQ = �Z

25%


AS

CT

xo –90 –60 –45 –30 0 30 45 60 90y –∞ –1.7 –1 –0.6 0 0.6 1 1.7 ∞

xo 105 120 135 150 180 210 225 240 270y –3.7 –1.7 –1 –0.6 0 0.6 1 1.7 –∞

�30o

�90o

5

0o

30o

60o

90o

270o

150o

240o

210o

180o

120o

4

3

�1

2

�5

�3

�2

�4

1

x

y

�60o

Yes, because each value of x only gives one value of y.

AS

CT

60o

tan tan ( )x x= ⇒ = =−3 3 601 o

Second quadrant: x = 180o − 60o = 120o

Fourth quadrant: x = 360o − 60o = 300o


Question 3 (b) (i)

Question 3 (a) (ii)

tantan( )

tan

330360 3030

13

o

o o

o

= −

= −

= −

Question 3 (b) (ii)

Period = 180o (function repeats itself every 180o)Range = [ , ]−∞ ∞

Question 3 (b) (iii)

Question 3 (b) (iv)



z x

xx

= − =−

− = −∴ =

0 68 104

2 72 107 28

.

.. ( )Lower Quartile

Interquartile range = UQ − LQ = 12.72 − 7.28 = 5.44

146 µ = 10

68%

Standard deviation = 4Median = 10 x z

x z

x z

= =−

=

= =−

=

= =−

= −

10 10 104

0

14 14 104

1

6 6 104

1

:

:

:

The standard normal distribution has a mean of 0 and standard deviation of 1 and has the same shape and area distribution as the original normal distribution.

P z Z

z x

xx

( ) .

.

.. ( )

< =

∴ = =−

= −∴ =

0 75

0 68 104

2 72 1012 72 Upper Quartile




z x=

− µσ

Question 4 (b)



UQ = Z

75%

zz Z=LQ = �Z

25%


AS

CT

xo –90 –60 –45 –30 0 30 45 60 90y –∞ –1.7 –1 –0.6 0 0.6 1 1.7 ∞

xo 105 120 135 150 180 210 225 240 270y –3.7 –1.7 –1 –0.6 0 0.6 1 1.7 –∞

�30o

�90o

5

0o

30o

60o

90o

270o

150o

240o

210o

180o

120o

4

3

�1

2

�5

�3

�2

�4

1

x

y

�60o

Yes, because each value of x only gives one value of y.

AS

CT

60o

tan tan ( )x x= ⇒ = =−3 3 601 o

Second quadrant: x = 180o − 60o = 120o

Fourth quadrant: x = 360o − 60o = 300o


Question 3 (b) (i)

Question 3 (a) (ii)

tantan( )

tan

330360 3030

13

o

o o

o

= −

= −

= −

Question 3 (b) (ii)

Period = 180o (function repeats itself every 180o)Range = [ , ]−∞ ∞


Question 3 (b) (iv)

Sample 1

Paper 2



B

A D

C

G

F8 cm

12 cm

12

4 cm

4 cm


AD BC

BF FC

BF FC

= =

=

∴ = =

12

2

8 4

cm

cm, cm

Triangle FCD: A = =12

24 4 8( )( ) cm

Triangle AFD: A = =12

212 4 24( )( ) cm

Parallelogram ABCD: A = =( )( )12 4 48 2cmTriangle BFA: A = − − =48 8 24 16 2cm

Question 6 (a) Question 6 (b)

Question 6 (c)

AF AF

BG

BG

2 2 2

12

4 12 16 144 160 160 4 10

4 10 16

4 105

= + = + = ⇒ = =

=

∴ =

( )

cm


µσ

µ σµ σ

=== + = += − = −

12015

135 120 15 1105 120 15 1

empiriCaL ruLe: 68% of young adults have a blood pressure reading of between 105 mm of Hg and 135 mm of Hg.

µσ==

= =−

= −

= =−

=

12015

116 116 12015

0 27

123 123 12015

0 2

1 1

2 1

x z

x z

: .

: .

Question 7 (a) (i)

Question 7 (a) (ii)

z

z1= 0.27� z

2= 0.2


9999 yx

Possibilities for arranging the numbers in ascending order:Median = +

=9 9

29

(ii) Mode = 9

(iii) Range cannot be worked out without knowing the values of x and y.

9, 9, 9, 9, x, yx, y, 9, 9, 9, 9x, 9, 9, 9, 9, yy, 9, 9, 9, 9, x9, 9, 9, 9, 9, 9

Mean = 9Therefore sum = 9 6 54× =x y+ = − =54 36 18Possibilities for x and y:x 1 2 3 4 5 6 7 8 9y 17 16 15 14 13 12 11 10 9

or

x 17 16 15 14 13 12 11 10 9y 1 2 3 4 5 6 7 8 9

Greatest range = [1, 17]R = 17 − 1 = 16

R x yx xy

= − =∴ = ⇒ =∴ =

62 24 12

6


Question 5 (a)

(i)

Question 5 (b)

Question 5 (c)



B

A D

C

G

F8 cm

12 cm

12

4 cm

4 cm


AD BC

BF FC

BF FC

= =

=

∴ = =

12

2

8 4

cm

cm, cm

Triangle FCD: A = =12

24 4 8( )( ) cm

Triangle AFD: A = =12

212 4 24( )( ) cm

Parallelogram ABCD: A = =( )( )12 4 48 2cmTriangle BFA: A = − − =48 8 24 16 2cm


Question 6 (c)

AF AF

BG

BG

2 2 2

12

4 12 16 144 160 160 4 10

4 10 16

4 105

= + = + = ⇒ = =

=

∴ =

( )

cm


µσ

µ σµ σ

=== + = += − = −

12015

135 120 15 1105 120 15 1

empiriCaL ruLe: 68% of young adults have a blood pressure reading of between 105 mm of Hg and 135 mm of Hg.

µσ==

= =−

= −

= =−

=

12015

116 116 12015

0 27

123 123 12015

0 2

1 1

2 1

x z

x z

: .

: .

Question 7 (a) (i)

Question 7 (a) (ii)

z

z1= 0.27� z

2= 0.2


9999 yx

Possibilities for arranging the numbers in ascending order:Median = +

=9 9

29

(ii) Mode = 9

(iii) Range cannot be worked out without knowing the values of x and y.

9, 9, 9, 9, x, yx, y, 9, 9, 9, 9x, 9, 9, 9, 9, yy, 9, 9, 9, 9, x9, 9, 9, 9, 9, 9

Mean = 9Therefore sum = 9 6 54× =x y+ = − =54 36 18Possibilities for x and y:x 1 2 3 4 5 6 7 8 9y 17 16 15 14 13 12 11 10 9

or

x 17 16 15 14 13 12 11 10 9y 1 2 3 4 5 6 7 8 9

Greatest range = [1, 17]R = 17 − 1 = 16

R x yx xy

= − =∴ = ⇒ =∴ =

62 24 12

6


Question 5 (a)

(i)

Question 5 (b)

Question 5 (c)

Sample 1

Paper 2



BA

CD

F230 m

45o

230 m

230 m

230 m

45o

cos

cos

45230

230 45 115 2

o

o m

=

= =

FC

FC

F

E

C

149 m

115 2 m

EC = + =149 115 2 2212 2( ) m

Question 8 (a) (ii)

Question 8 (b)

Question 8 (c)

h

A

x

16.3o

100 m

20o

160o

3.7o

p


1003 7 16 3

100 16 33 7

434 92

204

sin . sin .sin .

sin ..

sin

o

o

o

o

o

m

=

∴ = =

=

p

p

h334 92

434 92 20 148 75 149.

. sin .∴ = = ≈h o m

aA

bBsin sin

=


H

Sc

Start

S

Hc

H

Hc

0.9

0.1

0.4

0.6

0.2

0.8P S Hc c( ) .∩ = 0 72

P S Hc( ) .∩ = 0 18

P S H c( ) .∩ = 0 06

P S H( ) .∩ = 0 04Question 7 (d)

P P PP P zP

= −∞ − −∞ −= −∞ − >= −∞

( . ) ( . )( . ) ( . )(

to to to to

0 2 0 270 2 0 270.. ) { ( . )}

. .

.

2 1 0 270 5793 1 0 60640 1857

− − <= − +=

P z

From a sample of 100 people the number will be 100 0 1857× =. 18 or 19 people

Question 7 (b)(i) P(S) = 0.1(ii) P(H|S) = 0.4(iii) P(H|S

c) = 0.2

P S H P H SP H

( | ) ( )( )

=∩

Question 7 (c)

P H P S H P S Hc( ) ( ) ( ). ..

= ∩ + ∩= +=

0 04 0 180 22

P S H P S HP H

( | ) ( )( )

.

..=

∩= =

0 040 22

0 182

PPP

( ) .( ) .(

Having a strokeNot having a strokeAt le

==

0 1820 818

aast 2 out of 10 people will suffer a stroke2 out of

)(= −1 P 10 people will suffer a stroke)

{ ( . ) ( . )= −1 0 182 0 818100

0 10C ++=

101

1 90 182 0 8180 57

C ( . ) ( . ) }.

Question 7 (e) Question 7 (f)

Question 7 (g)



BA

CD

F230 m

45o

230 m

230 m

230 m

45o

cos

cos

45230

230 45 115 2

o

o m

=

= =

FC

FC

F

E

C

149 m

115 2 m

EC = + =149 115 2 2212 2( ) m

Question 8 (a) (ii)

Question 8 (b)

Question 8 (c)

h

A

x

16.3o

100 m

20o

160o

3.7o

p


1003 7 16 3

100 16 33 7

434 92

204

sin . sin .sin .

sin ..

sin

o

o

o

o

o

m

=

∴ = =

=

p

p

h334 92

434 92 20 148 75 149.

. sin .∴ = = ≈h o m

aA

bBsin sin

=


H

Sc

Start

S

Hc

H

Hc

0.9

0.1

0.4

0.6

0.2

0.8P S Hc c( ) .∩ = 0 72

P S Hc( ) .∩ = 0 18

P S H c( ) .∩ = 0 06

P S H( ) .∩ = 0 04Question 7 (d)

P P PP P zP

= −∞ − −∞ −= −∞ − >= −∞

( . ) ( . )( . ) ( . )(

to to to to

0 2 0 270 2 0 270.. ) { ( . )}

. .

.

2 1 0 270 5793 1 0 60640 1857

− − <= − +=

P z

From a sample of 100 people the number will be 100 0 1857× =. 18 or 19 people

Question 7 (b)(i) P(S) = 0.1(ii) P(H|S) = 0.4(iii) P(H|S

c) = 0.2

P S H P H SP H

( | ) ( )( )

=∩

Question 7 (c)

P H P S H P S Hc( ) ( ) ( ). ..

= ∩ + ∩= +=

0 04 0 180 22

P S H P S HP H

( | ) ( )( )

.

..=

∩= =

0 040 22

0 182

PPP

( ) .( ) .(

Having a strokeNot having a strokeAt le

==

0 1820 818

aast 2 out of 10 people will suffer a stroke2 out of

)(= −1 P 10 people will suffer a stroke)

{ ( . ) ( . )= −1 0 182 0 818100

0 10C ++=

101

1 90 182 0 8180 57

C ( . ) ( . ) }.


Question 7 (g)

Sample 1

Paper 2



a

a

a

a

b

b

bb

b

x

bb

2

a

2

a

b

a x b x b a b a b a2 4

44

42

2 2 22 2 2

12

2 2

+ = ⇒ = − =

−= −

Area of 4 triangles:

230 m

221 m

115 m

221 m

115 m

188.7 m

Surface area: A a b a= − = − =4 230 4 221 230 86 8122 2 2 2 2( ) m

A

a b a a b a

= × × = × ×

= × − = −

4 2

2 4 4

12

12

2 2 2 2

( )base height base height

Area of a casing stone = (0.86 m)(0.86 m) = 0.7396 m2

Number of stones = ≈86 8120 7396

117 000.

Question 8 (e)

Question 8 (f) (i)

Question 8 (f) (ii)


188.7 m 230 m 188.7 m

230 m

221 m

115 m

221 m

115 m

x

115 221

221 115 188 7

2 2 2

2 2

+ =

= − =

x

x . m

Length of square =+

=2 188 7 230

5012 15( . ) . m

Question 8 (d)



a

a

a

a

b

b

bb

b

x

bb

2

a

2

a

b

a x b x b a b a b a2 4

44

42

2 2 22 2 2

12

2 2

+ = ⇒ = − =

−= −

Area of 4 triangles:

230 m

221 m

115 m

221 m

115 m

188.7 m

Surface area: A a b a= − = − =4 230 4 221 230 86 8122 2 2 2 2( ) m

A

a b a a b a

= × × = × ×

= × − = −

4 2

2 4 4

12

12

2 2 2 2

( )base height base height

Area of a casing stone = (0.86 m)(0.86 m) = 0.7396 m2

Number of stones = ≈86 8120 7396

117 000.

Question 8 (e)

Question 8 (f) (i)

Question 8 (f) (ii)


188.7 m 230 m 188.7 m

230 m

221 m

115 m

221 m

115 m

x

115 221

221 115 188 7

2 2 2

2 2

+ =

= − =

x

x . m

Length of square =+

=2 188 7 230

5012 15( . ) . m

Question 8 (d)

Sample 1

Paper 2



Im

Re

1

2

3

4

0 1 2 3 4

w

z

30o

60o

z i

z

z

= +

= + = + = =

= = ⇒ = = =

∴ =

−

2 2 3

2 2 3 4 12 16 4

2 32

3 3 603

2 2

1

( )

tan tan ( )θ θπo

443 3

cos sinπ π+

i

w i

w

w

= +

= + = + = =

= ⇒ =

= =

∴ =

−

3

3 1 3 1 4 2

13

13

306

2

2 2

1

( ) ( )

tan tanθ θπo

ccos sinπ π6 6+

i

zw i i

i

= +

+

= +

43 3

26 6

82 2

cos sin cos sin

cos sin

π π π π

π π

= +=

8 08( )ii

‘According to De Moivre’s theorem, when two complex numbers are multiplied together, the arguments are added and the moduli are multiplied to form the new complex number.’

zw

i

ii=

+

+

= +

43 3

26 6

26 6

cos sin

cos sincos sin

π π

π ππ π

= +

= +2 3

212

3i i

‘According to De Moivre’s theorem, when two complex numbers are divided, the arguments are subtracted and the moduli are divided to form the new complex number.’

zw

i

i

= +

= +

1414

14

14

26 6

2 146

146

cos sin

cos sin

π π

π π

= +

= +

= +2 7

373

2 12

32

2 1 314 14 13cos sin ( )π πi i i

z r z= =, arg θ

z r i= +(cos sin )θ θ


Question 2 (b) (i) Question 2 (b) (ii)



(De Moivre’s Theorem)[ (cos sin )] (cos sin )r i r n i nn nq q q q+ = +


log ( ) log

log ( ) loglog

log ( ) log

9 3

99

9

99

12

2 0

23

0

2 0

x x

x x

x x

+ − =

+ − =

+ − =

llog ( ) log

log

(

9 9

9 2

20

2

2

2 2 02 0

2 9 1

22 0

2

x xxx

xx

x xx xx

+ − =+

=

+= =

+ =

− − =− ))( )x

x+ =

=1 0

2



log log loga a axy

x y

= −

a y y xxa= ⇔ =log

log loglogb

a

a

x xb

=

Question 1 (b) (i)

[x = −1 is not allowed as x > 0.]

53

5 4 33 2 1

10

52

5 42 1

10

=

× ×× ×

=

=

××

=

( )p q p q p q p q p q+ =

+

+

+

5 5 0 4 1 3 2 2 350

51

52

53

++

+

= + + + + +

54

55

5 10 10 5

1 4 0 5

5 4 1 3 2 2 3 1 4

p q p q

p p q p q p q p q q55

T p q33 2 3 210 10 0 6 0 4 0 3456= = =( . ) ( . ) .

Question 1 (b) (ii)




Im

Re

1

2

3

4

0 1 2 3 4

w

z

30o

60o

z i

z

z

= +

= + = + = =

= = ⇒ = = =

∴ =

−

2 2 3

2 2 3 4 12 16 4

2 32

3 3 603

2 2

1

( )

tan tan ( )θ θπo

443 3

cos sinπ π+

i

w i

w

w

= +

= + = + = =

= ⇒ =

= =

∴ =

−

3

3 1 3 1 4 2

13

13

306

2

2 2

1

( ) ( )

tan tanθ θπo

ccos sinπ π6 6+

i

zw i i

i

= +

+

= +

43 3

26 6

82 2

cos sin cos sin

cos sin

π π π π

π π

= +=

8 08( )ii

‘According to De Moivre’s theorem, when two complex numbers are multiplied together, the arguments are added and the moduli are multiplied to form the new complex number.’

zw

i

ii=

+

+

= +

43 3

26 6

26 6

cos sin

cos sincos sin

π π

π ππ π

= +

= +2 3

212

3i i

‘According to De Moivre’s theorem, when two complex numbers are divided, the arguments are subtracted and the moduli are divided to form the new complex number.’

zw

i

i

= +

= +

1414

14

14

26 6

2 146

146

cos sin

cos sin

π π

π π

= +

= +

= +2 7

373

2 12

32

2 1 314 14 13cos sin ( )π πi i i

z r z= =, arg θ






(De Moivre’s Theorem)[ (cos sin )] (cos sin )r i r n i nn nq q q q+ = +


log ( ) log

log ( ) loglog

log ( ) log

9 3

99

9

99

12

2 0

23

0

2 0

x x

x x

x x

+ − =

+ − =

+ − =

llog ( ) log

log

(

9 9

9 2

20

2

2

2 2 02 0

2 9 1

22 0

2

x xxx

xx

x xx xx

+ − =+

=

+= =

+ =

− − =− ))( )x

x+ =

=1 0

2



log log loga a axy

x y

= −

a y y xxa= ⇔ =log

log loglogb

a

a

x xb

=

Question 1 (b) (i)

[x = −1 is not allowed as x > 0.]

53

5 4 33 2 1

10

52

5 42 1

10

=

× ×× ×

=

=

××

=

( )p q p q p q p q p q+ =

+

+

+

5 5 0 4 1 3 2 2 350

51

52

53

++

+

= + + + + +

54

55

5 10 10 5

1 4 0 5

5 4 1 3 2 2 3 1 4

p q p q

p p q p q p q p q q55

T p q33 2 3 210 10 0 6 0 4 0 3456= = =( . ) ( . ) .

Question 1 (b) (ii)


Sample 2

Paper 1



Question 5 (25 marks)Question 5 (a)Yes. Each value of y corresponds to only one value of x.

It is not injective because certain values of y correspond to two values of x.

Local maximum is at y = 2.

Question 5 (b)

Question 5 (c)

Question 5 (d)

Question 5 (e)

B

A

x

y

f x( ) = 2x

h x( )

g x x( ) = log2

x

y

f x( ) = 2x

y x=

g(x) is the inverse function of f (x).Reflect the curve for f (x) by an axial symmetry through the line y = x to get its inverse function.

f x yyy xy x

f x y g x

x

x

( )log loglog loglog

( ) log ( )

= =

===

∴ = =−

22

22 2

2 2

21

2

x yA

h x aa

h x x

= ⇒ = =∴

∈ ⇒ = − − +∴ =

= − −

0 2 10 1

0 1 1 122 1

0

2

2

( , )( , ) ( ) ( )

( ) ( )

h x xx

xxx

B

( ) ( )( )

( )

( ,

= − −

∴ = − −

− =− =

∴ =

2 12 2 1

1 01 0

11 2

2

2

2

Local Maximum ))( , ) ( )?( )

( )

B f xf x x

1 22

2 21

∈

=

= True


100, 90, 81,...... a = 100, r = 0.9Which term is equal to 50, half of its original mass?

50 100 0 90 5 0 9

0 5 1 0 9

1

1

10 10

10

=

== −

∴ =

−

−

( . ). .

log ( . ) ( ) log ( . )log

n

n

n

n (( . )log ( . )

.0 50 9

1 7 5810

+ = days

S

S

S S

n

nn

n

=−

= −

= =

− =

−

∞

∞

108 1 162 1

108 162

0 05

162 1

13

23

13

23

( ( ) ) ( ( ) )

.

( (( ) ) .

( ) .

log loglo

13

13

10 10

162 0 05

162 0 05

3 32403 3240

n

n

n

n

n

− =

=

==

∴ =gglog

.10

10

32403

7 36=

Question 3 (25 marks)Question 3 (a) Question 3 (b)

Seven days must elapse before it is less than half its original mass.

Eight terms must be added together so that the sum differs from the sum to infinity by less than 0.05.


2

1

3

4

y

�1�2�3�4�5

�1

5

x

54321

y = 4

( 3, 4)�

(5, 4)

(1, 0)

(0, 1)

( 1)�0,

y x y xy xy x

= − ⇒ = ± −

= − → −= − + →

1 11 0 1 1 0

1

( ): ( , ), ( , )Intercepts

Interceptts: ( , ), ( , )0 1 1 0

Question 4 (a)Draw each of the straight line graphs by plotting their intercepts.

Question 4 (b)

Point of intersection: (1, 0)

Points of intersection: (−3, 4), (5, 4)Question 4 (c)x < –3, x > 5



Question 5 (25 marks)Question 5 (a)Yes. Each value of y corresponds to only one value of x.

It is not injective because certain values of y correspond to two values of x.

Local maximum is at y = 2.

Question 5 (b)

Question 5 (c)

Question 5 (d)

Question 5 (e)

B

A

x

y

f x( ) = 2x

h x( )

g x x( ) = log2

x

y

f x( ) = 2x

y x=

g(x) is the inverse function of f (x).Reflect the curve for f (x) by an axial symmetry through the line y = x to get its inverse function.

f x yyy xy x

f x y g x

x

x

( )log loglog loglog

( ) log ( )

= =

===

∴ = =−

22

22 2

2 2

21

2

x yA

h x aa

h x x

= ⇒ = =∴

∈ ⇒ = − − +∴ =

= − −

0 2 10 1

0 1 1 122 1

0

2

2

( , )( , ) ( ) ( )

( ) ( )

h x xx

xxx

B

( ) ( )( )

( )

( ,

= − −

∴ = − −

− =− =

∴ =

2 12 2 1

1 01 0

11 2

2

2

2

Local Maximum ))( , ) ( )?( )

( )

B f xf x x

1 22

2 21

∈

=

= True


100, 90, 81,...... a = 100, r = 0.9Which term is equal to 50, half of its original mass?

50 100 0 90 5 0 9

0 5 1 0 9

1

1

10 10

10

=

== −

∴ =

−

−

( . ). .

log ( . ) ( ) log ( . )log

n

n

n

n (( . )log ( . )

.0 50 9

1 7 5810

+ = days

S

S

S S

n

nn

n

=−

= −

= =

− =

−

∞

∞

108 1 162 1

108 162

0 05

162 1

13

23

13

23

( ( ) ) ( ( ) )

.

( (( ) ) .

( ) .

log loglo

13

13

10 10

162 0 05

162 0 05

3 32403 3240

n

n

n

n

n

− =

=

==

∴ =gglog

.10

10

32403

7 36=


Seven days must elapse before it is less than half its original mass.

Eight terms must be added together so that the sum differs from the sum to infinity by less than 0.05.


2

1

3

4

y

�1�2�3�4�5

�1

5

x

54321

y = 4

( 3, 4)�

(5, 4)

(1, 0)

(0, 1)

( 1)�0,

y x y xy xy x

= − ⇒ = ± −

= − → −= − + →

1 11 0 1 1 0

1

( ): ( , ), ( , )Intercepts

Interceptts: ( , ), ( , )0 1 1 0

Question 4 (a)Draw each of the straight line graphs by plotting their intercepts.

Question 4 (b)

Point of intersection: (1, 0)

Points of intersection: (−3, 4), (5, 4)Question 4 (c)x < –3, x > 5

Sample 2

Paper 1



V x x xdVdx

x x

dVdx

x x

= − +

= − +

= ⇒ − + =

4 130 1000

12 260 1000

0 12 260 1000

3 2

2

2 00

3 65 250 03 50 5 0

5

2

503

x xx xx

− + =− − ==

( )( ), cm

VMax. ( ) ( ) ( )= − + = − + =4 5 130 5 1000 5 500 3250 5000 22503 2 3cm

l = 40 − 2(5) = 30 cm, b = 25 − 2(5) = 15 cm, h = 5 cm

Maximum Volume: dVdx

= 0

[Ignore the second solution as it will cause one of the sides to be negative.]

Question 7 (d)

dVdt

t

V t dt t t c

V t cV t

= − +

= − + = − − +

= = =

∴ = −

∫

( )

( )

:

2 5

2 5 5

2250 0 2250

2

when22 5 2250− +t

− − + =

+ − =− + =

∴ =

t tt tt tt

2

2

5 2250 05 2250 045 50 0

45( )( )

s

V rdVdt

r drdt

drdt r

A rdAdt

r drdt

rr

=

= × = ⇒ =

=

= × = ×

13

3

22

2

3 3

2 2 3

π

ππ

π

π ππ 22

2

6

62

3

=

= =

=

rdAdt r

cm /s2

Question 7 (e)

h

r

45o

Area A

Question 7 (f) (i)

Question 7 (f) (ii)

tan 45 1

13

2 13

3

o = ⇒ =

∴ =

= =

rh

rh

r hV r h rπ π

V x x xx x xx x xx

= − +

− + =

− + − =

=

4 130 10004 130 1000 15122 65 500 756 0

1

3 2

3 2

3

:: ( ) ( ) ( ): ( ) ( ) ( )2 1 65 1 500 1 756 319 0

2 2 2 65 2 500 2 756

3

3

− + − = − ≠

= − + − =x 00

Question 7 (c)



f x g xx x xx xx x

xgg

( ) ( )

( )( ),

( )(

=

− = +

= − += − −

∴ == + =

5 30 4 30 1 3

1 31 1 3 43

2

2

))( , ), ( , )

= + =3 3 61 4 3 6A B

Question 6 (b) Question 6 (c)

A f x g x dx

x x dx

x x x

= −

= − + −

= − + −

= −

∫

∫

( ( ) ( ))

( )

[ ]

{ (

1

3

2

1

3

13

3 213

13

4 3

2 3

3)) ( ) ( )} { ( ) ( ) ( )}3 2 13

3 2

13

43

2 3 3 3 1 2 1 3 19 18 9 2 3

+ − − − + −

= − + − + − +

=

C D

f x dx

x x dx

( , ), ( , )

( )

( )

0 0 5 01

5 0

5

0

5

15

2

0

5

15

Average value =−

= −

=

∫

∫[[ ]

{ ( ) ( ) } { }( )

52

2 13

305

15

52

2 13

3

15

1252

1253

256

5 5 0

x x−

= − −

= − =

V x x xx x x xx x x

= − −

= − − +

= − +

=

( )( )( )( )

40 2 25 21000 80 50 41000 130 4

100

2

2

00 130 42 3x x x− +

S x x x x x xx x x x

= − + − + − −

= − + − + −

2 40 2 2 25 2 40 2 25 280 4 50 4 10002 2

( ) ( ) ( )( )880 50 4

1000 4

2

2

x x xx

− +

= −

600 1000 44 400

10010

2

2

2

= −

=

=∴ =

xx

xx cm

(40 2 ) cm� x

(25 2 ) cm� x

x x

x

x

x x

x

x


l = 40 – 2x, b = 25 – 2x, h = x

Question 7 (a) (ii)


A

x

y

DC

f x( )B

g x( )



V x x xdVdx

x x

dVdx

x x

= − +

= − +

= ⇒ − + =

4 130 1000

12 260 1000

0 12 260 1000

3 2

2

2 00

3 65 250 03 50 5 0

5

2

503

x xx xx

− + =− − ==

( )( ), cm

VMax. ( ) ( ) ( )= − + = − + =4 5 130 5 1000 5 500 3250 5000 22503 2 3cm

l = 40 − 2(5) = 30 cm, b = 25 − 2(5) = 15 cm, h = 5 cm

Maximum Volume: dVdx

= 0

[Ignore the second solution as it will cause one of the sides to be negative.]

Question 7 (d)

dVdt

t

V t dt t t c

V t cV t

= − +

= − + = − − +

= = =

∴ = −

∫

( )

( )

:

2 5

2 5 5

2250 0 2250

2

when22 5 2250− +t

− − + =

+ − =− + =

∴ =

t tt tt tt

2

2

5 2250 05 2250 045 50 0

45( )( )

s

V rdVdt

r drdt

drdt r

A rdAdt

r drdt

rr

=

= × = ⇒ =

=

= × = ×

13

3

22

2

3 3

2 2 3

π

ππ

π

π ππ 22

2

6

62

3

=

= =

=

rdAdt r

cm /s2

Question 7 (e)

h

r

45o

Area A

Question 7 (f) (i)

Question 7 (f) (ii)

tan 45 1

13

2 13

3

o = ⇒ =

∴ =

= =

rh

rh

r hV r h rπ π

V x x xx x xx x xx

= − +

− + =

− + − =

=

4 130 10004 130 1000 15122 65 500 756 0

1

3 2

3 2

3

:: ( ) ( ) ( ): ( ) ( ) ( )2 1 65 1 500 1 756 319 0

2 2 2 65 2 500 2 756

3

3

− + − = − ≠

= − + − =x 00

Question 7 (c)



f x g xx x xx xx x

xgg

( ) ( )

( )( ),

( )(

=

− = +

= − += − −

∴ == + =

5 30 4 30 1 3

1 31 1 3 43

2

2

))( , ), ( , )

= + =3 3 61 4 3 6A B


A f x g x dx

x x dx

x x x

= −

= − + −

= − + −

= −

∫

∫

( ( ) ( ))

( )

[ ]

{ (

1

3

2

1

3

13

3 213

13

4 3

2 3

3)) ( ) ( )} { ( ) ( ) ( )}3 2 13

3 2

13

43

2 3 3 3 1 2 1 3 19 18 9 2 3

+ − − − + −

= − + − + − +

=

C D

f x dx

x x dx

( , ), ( , )

( )

( )

0 0 5 01

5 0

5

0

5

15

2

0

5

15

Average value =−

= −

=

∫

∫[[ ]

{ ( ) ( ) } { }( )

52

2 13

305

15

52

2 13

3

15

1252

1253

256

5 5 0

x x−

= − −

= − =

V x x xx x x xx x x

= − −

= − − +

= − +

=

( )( )( )( )

40 2 25 21000 80 50 41000 130 4

100

2

2

00 130 42 3x x x− +

S x x x x x xx x x x

= − + − + − −

= − + − + −

2 40 2 2 25 2 40 2 25 280 4 50 4 10002 2

( ) ( ) ( )( )880 50 4

1000 4

2

2

x x xx

− +

= −

600 1000 44 400

10010

2

2

2

= −

=

=∴ =

xx

xx cm

(40 2 ) cm� x

(25 2 ) cm� x

x x

x

x

x x

x

x


l = 40 – 2x, b = 25 – 2x, h = x

Question 7 (a) (ii)


A

x

y

DC

f x( )B

g x( )

Sample 2

Paper 1




Question 9 (a)A

s

B

: ..

: ..

.

s ut tt

s t tt tt

= fi =\ =

= += +

\ +

100 9 110 99

5 0 56100 5 0 56

0 56

2

2

2 55 100 0

5 25 2242 0 56

9 62

t

t

- =

= - ± +¥

=.

. s

91 5 0 560 56 4 1 0

0 56 4 1 04 1

0 567 32

2

2

. .

. .( . . )

..

.

t t tt t

t t

t

= +- =- =

= = s

s t= = =9 1 9 1 7 32 66 6. . ( . ) . m

P l r= +2 2p

400 2 2200

200

= += +

= -

l rl r

r l

pp

p

A rl l l l l= = - = -2 2 200 400 2 2( )p p p

dAdl

l

dAdl

l

l

l

= -

= fi - =

=

\ =

400 4

0 400 4 0

400 4

100

p p

p p

p pm

r

r r

r

l

l

D

A B

C

Question 9 (b)

Question 9 (c)

Question 9 (d)

Question 9 (e) (i)

Question 9 (e) (ii)B wins the race.

Question 9 (e) (iii)



M ss

=

= =log log10 10 1 0

M As

M As

M M As

1 101

2 101

2 1 101

8 3

4

4

=

=

=

− =

log .

log

log −

− = ×

− =∴

log

. log

. log

101

2 101

1

2 10

8 3 4

8 3 4

As

M As

sA

MMM 2 10 4 8 3 8 9= + =log . .

Question 8 (a)

Question 8 (b)

M As

M As

M M As

1 101

2 102

1 2 101

8 3

4

=

=

=

=

− =

log .

log

log −

= ×

=

∴

log

. log log

102

101

210

1

2

4 3

As

As

sA

AA

A11

2

4 310 19953A

= =.

Question 8 (c)

M A A

M As

As

M A A

M

1 1

1 101 1 5 9

2 2

2

5 9

5 9 10

6 5

= =

=

= ⇒ =

= =

=

. ,

log .

. ,

lo

.

gg . .10

2 2 6 56 5 10As

As

A

= ⇒ =

=Fractionalchange in amplitude 22 1

1

2

16 5

5 90 6

1

1010

1 10 1 2 98

298

−= −

=××

− = − =

=

AA

AA

ss

.

.. .

% %change

Modified Mercalli Scale

II

I

IV

III

VI

V

VIII

VII

X

IX

XII

XI

4.5

8

2.5

3.5

7.5

5.5

6.5

1.5

2

3

4

5

6

7

Felt by few persons at rest, especially on

upper floors; delicately suspended objects

may swing.

Detected only by sensitive instruments

Felt noticeably indoors, but not always recognised

as earthquake; standing autos rock slightly,

vibration like passing truck.

Felt indoors by many, outdoors by few, at night

some may awaken; dishes, windows, doors

disturbed; autos rock noticeably

Felt by most people; some breakage of dishes,

windows, and plaster; disturbance of tall objects

Felt by all, many frightened and run outdoors;

falling plaster and chimneys; sand and mud

ejected; drivers of autos disturbed

Everybody runs outdoors; damage to buildings

varies depending on quality of construction;

noticed by drivers of autos

Panel walls thrown out of frames; fall of walls,

monuments, chimneys; sand and mud ejected;

drivers of autos disturbed

Buildings shifted off foundations, cracked, thrown

out of plumb; ground cracked; underground pipes

broken

Most masonry and frame structures destroyed;

ground cracked, rails bent, landslides

Few structures remain standing; bridges destroyed,

fissures in ground, pipes broken, landslides,

rails bent

Damage total; waves seen on ground surface,

lines of sight and level distorted, objects thrown

up in air

Richter

Magnitude

Scale

Question 8 (d)

6 510

1010

10 10 10 316

10 4

6 54

6 5 4 2 5

. log

.

. .

=

=

∴ = × = =

−

−

−

A

A

A cm

E = × = ×+ ×1 74 10 2 7 105 1 44 4 3 11. .( . . ) J

Question 8 (e)

Question 8 (f)




Question 9 (a)A

s

B

: ..

: ..

.

s ut tt

s t tt tt

= fi =\ =

= += +

\ +

100 9 110 99

5 0 56100 5 0 56

0 56

2

2

2 55 100 0

5 25 2242 0 56

9 62

t

t

- =

= - ± +¥

=.

. s

91 5 0 560 56 4 1 0

0 56 4 1 04 1

0 567 32

2

2

. .

. .( . . )

..

.

t t tt t

t t

t

= +- =- =

= = s

s t= = =9 1 9 1 7 32 66 6. . ( . ) . m

P l r= +2 2p

400 2 2200

200

= += +

= -

l rl r

r l

pp

p

A rl l l l l= = - = -2 2 200 400 2 2( )p p p

dAdl

l

dAdl

l

l

l

= -

= fi - =

=

\ =

400 4

0 400 4 0

400 4

100

p p

p p

p pm

r

r r

r

l

l

D

A B

C

Question 9 (b)

Question 9 (c)

Question 9 (d)

Question 9 (e) (i)

Question 9 (e) (ii)B wins the race.




M ss

=

= =log log10 10 1 0

M As

M As

M M As

1 101

2 101

2 1 101

8 3

4

4

=

=

=

− =

log .

log

log −

− = ×

− =∴

log

. log

. log

101

2 101

1

2 10

8 3 4

8 3 4

As

M As

sA

MMM 2 10 4 8 3 8 9= + =log . .

Question 8 (a)

Question 8 (b)

M As

M As

M M As

1 101

2 102

1 2 101

8 3

4

=

=

=

=

− =

log .

log

log −

= ×

=

∴

log

. log log

102

101

210

1

2

4 3

As

As

sA

AA

A11

2

4 310 19953A

= =.

Question 8 (c)

M A A

M As

As

M A A

M

1 1

1 101 1 5 9

2 2

2

5 9

5 9 10

6 5

= =

=

= ⇒ =

= =

=

. ,

log .

. ,

lo

.

gg . .10

2 2 6 56 5 10As

As

A

= ⇒ =

=Fractionalchange in amplitude 22 1

1

2

16 5

5 90 6

1

1010

1 10 1 2 98

298

−= −

=××

− = − =

=

AA

AA

ss

.

.. .

% %change

Modified Mercalli Scale

II

I

IV

III

VI

V

VIII

VII

X

IX

XII

XI

4.5

8

2.5

3.5

7.5

5.5

6.5

1.5

2

3

4

5

6

7

Felt by few persons at rest, especially on

upper floors; delicately suspended objects

may swing.

Detected only by sensitive instruments

Felt noticeably indoors, but not always recognised

as earthquake; standing autos rock slightly,

vibration like passing truck.

Felt indoors by many, outdoors by few, at night

some may awaken; dishes, windows, doors

disturbed; autos rock noticeably

Felt by most people; some breakage of dishes,

windows, and plaster; disturbance of tall objects

Felt by all, many frightened and run outdoors;

falling plaster and chimneys; sand and mud

ejected; drivers of autos disturbed

Everybody runs outdoors; damage to buildings

varies depending on quality of construction;

noticed by drivers of autos

Panel walls thrown out of frames; fall of walls,

monuments, chimneys; sand and mud ejected;

drivers of autos disturbed

Buildings shifted off foundations, cracked, thrown

out of plumb; ground cracked; underground pipes

broken

Most masonry and frame structures destroyed;

ground cracked, rails bent, landslides

Few structures remain standing; bridges destroyed,

fissures in ground, pipes broken, landslides,

rails bent

Damage total; waves seen on ground surface,

lines of sight and level distorted, objects thrown

up in air

Richter

Magnitude

Scale

Question 8 (d)

6 510

1010

10 10 10 316

10 4

6 54

6 5 4 2 5

. log

.

. .

=

=

∴ = × = =

−

−

−

A

A

A cm

E = × = ×+ ×1 74 10 2 7 105 1 44 4 3 11. .( . . ) J

Question 8 (e)

Question 8 (f)

Sample 2

Paper 1



The perpendicular line from O bisects the chord.The distance from the centre to the chord is 3.

l mx y kQ l m k k ml mx y m

:( , ) :

:

− + =− − ∈ − + + = ⇒ = −− + − =

01 3 3 0 3

3 0

The chord passing through Q is called l.

dax by c

a bd x yl mx y m

m m

=+ +

+= =

− + − =

=− + −

1 1

2 2

1 13 3 03 0

33 0

, ( , ) ( , ):

( ) ( ) 33

1

3 1 3 3

3 1 4 3

9 9 16 24 97 24 0

7

2

2

2

2 2

2

m

m m m

m m

m m mm mm m

+

+ = + −

+ = −

+ = − +

− =−( 224 0

0 247

),

==m

Q( 1, 3)� �

4l

r = 5r = 53

8

4 Slope m

O(3, 0)

s

[The perpendicular distance of the chord l from the centre O is 3.]

Slope of t: m = − 43

Circle s: Centre O(3, 0), r = 5

Equation of OQ: Point O(3, 0), m = 34

my xy xx y

=

− = −

= −− − =

34

340 3

4 3 93 4 9 0

( )

Find the point of intersection of lines t and OQ.

4 3 13 0 33 4 9 0 4

12 9 39 012 16 36 0

25 75

x yx y

x yx y

y

+ + = ×− − = ×−

+ + =− + + =

+

( )( )

== ⇒ = −

= − + − + = ⇒ = −∴ = −

0 3

3 4 3 3 13 0 4 41

y

y x xx

: ( )

Point of contact Q(−1, −3)


O(3, 0)

t

r = 5 Q

s

4 3 13 0 33 4 9 0 4

12 9 39 012 16 36 0

25 75

x yx y

x yx y

y

+ + = ×− − = ×−

+ + =− + + =

+

( )( )

== ⇒ = −

= − + − + = ⇒ = −∴ = −

0 3

3 4 3 3 13 0 4 41

y

y x xx

: ( )


The diagonal bisects the area of a parallelogram. Find the area of triangle ABD and multiply the answer by 2.

ABD

( , ) ( , )( , ) ( , )( , ) ( , )

1 3 0 04 4 3 1

1 7 2 4

→→

− → −

Area of parallelogram ABCD = 14

A(6, 0)O(0, 0)

B(0, 8)

y

x

Intercepts of 2x + 3y = c: D c E c( , ), ( , )12

130 0

O D c E c

ODE c c c

( , ), ( , ), ( , )

( )( ) ( )( )

0 0 0 0

0 0

12

13

12

12

13

12

16Area ∆ = − = 22

12

12

0 0 6 0 0 80 0 6 8 48

O A BOAB

OD

( , ), ( , ), ( , )( )( ) ( )( )Area

Area

∆ = − =

∆ EE OAB

c

c

c

c

= ∆

∴ = ×

=

=

∴ = ± = ±

12

12

16

2 12

12

16

2

2

48

24

144

144 12

Area



A x y x y= −

= − −

= +

=

12 1 2 2 1

12

12

3 4 2 1

12 27

( ) ( )( )

Question 1 (b)

B(4, 4)

A(1, 3)

C(2, 8)

D( 1, 7)�

note: The diagram in the question is a rough sketch showing the relative positions of the points. A grid is drawn whenever we wish to display the absolute positions of the points.



The perpendicular line from O bisects the chord.The distance from the centre to the chord is 3.

l mx y kQ l m k k ml mx y m

:( , ) :

:

− + =− − ∈ − + + = ⇒ = −− + − =

01 3 3 0 3

3 0

The chord passing through Q is called l.

dax by c

a bd x yl mx y m

m m

=+ +

+= =

− + − =

=− + −

1 1

2 2

1 13 3 03 0

33 0

, ( , ) ( , ):

( ) ( ) 33

1

3 1 3 3

3 1 4 3

9 9 16 24 97 24 0

7

2

2

2

2 2

2

m

m m m

m m

m m mm mm m

+

+ = + −

+ = −

+ = − +

− =−( 224 0

0 247

),

==m

Q( 1, 3)� �

4l

r = 5r = 53

8

4 Slope m

O(3, 0)

s

[The perpendicular distance of the chord l from the centre O is 3.]

Slope of t: m = − 43

Circle s: Centre O(3, 0), r = 5

Equation of OQ: Point O(3, 0), m = 34

my xy xx y

=

− = −

= −− − =

34

340 3

4 3 93 4 9 0

( )

Find the point of intersection of lines t and OQ.

4 3 13 0 33 4 9 0 4

12 9 39 012 16 36 0

25 75

x yx y

x yx y

y

+ + = ×− − = ×−

+ + =− + + =

+

( )( )

== ⇒ = −

= − + − + = ⇒ = −∴ = −

0 3

3 4 3 3 13 0 4 41

y

y x xx

: ( )

Point of contact Q(−1, −3)


O(3, 0)

t

r = 5 Q

s

4 3 13 0 33 4 9 0 4

12 9 39 012 16 36 0

25 75

x yx y

x yx y

y

+ + = ×− − = ×−

+ + =− + + =

+

( )( )

== ⇒ = −

= − + − + = ⇒ = −∴ = −

0 3

3 4 3 3 13 0 4 41

y

y x xx

: ( )


The diagonal bisects the area of a parallelogram. Find the area of triangle ABD and multiply the answer by 2.

ABD

( , ) ( , )( , ) ( , )( , ) ( , )

1 3 0 04 4 3 1

1 7 2 4

→→

− → −

Area of parallelogram ABCD = 14

A(6, 0)O(0, 0)

B(0, 8)

y

x

Intercepts of 2x + 3y = c: D c E c( , ), ( , )12

130 0

O D c E c

ODE c c c

( , ), ( , ), ( , )

( )( ) ( )( )

0 0 0 0

0 0

12

13

12

12

13

12

16Area ∆ = − = 22

12

12

0 0 6 0 0 80 0 6 8 48

O A BOAB

OD

( , ), ( , ), ( , )( )( ) ( )( )Area

Area

∆ = − =

∆ EE OAB

c

c

c

c

= ∆

∴ = ×

=

=

∴ = ± = ±

12

12

16

2 12

12

16

2

2

48

24

144

144 12

Area



A x y x y= −

= − −

= +

=

12 1 2 2 1

12

12

3 4 2 1

12 27

( ) ( )( )

Question 1 (b)

B(4, 4)

A(1, 3)

C(2, 8)

D( 1, 7)�

note: The diagram in the question is a rough sketch showing the relative positions of the points. A grid is drawn whenever we wish to display the absolute positions of the points.

Sample 2

Paper 2



30–35 2 < 35 2

35–40 3 < 40 5

40–45 9 < 45 14

45–50 12 < 50 26

50–55 22 < 55 48

55–60 (LQ) 46 < 60 94

60–65 (M) 58 < 65 152

65–70 50 < 70 202

70–75 (UQ) 40 < 75 242

75–80 32 < 80 274

80–85 16 < 85 290

290

Life expectancy (years)

Number of countries

Cumulative number of countries


Number of countries = 290Median class: 290(0.5) + 0.5 = 145.5 (Class 60−65)Upper quartile class: 290(0.75) + 0.75 = 218.25 (Class 70−75)Lower quartile class: 290(0.25) + 0.25 = 72.75 (Class 55−60)

Question 5 (a)Question 5 (25 marks)

Question 5 (b)

Nu

mb

er

of

Co

un

trie

sLife Expectancy (years)

5030 6055

[3]

4540 65 7570 8580

40

30

20

10

60

50

Median = 145.5M

35

[2]

[9][12]

[22]

[46]

[58]

(152)(94)

Modal class: 60−65Shape: Skewed left


M = +−

=60 145 5 94

585 64 4. .


Question 3 (a)


α

Ladder

3 m

10 m

30o

h

x

d

y

sin sin3010

10 30 5o o m= ⇒ = =h h

tantan

.

30 5 530

5 3

5 3 3 5 66

oo m

m

= ⇒ = =

= − =x

x

d

tan.

tan.

.α α= ⇒ =

=

−55 66

55 66

41 461 o

Question 3 (b)

Question 3 (c)

sin

sin .

sin ..

α =

=

= =

5

41 46 5

541 46

7 55

y

y

y

o

o m

Length of ladder protruding beyond the top of the wall = 10 − 7.55 = 2.45 m

Question 3 (d)

Condition: There are only two possible outcomes (success or failure) in each trial.Condition: There is a fixed number of trials n.Condition: The probability of success p is fixed from trial to trial.Condition: The trials are independent.Condition: The binomial random variable is the number of successes in n trials.

P P

P x C

( ) , ( )

( ) ( ) (

Income Income< = > =

≥ =

€ €32000 32000

8

12

12

108

12

2 12 )) ( ) ( ) ( ) ( ) .8 10

912

1 12

9 1010

12

0 12

10 0 055+ + =C C


Question 4 (b) (i)

P x P C C( ) ( ) ( ) ( ) ( ) ( ) .≥ = − = − − =2 1 1 1 0 99100

12

10 12

0 101

12

9 12

1None or

Question 4 (b) (ii)



30–35 2 < 35 2

35–40 3 < 40 5

40–45 9 < 45 14

45–50 12 < 50 26

50–55 22 < 55 48

55–60 (LQ) 46 < 60 94

60–65 (M) 58 < 65 152

65–70 50 < 70 202

70–75 (UQ) 40 < 75 242

75–80 32 < 80 274

80–85 16 < 85 290

290


Number of countries

Cumulative number of countries


Number of countries = 290Median class: 290(0.5) + 0.5 = 145.5 (Class 60−65)Upper quartile class: 290(0.75) + 0.75 = 218.25 (Class 70−75)Lower quartile class: 290(0.25) + 0.25 = 72.75 (Class 55−60)


Question 5 (b)

Nu

mb

er

of

Co

un

trie

s

Life Expectancy (years)

5030 6055

[3]

4540 65 7570 8580

40

30

20

10

60

50

Median = 145.5M

35

[2]

[9][12]

[22]

[46]

[58]

(152)(94)

Modal class: 60−65Shape: Skewed left


M = +−

=60 145 5 94

585 64 4. .


Question 3 (a)


α

Ladder

3 m

10 m

30o

h

x

d

y

sin sin3010

10 30 5o o m= ⇒ = =h h

tantan

.

30 5 530

5 3

5 3 3 5 66

oo m

m

= ⇒ = =

= − =x

x

d

tan.

tan.

.α α= ⇒ =

=

−55 66

55 66

41 461 o

Question 3 (b)

Question 3 (c)

sin

sin .

sin ..

α =

=

= =

5

41 46 5

541 46

7 55

y

y

y

o

o m

Length of ladder protruding beyond the top of the wall = 10 − 7.55 = 2.45 m

Question 3 (d)

Condition: There are only two possible outcomes (success or failure) in each trial.Condition: There is a fixed number of trials n.Condition: The probability of success p is fixed from trial to trial.Condition: The trials are independent.Condition: The binomial random variable is the number of successes in n trials.

P P

P x C

( ) , ( )

( ) ( ) (

Income Income< = > =

≥ =

€ €32000 32000

8

12

12

108

12

2 12 )) ( ) ( ) ( ) ( ) .8 10

912

1 12

9 1010

12

0 12

10 0 055+ + =C C


Question 4 (b) (i)

P x P C C( ) ( ) ( ) ( ) ( ) ( ) .≥ = − = − − =2 1 1 1 0 99100

12

10 12

0 101

12

9 12

1None or

Question 4 (b) (ii)

Sample 2

Paper 2




R I

R II

I

=

= = =

=

∴ =×

100

10 1 159

10 1 100159

159 10

GDPGDP billion. %, , ?

.

.

€

11100

16 1= € . billion

GDP per capita = =€

€159000000000

458825234654

7

12 13 14 15 16 1711

3

4

5

6

11

8

9

10

Unem

plo

ym

ent

Rate

Investment to GDP ratio

Run = 16 14�


Question 7 (b)

The data shows that the most effective way to reduce the unemployment rate is to increase investment to GDP ratio.

A. Linear, B. Negative shape, C. Very strong correlation

Question 7 (c) (i)

Question 7 (c) (ii)


7 cm

Question 6 (a)


Draw a line of length 7 cm with your ruler.Open your compass to a radius of 7 cm using this line. Placing the point of the compass on each end point of the line draw arcs of radius 7 cm.Draw lines from the end points of the line to the point of intersection of the two arcs.An equilateral triangle of side 7 cm has been constructed.

BE BE

AD

2 2 2

2 2 2

10 16 2 10 16 60 196 14

6 16 2 6 16

= + − = ⇒ =

= + −

( )( ) cos

( )( ) co

o

ss 60 196 14o = ⇒ =

∴⇒ =

AD

BE AD

E

B

6

6

60o

10 10

60o

CA

D

6

10

60o

CA

D

6

10

120o

E

B

6

10

C

120o

Triangles ACD and BCE are congruent (SAS).

Question 6 (b)

Question 6 (c)




R I

R II

I

=

= = =

=

∴ =×

100

10 1 159

10 1 100159

159 10

GDPGDP billion. %, , ?

.

.

€

11100

16 1= € . billion

GDP per capita = =€

€159000000000

458825234654

7

12 13 14 15 16 1711

3

4

5

6

11

8

9

10

Unem

plo

ym

ent

Rate


Run = 16 14�


Question 7 (b)

The data shows that the most effective way to reduce the unemployment rate is to increase investment to GDP ratio.

A. Linear, B. Negative shape, C. Very strong correlation

Question 7 (c) (i)

Question 7 (c) (ii)


7 cm

Question 6 (a)


Draw a line of length 7 cm with your ruler.Open your compass to a radius of 7 cm using this line. Placing the point of the compass on each end point of the line draw arcs of radius 7 cm.Draw lines from the end points of the line to the point of intersection of the two arcs.An equilateral triangle of side 7 cm has been constructed.

BE BE

AD

2 2 2

2 2 2

10 16 2 10 16 60 196 14

6 16 2 6 16

= + − = ⇒ =

= + −

( )( ) cos

( )( ) co

o

ss 60 196 14o = ⇒ =

∴⇒ =

AD

BE AD

E

B

6

6

60o

10 10

60o

CA

D

6

10

60o

CA

D

6

10

120o

E

B

6

10

C

120o

Triangles ACD and BCE are congruent (SAS).

Question 6 (b)

Question 6 (c)

Sample 2

Paper 2



O

B

P

A

C

r =1 16 cm

r = 9 cm2

r

Q

D

7 cm 9 cm

16 cm

r1 = 16 cmr2 = 9 cm

|PQ| = |PC| + |CQ| = r1 + r2 = 16 cm + 9 cm = 25 cm

725

x

x

x

2 2 2

2 2

7 25

25 7 24

+ =

= − = cm

tan( )

tan

. .

∠ =

∴ ∠ =

= =

−

APQ

APQ

247

247

73 74 1 287

1

o rads

A r= = =12

2 12

2 216 1 287 164 7θ ( ) ( . ) . cm





CaSio CaLCuLator (fx-85GT PLUS)Steps to find r:Press Mode.Press 2: StatPress 2: A + BxInput your x and y valuesPress AC ButtonPress Shift followed by the Number 1Press 5: RegPress 3: rPress =

r = –0.9767

( , . ), ( , . ). . .

14 8 4 16 6 38 4 6 314 16

1 1m =−−

= −

m x yy xy x

= − =− = − −− = − +

1 1 14 8 48 4 1 1 14

8 4 1 1 15 41

1 2. , ( , ) ( , . )( . ) . ( )

. . ... .1 23 8 0x y+ − =

y xx= + − =

∴ =3 1 1 3 23 8 0

18 9: . ( ) .

.

7

12 13 14 15 16 1711

3

4

5

6

11

8

9

10

Unem

plo

ym

ent

Rate

From 1948 to Present


Question 7 (d) Question 7 (e) (i)

Question 7 (e) (ii)


The 11 year plot is misleading as the 64 year plot shows little correlation between the unemployment rate and investment to GDP ratio. It is a prime example of someone using statistics to support their own position.

Question 7 (f) (i)

Question 7 (f) (ii)Yes



O

B

P

A

C

r =1 16 cm

r = 9 cm2

r

Q

D

7 cm 9 cm

16 cm

r1 = 16 cmr2 = 9 cm

|PQ| = |PC| + |CQ| = r1 + r2 = 16 cm + 9 cm = 25 cm

725

x

x

x

2 2 2

2 2

7 25

25 7 24

+ =

= − = cm

tan( )

tan

. .

∠ =

∴ ∠ =

= =

−

APQ

APQ

247

247

73 74 1 287

1

o rads

A r= = =12

2 12

2 216 1 287 164 7θ ( ) ( . ) . cm






r = –0.9767

( , . ), ( , . ). . .

14 8 4 16 6 38 4 6 314 16

1 1m =−−

= −

m x yy xy x

= − =− = − −− = − +

1 1 14 8 48 4 1 1 14

8 4 1 1 15 41

1 2. , ( , ) ( , . )( . ) . ( )

. . ... .1 23 8 0x y+ − =

y xx= + − =

∴ =3 1 1 3 23 8 0

18 9: . ( ) .

.

7

12 13 14 15 16 1711

3

4

5

6

11

8

9

10

Unem

plo

ym

ent

Rate

From 1948 to Present


Question 7 (d) Question 7 (e) (i)

Question 7 (e) (ii)


The 11 year plot is misleading as the 64 year plot shows little correlation between the unemployment rate and investment to GDP ratio. It is a prime example of someone using statistics to support their own position.

Question 7 (f) (i)

Question 7 (f) (ii)Yes

Sample 2

Paper 2



Area of space between wheels = Area of trapezium − Area of sector APC − Area of sector CQB − Area of smallest wheel

Area of sector CQB:

O

B

P

A

C

r =1 16 cm

r = 9 cm2

r

Q

D

�

α = − =

∠ = + = =

=

90 73 74 16 2616 26 90 106 26 1 855

12

2

o o o

o o o rads. .. . .CQB

A r θθ = =12

2 29 1 855 75 13( ) ( . ) . cm

Area of space between wheels = − − −

≈300 164 7 75 13 144

4933

22. . π cm

% of air space to area of three wheels =+ +

× ≈33

16 9 14449

100 32 2

2

π

% %

Question 8 (f)


O

B

y

A

C

(16 - r) (9 + r)

r

Q

D

(16 + r)

(9 - r)

P

z

y r ry r ry r r r

2 2 2

2 2 2

2

16 1616 1616 16 16 16

+ − = +

= + − −

= + + − + −

( ) ( )( ) ( )( )( ++

= =

=

ry r r

y r

)( )( )2 32 2 64

8

z r rz r rz r r r rz

2 2 2

2 2 2

2

2

9 99 99 9 9 91

+ − = +

= + − −

= + + − + − +

=

( ) ( )( ) ( )( )( )( 88 2 36

6

)( )r r

z r

=

=

y z

r r

r

r

r

+ =

+ =

=

= =

∴ =

24

8 6 24

14 242414

127

14449

Area of trapezium ABQP: A = + = =12

216 9 24 25 12 300( )( ) ( ) cm


Question 8 (d)

Question 8 (e)



Area of space between wheels = Area of trapezium − Area of sector APC − Area of sector CQB − Area of smallest wheel

Area of sector CQB:

O

B

P

A

C

r =1 16 cm

r = 9 cm2

r

Q

D

�

α = − =

∠ = + = =

=

90 73 74 16 2616 26 90 106 26 1 855

12

2

o o o

o o o rads. .. . .CQB

A r θθ = =12

2 29 1 855 75 13( ) ( . ) . cm

Area of space between wheels = − − −

≈300 164 7 75 13 144

4933

22. . π cm

% of air space to area of three wheels =+ +

× ≈33

16 9 14449

100 32 2

2

π

% %

Question 8 (f)


O

B

y

A

C

(16 - r) (9 + r)

r

Q

D

(16 + r)

(9 - r)

P

z

y r ry r ry r r r

2 2 2

2 2 2

2

16 1616 1616 16 16 16

+ − = +

= + − −

= + + − + −

( ) ( )( ) ( )( )( ++

= =

=

ry r r

y r

)( )( )2 32 2 64

8

z r rz r rz r r r rz

2 2 2

2 2 2

2

2

9 99 99 9 9 91

+ − = +

= + − −

= + + − + − +

=

( ) ( )( ) ( )( )( )( 88 2 36

6

)( )r r

z r

=

=

y z

r r

r

r

r

+ =

+ =

=

= =

∴ =

24

8 6 24

14 242414

127

14449

Area of trapezium ABQP: A = + = =12

216 9 24 25 12 300( )( ) ( ) cm


Question 8 (d)

Question 8 (e)

Sample 2

Paper 2




log ( ) loglog ( ) log

log

4 4

4 4

4 2

6 1 2 26 1 2 26 1 2

6

x xx xxx

x

+ − =+ − =

+

=

++=

+ =

− − =+ − =

=

1 4

6 1 1616 6 1 08 1 2 1 0

22

2

2

12

xx xx xx x

x( )( )

log log loga a axy

x y

= −

a y y xxa= ⇔ =log

F ab

F a F b

F a bF F a F a

= −

⇒ == − =

( ) ( )

( )( ) ( ) ( )11 0

F y F y F y Fy

F y F F yF y

y

( ) ( )

( ) [ ( ) ( )]( )

21

1

10

=

= −

= − −= − + FF yF y

( )( )= 2

Fx

F F x

F xF x

1 1

0

= −

= −= −

( ) ( )

( )( )

Question 2 (b) (i) Question 2 (b) (ii) Question 2 (b) (iii)

[ x = − 18 is not allowed as log ( )4

18− is not defined.]


x x xxxx x

3 2

3 2

3

3 2 02 2 3 2 2 2 023

+ + − =

= − − + − − − =∴ +

∴ +

: ( ) ( )( ) is a factor

22 2

3 2

3

2 2 12 2 1 2

3 2 13

+ − = + + −

= + + + − −∴ = + ⇒ =

∴ +

x x x kxx k x k x

k kx x

( )( )( ) ( )

22 22 2 1 0+ − = + + − =x x x x( )( )



Rationalise the denominators by multiplying above and below by the conjugate of the denominator.

Question 1 (b)

x x

x

x

2

2

1 0

1 1 4 1 12 1

1 1 42

1 52

2 1 52

+ − =

=− ± − −

=− ± +

=− ±

∴ = −− ±

( )( )( )

,

[Line up the coefficients.]

[Solve the quadratic equation.]

11 2

1 21 2

1 21 2

2 1

11 2

1 21 2

1 21 2

2 1

( )( )( )

( )( )( )

+×

−−

=−−

= −

−×

++

=+−

= − −

1 1 2

1 1 2

1 2 1

11 2

11 2

2

−

=

− = ±

± =

=+ −

x

x

x

x ,




log ( ) loglog ( ) log

log

4 4

4 4

4 2

6 1 2 26 1 2 26 1 2

6

x xx xxx

x

+ − =+ − =

+

=

++=

+ =

− − =+ − =

=

1 4

6 1 1616 6 1 08 1 2 1 0

22

2

2

12

xx xx xx x

x( )( )

log log loga a axy

x y

= −

a y y xxa= ⇔ =log

F ab

F a F b

F a bF F a F a

= −

⇒ == − =

( ) ( )

( )( ) ( ) ( )11 0

F y F y F y Fy

F y F F yF y

y

( ) ( )

( ) [ ( ) ( )]( )

21

1

10

=

= −

= − −= − + FF yF y

( )( )= 2

Fx

F F x

F xF x

1 1

0

= −

= −= −

( ) ( )

( )( )


[ x = − 18 is not allowed as log ( )4

18− is not defined.]


x x xxxx x

3 2

3 2

3

3 2 02 2 3 2 2 2 023

+ + − =

= − − + − − − =∴ +

∴ +

: ( ) ( )( ) is a factor

22 2

3 2

3

2 2 12 2 1 2

3 2 13

+ − = + + −

= + + + − −∴ = + ⇒ =

∴ +

x x x kxx k x k x

k kx x

( )( )( ) ( )

22 22 2 1 0+ − = + + − =x x x x( )( )



Rationalise the denominators by multiplying above and below by the conjugate of the denominator.

Question 1 (b)

x x

x

x

2

2

1 0

1 1 4 1 12 1

1 1 42

1 52

2 1 52

+ − =

=− ± − −

=− ± +

=− ±

∴ = −− ±

( )( )( )

,

[Line up the coefficients.]

[Solve the quadratic equation.]

11 2

1 21 2

1 21 2

2 1

11 2

1 21 2

1 21 2

2 1

( )( )( )

( )( )( )

+×

−−

=−−

= −

−×

++

=+−

= − −

1 1 2

1 1 2

1 2 1

11 2

11 2

2

−

=

− = ±

± =

=+ −

x

x

x

x , Sample 3

Paper 1



S = {6, 12, 18, 24,.....}S3 = {12, 24, 36, 48,...}Yes, all elements of S3 are elements of S.

Question 4 (b) (i)

S = {6, 12, 18, 24,.....}S1 = {3, 6, 9, 12,...}No. 9 is an element of S1 but not an element of S.


Question 4 (b) (ii)


Question 4 (c)

z r z= =, arg θ


Z

N

Q

A

70

−8

− 56

23

Question 4 (25 marks)Question 4 (a) (i) & (ii)

θ

z

n z i

n z

= =

+

= =

0 2 518

518

1 2 1718

1

2

: cos sin

: cos

p p

p

+

= =

+

i

n z i

sin

: cos sin

1718

2 2 2918

293

p

p pp18

z i

z

= − +

= − + = + = =

=−

= =

∴ = −

4 3 4

4 3 4 48 16 64 8

44 3

13

13

2 2

1

( )

tan tan ,

tan

θ α

α

= =

= − =

∴ = +

306

656

8 56

2

o

Second quadrant:

p

p p p

p p

θ

z ncos + +

=

+

+

+i n n i nsin cos sin56

2 8 5 126

5 12p p p p p p66

8 5 126

5 126

13

13

=+

+

+

z n i ncos sinp p p p

=

+

+

+

13

2 5 1218

5 1218

cos sinp p p pn i n

where α is the related angle in the first quadrant.


Three consecutive numbers: n − 1, n, n + 1

S n n nn n n n nn

= − + + +

= − + + + + +

= +

( ) ( )1 12 1 2 1

3 2

2 2 2

2 2 2

2


Question 3 (b)

3 23

22 2

3n n+

= +



1. Prove true for n = 1: n = + + =1 7 3 1 8 181: ( ) [Therefore, true for n = 1.]

2. Assume true for n = k: Assume n k k a ak= + + = ∈: ,7 3 8 9 �

3. Prove true for n = k + 1: Prove 7 3 1 8 91k k b b+ + + + = ∈( ) , �


Proof:7 3 1 8

7 7 3 3 87 9 3 8 3 1163 21

1k

k

kk

a k ka k

+ + + +

= + + += − − + += − −

( )( )( )

556 3 1163 18 459 7 2 59

+ += − −= − −= ∈

ka ka k

b b[ ]

, �



S = {6, 12, 18, 24,.....}S3 = {12, 24, 36, 48,...}Yes, all elements of S3 are elements of S.

Question 4 (b) (i)



Question 4 (b) (ii)


Question 4 (c)

z r z= =, arg θ


Z

N

Q

A

70

−8

− 56

23

Question 4 (25 marks)Question 4 (a) (i) & (ii)

θ

z

n z i

n z

= =

+

= =

0 2 518

518

1 2 1718

1

2

: cos sin

: cos

p p

p

+

= =

+

i

n z i

sin

: cos sin

1718

2 2 2918

293

p

p pp18

z i

z

= − +

= − + = + = =

=−

= =

∴ = −

4 3 4

4 3 4 48 16 64 8

44 3

13

13

2 2

1

( )

tan tan ,

tan

θ α

α

= =

= − =

∴ = +

306

656

8 56

2

o

Second quadrant:

p

p p p

p p

θ

z ncos + +

=

+

+

+i n n i nsin cos sin56

2 8 5 126

5 12p p p p p p66

8 5 126

5 126

13

13

=+

+

+

z n i ncos sinp p p p

=

+

+

+

13

2 5 1218

5 1218

cos sinp p p pn i n

where α is the related angle in the first quadrant.


Three consecutive numbers: n − 1, n, n + 1

S n n nn n n n nn

= − + + +

= − + + + + +

= +

( ) ( )1 12 1 2 1

3 2

2 2 2

2 2 2

2


Question 3 (b)

3 23

22 2

3n n+

= +



1. Prove true for n = 1: n = + + =1 7 3 1 8 181: ( ) [Therefore, true for n = 1.]

2. Assume true for n = k: Assume n k k a ak= + + = ∈: ,7 3 8 9 �

3. Prove true for n = k + 1: Prove 7 3 1 8 91k k b b+ + + + = ∈( ) , �


Proof:7 3 1 8

7 7 3 3 87 9 3 8 3 1163 21

1k

k

kk

a k ka k

+ + + +

= + + += − − + += − −

( )( )( )

556 3 1163 18 459 7 2 59

+ += − −= − −= ∈

ka ka k

b b[ ]

, �

Sample 3

Paper 1



y

x

0.5

0.3

0.2

20

0.4

0.1

1510c = 1.4 5

y

x

0.5

0.3

0.2

20

0.4

0.1

15105e = 2.7

The maximum value of f (x) as calculated in part (b) is e.∴ ≈e 2 7.

Question 5 (e) (ii)

10 1010

10 1010

10

10c c cc

f x xx

f

f c cc

f f

= ⇒ =

=

=

=

∴ =

ln ln

( ) ln

( ) ln

( ) ln

( ) (cc c) .⇒ ≈1 4




x2

y

x

x1

x3

BA C

f x y xx

dydx

x xx

xx

dydx

xx

x

( ) ln

( ) (ln )( ) ln

ln

= =

=−

=−

= ⇒−

=

−

1

2 2

2

1 1

0 1 0

1 llnln

( ) ln

: ,

xx x e

f e ee e

ee

== ⇒ =

= =

01

1

1TP

dydx

xx

d ydx

x x xx

x x x xx

x

=−

=− − −

=− − +

=

1

1 2

2 2

2

2

2

2 1

2 2

4

ln

( ) ( ln )( )( )

ln

−− +

=

− +=− +

=−

= −=

3 2

3 2 3 2 1

4

2

2 4 4 4 3

x x xx

d ydx

e e ee

e ee

ee ex e

ln

ln<<

∴

0

1ee

, is a local maximum

a ba b

b a a baa

bb

b a

b a

=

==

∴ =

ln lnln lnln ln

It is not injective because there is more than one x value mapping on to the same y value.

It is surjective because every y value has a corresonding x value.


Question 5 (d)

y

x

0.5

0.3

0.2

20

0.4

0.1

15102 54

0.35

Calculator: ln ln .22

44

0 3466= =

Question 5 (e) (i)

2 4 22

44

2 22

4 44

2 4 0 35

4 2= ⇒ =

=

=

=

∴ = =

ln ln

( ) ln

( ) ln

( ) ln

( ) ( ) .

f x xx

f

f

f f



y

x

0.5

0.3

0.2

20

0.4

0.1

1510c = 1.4 5

y

x

0.5

0.3

0.2

20

0.4

0.1

15105e = 2.7

The maximum value of f (x) as calculated in part (b) is e.∴ ≈e 2 7.

Question 5 (e) (ii)

10 1010

10 1010

10

10c c cc

f x xx

f

f c cc

f f

= ⇒ =

=

=

=

∴ =

ln ln

( ) ln

( ) ln

( ) ln

( ) (cc c) .⇒ ≈1 4




x2

y

x

x1

x3

BA C

f x y xx

dydx

x xx

xx

dydx

xx

x

( ) ln

( ) (ln )( ) ln

ln

= =

=−

=−

= ⇒−

=

−

1

2 2

2

1 1

0 1 0

1 llnln

( ) ln

: ,

xx x e

f e ee e

ee

== ⇒ =

= =

01

1

1TP

dydx

xx

d ydx

x x xx

x x x xx

x

=−

=− − −

=− − +

=

1

1 2

2 2

2

2

2

2 1

2 2

4

ln

( ) ( ln )( )( )

ln

−− +

=

− +=− +

=−

= −=

3 2

3 2 3 2 1

4

2

2 4 4 4 3

x x xx

d ydx

e e ee

e ee

ee ex e

ln

ln<<

∴

0

1ee

, is a local maximum

a ba b

b a a baa

bb

b a

b a

=

==

∴ =

ln lnln lnln ln

It is not injective because there is more than one x value mapping on to the same y value.

It is surjective because every y value has a corresonding x value.


Question 5 (d)

y

x

0.5

0.3

0.2

20

0.4

0.1

15102 54

0.35

Calculator: ln ln .22

44

0 3466= =

Question 5 (e) (i)

2 4 22

44

2 22

4 44

2 4 0 35

4 2= ⇒ =

=

=

=

∴ = =

ln ln

( ) ln

( ) ln

( ) ln

( ) ( ) .

f x xx

f

f

f f

Sample 3

Paper 1



Solve equations (1), (2) and (3) simultaneously.

′ = − −

⇒ ′ = ⇒ − − =

−

f x x xD f x x xx

( )( )

6 4 100 6 4 10 0

3

2

2

2

is a turning point22 5 0

3 5 1 01 5

353

xx xx

x D

− =− + =

∴ = −

=

( )( ),

-coordinate of

B C g x

m

g x x y

( , ) ( , ) ( )

( ):

3 0 0 60 63 0

2

2

and

Slope

Equation of

− ∈

=+−

=

− ++ =∈ − + = ⇒ = −

− − =

kB g x k kx y

g x g x

03 0 6 0 0 6

2 6 0( , ) ( ) :

( ): ( )Equation of == = −y x2 6

A g x f x dx

x x x x dx

x x x

= −

= − − + + +

= − + +

∫

∫

( ( ) ( ))

( )

(

0

3

3 2

0

3

3 2

2 6 2 2 10 6

2 2 12 ))

{

0

3

4 3 2

0

3

12

4 23

3 2

0

3

12

24

23

122

6

∫

= − + +

= − + +

= −

dx

x x x

x x x

(( ) ( ) ( ) }3 3 6 3 04 23

3 2

632

+ + −

=


Question 6 (d)

− + − =+ + =− + =

+ +

a b ca b ca b c

a

69 3 23 2 0

8 4

.....( )...( )...( )

( ) ( ) :

123

1 2 bb a ba b

aa

= ⇒ + =+ − =

∴ = +∴ =

8 2 22 6

4 82

...( )( ) ( ) : .....( )

...( ) ( )

41 3 5

4 5

SSubstitute into equation Substitute into

( ) : ( )4 2 2 2 2+ = ⇒ = −b b equation ( ) :

( )1 − − − = ⇒ = −

∴ = − − −

2 2 6 102 2 10 63 2

c cf x x x x



A( 1, 0)�

B(3, 0)

C(0, 6)�

y

D

y g x= ( )

y f x= ( )

x

f x ax bx cx dA f x f a b c d

( )( , ) ( ) : ( ) ( ) ( ) ( )

= + + +

− ∈ − = − + − + − + =

3 2

3 21 0 1 1 1 1 000

3 0 3 3 3 3 02

3 2

∴− + − + =

∈ = + + + =∴

a b c d

B f x f a b c d

...( )

( , ) ( ) : ( ) ( ) ( ) ( )

1

77 9 3 0

0 6 0 0 0 0 63 2

a b c d

C f x f a b c d

+ + + =

− ∈ = + + + = −

...( )

( , ) ( ) : ( ) ( ) ( ) ( )

2

∴∴ = −

∴− + − =∴ + + = ⇒ + + =

d

a b ca b c a b c

6

627 9 3 6 9 3 2

...( )...( )

12

f x ax bx cx df x ax bx cA

( )( )( , )

= + + +

′ = + +

−

3 2

23 21 0 is a turning point ⇒⇒ ′ − = − + − + =

∴ − + =f a b c

a b c( ) ( ) ( )

...( )1 3 1 2 1 0

3 2 0

2

3



Solve equations (1), (2) and (3) simultaneously.

′ = − −

⇒ ′ = ⇒ − − =

−

f x x xD f x x xx

( )( )

6 4 100 6 4 10 0

3

2

2

2

is a turning point22 5 0

3 5 1 01 5

353

xx xx

x D

− =− + =

∴ = −

=

( )( ),

-coordinate of

B C g x

m

g x x y

( , ) ( , ) ( )

( ):

3 0 0 60 63 0

2

2

and

Slope

Equation of

− ∈

=+−

=

− ++ =∈ − + = ⇒ = −

− − =

kB g x k kx y

g x g x

03 0 6 0 0 6

2 6 0( , ) ( ) :

( ): ( )Equation of == = −y x2 6

A g x f x dx

x x x x dx

x x x

= −

= − − + + +

= − + +

∫

∫

( ( ) ( ))

( )

(

0

3

3 2

0

3

3 2

2 6 2 2 10 6

2 2 12 ))

{

0

3

4 3 2

0

3

12

4 23

3 2

0

3

12

24

23

122

6

∫

= − + +

= − + +

= −

dx

x x x

x x x

(( ) ( ) ( ) }3 3 6 3 04 23

3 2

632

+ + −

=


Question 6 (d)

− + − =+ + =− + =

+ +

a b ca b ca b c

a

69 3 23 2 0

8 4

.....( )...( )...( )

( ) ( ) :

123

1 2 bb a ba b

aa

= ⇒ + =+ − =

∴ = +∴ =

8 2 22 6

4 82

...( )( ) ( ) : .....( )

...( ) ( )

41 3 5

4 5

SSubstitute into equation Substitute into

( ) : ( )4 2 2 2 2+ = ⇒ = −b b equation ( ) :

( )1 − − − = ⇒ = −

∴ = − − −

2 2 6 102 2 10 63 2

c cf x x x x



A( 1, 0)�

B(3, 0)

C(0, 6)�

y

D

y g x= ( )

y f x= ( )

x

f x ax bx cx dA f x f a b c d

( )( , ) ( ) : ( ) ( ) ( ) ( )

= + + +

− ∈ − = − + − + − + =

3 2

3 21 0 1 1 1 1 000

3 0 3 3 3 3 02

3 2

∴− + − + =

∈ = + + + =∴

a b c d

B f x f a b c d

...( )

( , ) ( ) : ( ) ( ) ( ) ( )

1

77 9 3 0

0 6 0 0 0 0 63 2

a b c d

C f x f a b c d

+ + + =

− ∈ = + + + = −

...( )

( , ) ( ) : ( ) ( ) ( ) ( )

2

∴∴ = −

∴− + − =∴ + + = ⇒ + + =

d

a b ca b c a b c

6

627 9 3 6 9 3 2

...( )...( )

12

f x ax bx cx df x ax bx cA

( )( )( , )

= + + +

′ = + +

−

3 2

23 21 0 is a turning point ⇒⇒ ′ − = − + − + =

∴ − + =f a b c

a b c( ) ( ) ( )

...( )1 3 1 2 1 0

3 2 0

2

3

Sample 3

Paper 1



Let x = Mass of Sterling silverLet y = Mass of Britannia silverMass of ring = 28 g

x yx y

+ = ×+ =28 0 925

0 925 0 958 26 3224( . )

. . .

1 31 10347681

31 10347681

4031 1034768

40

40 1 286

T g

T g

T g

g T

=

=

=

∴ =

.

.

.

.

S C T pC

T

pS

= × ×=

= =

== × × =

0 9428

31 10347680 9

260 94 0 9 26 22

.

..

$. . $

Question 8 (b)


10.00

15.00

0.00

5.00

30.00

35.00

20.00

25.00

45.00

50.00

40.00

US

D p

ertr

oy o

un

ce

Jul1

0

Jan11

Jul0

9

Jan10

Jul0

8

Jan09

Jul0

7

Jan08

Jan12

Jul1

2

Jul1

1

10.00

15.00

0.00

5.00

30.00

35.00

20.00

25.00

45.00

50.00

40.00

5 Year Pure SilverHigh 48.48 Low 8.92

0 925 0 925 25 90 925 0 958 26 3224

0 033 0 4224 12

. . .

. . .. . .

x yx y

y y

+ =+ =

= ⇒ = 88

28 12 8 15 2

g

g∴ = − =x . .

Question 8 (e)

Pure silver: 14 gMass of coin = 15 g

Purity = =1415

93 3gg

. %


Question 8 (a)

Pure silver is, of course, 100% pure silver.Britannia silver contains 95.8% pure silver.Sterling silver contains 92.5% pure silver.


P Fr n=

+

1

100

P =( )

=500001 03

418746.

P =( )

=200

1 061337.

million


The exploration company should sell in 2020 because you would need 133 million now to make this money. However, the multinational company is only offering you 120 million now.

Question 7 (c) (i)P =

( )=

2001 08

116 77.. million

The exploration company should take the offer now because you would need 116.7 million now to make this money. However, the multinational company is offering you 120 million now.

Question 7 (c) (ii)

Year Reduction in Billions

1 3 2 3 3 3 4 3 5 3 6 2 7 2 8 2 9 2 10 2

120 2001

1 200120

53

1 53

53

1

7

7

17

17

=+

+ = =

+ =

∴ =

−

( )

( )

i

i

i

i == =0 076 7 6. . %

P = =3

1 062 83

.. billion

Question 7 (c) (iii) Question 7 (d) (i)

Question 7 (d) (ii)

P = + + + + + + + +3

1 063

1 063

1 063

1 063

1 062

1 062

1 062

1 062

2 3 4 5 6 7 8. . . . . . . . 11 062

1 063

1 061 1

1 061

1 061

1 061

1 062

1

9 10

2 3 4

. .

. . . . . .

+

= + + + +

+

0061 1

1 061

1 061

1 061

1 063

1 062

1 061

6 2 3 4

6

+ + + +

= +

. . . .

. .++ + + +

11 06

11 06

11 06

11 062 3 4. . . .

= +

−

−

3

1 062

1 06

1 1 11 06

1 11 06

6

5

. ..

.

= 18 9. billion



Let x = Mass of Sterling silverLet y = Mass of Britannia silverMass of ring = 28 g

x yx y

+ = ×+ =28 0 925

0 925 0 958 26 3224( . )

. . .

1 31 10347681

31 10347681

4031 1034768

40

40 1 286

T g

T g

T g

g T

=

=

=

∴ =

.

.

.

.

S C T pC

T

pS

= × ×=

= =

== × × =

0 9428

31 10347680 9

260 94 0 9 26 22

.

..

$. . $

Question 8 (b)


10.00

15.00

0.00

5.00

30.00

35.00

20.00

25.00

45.00

50.00

40.00

US

D p

ertr

oy o

un

ce

Jul1

0

Jan11

Jul0

9

Jan10

Jul0

8

Jan09

Jul0

7

Jan08

Jan12

Jul1

2

Jul1

1

10.00

15.00

0.00

5.00

30.00

35.00

20.00

25.00

45.00

50.00

40.00

5 Year Pure SilverHigh 48.48 Low 8.92

0 925 0 925 25 90 925 0 958 26 3224

0 033 0 4224 12

. . .

. . .. . .

x yx y

y y

+ =+ =

= ⇒ = 88

28 12 8 15 2

g

g∴ = − =x . .

Question 8 (e)

Pure silver: 14 gMass of coin = 15 g

Purity = =1415

93 3gg

. %


Question 8 (a)

Pure silver is, of course, 100% pure silver.Britannia silver contains 95.8% pure silver.Sterling silver contains 92.5% pure silver.


P Fr n=

+

1

100

P =( )

=500001 03

418746.

P =( )

=200

1 061337.

million


The exploration company should sell in 2020 because you would need 133 million now to make this money. However, the multinational company is only offering you 120 million now.

Question 7 (c) (i)P =

( )=

2001 08

116 77.. million

The exploration company should take the offer now because you would need 116.7 million now to make this money. However, the multinational company is offering you 120 million now.

Question 7 (c) (ii)

Year Reduction in Billions

1 3 2 3 3 3 4 3 5 3 6 2 7 2 8 2 9 2 10 2

120 2001

1 200120

53

1 53

53

1

7

7

17

17

=+

+ = =

+ =

∴ =

−

( )

( )

i

i

i

i == =0 076 7 6. . %

P = =3

1 062 83

.. billion

Question 7 (c) (iii) Question 7 (d) (i)

Question 7 (d) (ii)

P = + + + + + + + +3

1 063

1 063

1 063

1 063

1 062

1 062

1 062

1 062

2 3 4 5 6 7 8. . . . . . . . 11 062

1 063

1 061 1

1 061

1 061

1 061

1 062

1

9 10

2 3 4

. .

. . . . . .

+

= + + + +

+

0061 1

1 061

1 061

1 061

1 063

1 062

1 061

6 2 3 4

6

+ + + +

= +

. . . .

. .++ + + +

11 06

11 06

11 06

11 062 3 4. . . .

= +

−

−

3

1 062

1 06

1 1 11 06

1 11 06

6

5

. ..

.

= 18 9. billion

Sample 3

Paper 1




V r h hr

= = ⇒ =16 1622p p

A r rh

r rr

rr

= +

= +

= +

2 2

2 2 16

2 32

2

22

2

p p

p p

p p

A r rdAdr

r r

rr

r

r

= +

= ⇒ − =

− = ⇒ =

∴ =

−

−

2 32

0 4 32 0

4 32 0 8

2

2 1

2

23

p p

p p

p p

mm

A rr

A

= +

= + = + =

2 32

2 2 322

8 16 24

2

2 2

p p

p p p p pMin. mm( )

dmdt

e

dm e dt

m e ct m e c c

t

t

t

= −

= −

= +

= = = + ⇒

−

−

−

∫ ∫

110

110

11010

0 10 10 10 0, : ==

∴ = −

0

101

10m e t mg

t m e e= = = =− −5 10 10 6 0651

10 5 0 5: .( ) . mg

m e e= + =− −10 10 14 5361

106

10 . mg

2 10 10

0 2 10 2

10

110

110

110

12

110

5

1

= +

= +

=+

=

− − +

− −

−

e e

e e

e

T T

T

T

e

( )

. ( ). .112449

0 1244910 0 12449 20 835

110− =

∴ = − =

TT

ln( . )ln( . ) . hours

Question 9 (a) (i)

r (mm)

h (mm)

Question 9 (a) (ii)

Question 9 (b) (ii)

Question 9 (a) (iii) Question 9 (a) (iv)

Question 9 (b) (i)


Question 9 (b) (iv)


S C T pC

T

pS

= × ×=

= =

== × × =

0 9660

31 10347681 93

12 500 96 1 93 12 5

.

..

$ .. . . $223 16.

S C T pCTpS

= × ×==== × × =

0 961 93

27 500 96 1 93 27 5 50 95

..

$ .. . . $ .

Question 8 (e) (i) Question 8 (e) (ii)

Minimum value = $8.92Maximum value = $48.48

% change =−

× =( . . )

.% . %48 48 8 92

8 92100 443 5

Question 8 (f)

You thought you were not getting the full market value for pure silver. The scrap dealer was right.

S C T pCTpS

C

C

= × ×===== × ×

∴ = =

?

$$

. %

13025

25 1 302530

83 33

9025

3 6 3 6 31 1034768 111 97

0 83 83 3

= = × =

= =

. . . .

. . %

T g

93.2111.97

Question 8 (g)




V r h hr

= = ⇒ =16 1622p p

A r rh

r rr

rr

= +

= +

= +

2 2

2 2 16

2 32

2

22

2

p p

p p

p p

A r rdAdr

r r

rr

r

r

= +

= ⇒ − =

− = ⇒ =

∴ =

−

−

2 32

0 4 32 0

4 32 0 8

2

2 1

2

23

p p

p p

p p

mm

A rr

A

= +

= + = + =

2 32

2 2 322

8 16 24

2

2 2

p p

p p p p pMin. mm( )

dmdt

e

dm e dt

m e ct m e c c

t

t

t

= −

= −

= +

= = = + ⇒

−

−

−

∫ ∫

110

110

11010

0 10 10 10 0, : ==

∴ = −

0

101

10m e t mg

t m e e= = = =− −5 10 10 6 0651

10 5 0 5: .( ) . mg

m e e= + =− −10 10 14 5361

106

10 . mg

2 10 10

0 2 10 2

10

110

110

110

12

110

5

1

= +

= +

=+

=

− − +

− −

−

e e

e e

e

T T

T

T

e

( )

. ( ). .112449

0 1244910 0 12449 20 835

110− =

∴ = − =

TT

ln( . )ln( . ) . hours

Question 9 (a) (i)

r (mm)

h (mm)

Question 9 (a) (ii)

Question 9 (b) (ii)

Question 9 (a) (iii) Question 9 (a) (iv)

Question 9 (b) (i)


Question 9 (b) (iv)


S C T pC

T

pS

= × ×=

= =

== × × =

0 9660

31 10347681 93

12 500 96 1 93 12 5

.

..

$ .. . . $223 16.

S C T pCTpS

= × ×==== × × =

0 961 93

27 500 96 1 93 27 5 50 95

..

$ .. . . $ .


Minimum value = $8.92Maximum value = $48.48

% change =−

× =( . . )

.% . %48 48 8 92

8 92100 443 5

Question 8 (f)

You thought you were not getting the full market value for pure silver. The scrap dealer was right.

S C T pCTpS

C

C

= × ×===== × ×

∴ = =

?

$$

. %

13025

25 1 302530

83 33

9025

3 6 3 6 31 1034768 111 97

0 83 83 3

= = × =

= =

. . . .

. . %

T g

93.2111.97

Question 8 (g)

Sample 3

Paper 1



Q

O(2, 0)

10c

P a b( , )Oc y xP a b c b a

PO

a b

a b

( , ):( , )

( ) ( )

( )

2 03

3

10

2 0 10

2

2

2

2 2

2 2

=

∈ ⇒ =

=

∴ − + − =

− + ==

− + =

− + + − =

− − =+ − =

∴ = −

102 3 104 4 3 10 0

6 02 3 0

2 3

2

2

2

( )

( )( )

,

a aa a aa aa a

abb bP Q

2 3 3 9 33 3 3 3= = ⇒ = ±

−( )

( , ), ( , )

A

C

B

F

E

D

G H

A x y B x y Ca b

( , ) ( , ), ( , ) ( , ), ?: :

( ) (

− = − = ==

∴− +

1 2 5 41 2

2 1 1 5

1 1 2 2

Ratio:))

( ) ( )

( , )

2 12 53

1

2 2 1 42 1

4 43

0

1 0

+=− +

=

∴+ −+

=−

=

∴C

D is the midpoint of BC:

B C

D

( , ), ( , )

, ( , )

5 4 1 05 1

24 02

3 2

−

=+ − +

= −


Question 2 (b)

Question 2 (c)

Draw a line AC.On line AC mark out three arcs of equal radius with a compass.Draw the line FB.Using a set square and ruler draw lines EH and DG parallel to FB.The line AB has been divided into three equal parts.

C BA D


OC is divided in the ratio a:b = 1:2. bx axb a

by ayb a

1 2 1 2++

++

,O x y C x y A

a bx

( , ) ( , ), ( , ), ( , ): :

( )

2 3 6 61 2

2 2 12 1

1 1 2 2

2

− = −=

∴++

=

Ratio:

66 4 18 14

2 3 12 1

6 6 18 12

14 12

2 2

22 2

⇒ + = ⇒ =

∴− ++

= − ⇒ − + = − ⇒ = −

∴ −

x x

y y y

C

( )

( , ))

Equation of second circle: Centre C rx y

( , ),( ) ( )

14 12 1014 12 1002 2

− =

− + + =

s x y

O r

: ( ) ( )

( , ),

− + + =

− = =

2 3 25

2 3 25 5

2 2

Centre

10

5O(2, 3)�

B( 2, )��

A(6, )��

C x( , )2

y2

t

Slope of OA: m16 36 2

34

34

=− − −

−=−

= −( )

Slope of t: m243

=



Centre Radius( , ),( ) ( )

h k rx h y k r− + − =2 2 2

A s( , ) ?( ) ( )

6 66 2 6 316 925

2 2

− ∈

− + − += +=

A O B( , ) ( , ) ( , )6 6 2 3 2 0− → − → −

Substitute A into s to see if it satisfies the equation of the circle.

Question 1 (b)

Question 1 (c)

Pass A through O by a central symmetry to find B.

Equation of t:

[Perpendicular slope]


h k rx h y k r− + − =2 2 2

Point dividing [PQ] in the ratio a:b

m x y A

y xy xt x y

= = −

+ = −

+ = −− − =

43 1 1

43

6 6

6 63 18 4 24

4 3 42 0

, ( ) ( , )

( )

:

,



Q

O(2, 0)

10c

P a b( , )Oc y xP a b c b a

PO

a b

a b

( , ):( , )

( ) ( )

( )

2 03

3

10

2 0 10

2

2

2

2 2

2 2

=

∈ ⇒ =

=

∴ − + − =

− + ==

− + =

− + + − =

− − =+ − =

∴ = −

102 3 104 4 3 10 0

6 02 3 0

2 3

2

2

2

( )

( )( )

,

a aa a aa aa a

abb bP Q

2 3 3 9 33 3 3 3= = ⇒ = ±

−( )

( , ), ( , )

A

C

B

F

E

D

G H

A x y B x y Ca b

( , ) ( , ), ( , ) ( , ), ?: :

( ) (

− = − = ==

∴− +

1 2 5 41 2

2 1 1 5

1 1 2 2

Ratio:))

( ) ( )

( , )

2 12 53

1

2 2 1 42 1

4 43

0

1 0

+=− +

=

∴+ −+

=−

=

∴C

D is the midpoint of BC:

B C

D

( , ), ( , )

, ( , )

5 4 1 05 1

24 02

3 2

−

=+ − +

= −


Question 2 (b)

Question 2 (c)

Draw a line AC.On line AC mark out three arcs of equal radius with a compass.Draw the line FB.Using a set square and ruler draw lines EH and DG parallel to FB.The line AB has been divided into three equal parts.

C BA D


OC is divided in the ratio a:b = 1:2. bx axb a

by ayb a

1 2 1 2++

++

,O x y C x y A

a bx

( , ) ( , ), ( , ), ( , ): :

( )

2 3 6 61 2

2 2 12 1

1 1 2 2

2

− = −=

∴++

=

Ratio:

66 4 18 14

2 3 12 1

6 6 18 12

14 12

2 2

22 2

⇒ + = ⇒ =

∴− ++

= − ⇒ − + = − ⇒ = −

∴ −

x x

y y y

C

( )

( , ))

Equation of second circle: Centre C rx y

( , ),( ) ( )

14 12 1014 12 1002 2

− =

− + + =

s x y

O r

: ( ) ( )

( , ),

− + + =

− = =

2 3 25

2 3 25 5

2 2

Centre

10

5O(2, 3)�

B( 2, )��

A(6, )��

C x( , )2

y2

t

Slope of OA: m16 36 2

34

34

=− − −

−=−

= −( )

Slope of t: m243

=




h k rx h y k r− + − =2 2 2

A s( , ) ?( ) ( )

6 66 2 6 316 925

2 2

− ∈

− + − += +=

A O B( , ) ( , ) ( , )6 6 2 3 2 0− → − → −

Substitute A into s to see if it satisfies the equation of the circle.

Question 1 (b)

Question 1 (c)

Pass A through O by a central symmetry to find B.

Equation of t:

[Perpendicular slope]


h k rx h y k r− + − =2 2 2

Point dividing [PQ] in the ratio a:b

m x y A

y xy xt x y

= = −

+ = −

+ = −− − =

43 1 1

43

6 6

6 63 18 4 24

4 3 42 0

, ( ) ( , )

( )

:

,

Sample 3

Paper 2



Correlation 0.72 –0.90 0.96 –0.42Graph C B A D

Graph BGraph A Graph DGraph C

Graph A: Very tight clustering indicating r is close to 1 and a positive slope.Graph B: Mostly tight clustering indicating r is close enough to 1 and a negative slope.Graph C: Clustering is moderate indicating r is somewhere around 0.75 and a positive slope.Graph D: Weaker clustering indicating r is less than 0.5 and a negative slope.


1 2 3 4�4 �3 �2 �10

z1

z

�z1

Is P(z > z1) = P(z < –z1)? Yes

P(0) = 0

P(z > 0) = 0.5 P(z < 0) = 0.5

P z( )−∞ < < +∞ =1

P z( ) .≥ =0 0 5 P z( ) .≤ =0 0 5



Question 4 (b) (iv)P z z P z z( ) ( )> = − ≤1 11

Question 4 (b) (v)


r

8

h

12

4

8

h

12

4

r� �

124 4

34

12 3

=−

=−

∴ − =

hr

hrr h

V r h r rr r

r r

= = − =

− =

− + =

π π π2 2

2 3

3 2

12 3 2412 3 24

4 8 0

( ) f r r rffr

h rh

( )( )( )

= − += − + ≠= − + =

∴ == −

∴ = − =

3 24 81 1 4 8 02 8 16 8 0

212 3

12 6 6


The two highlighted triangles are similar triangles.

Question 3 (a) Question 3 (b) Question 3 (c)



Correlation 0.72 –0.90 0.96 –0.42Graph C B A D

Graph BGraph A Graph DGraph C

Graph A: Very tight clustering indicating r is close to 1 and a positive slope.Graph B: Mostly tight clustering indicating r is close enough to 1 and a negative slope.Graph C: Clustering is moderate indicating r is somewhere around 0.75 and a positive slope.Graph D: Weaker clustering indicating r is less than 0.5 and a negative slope.


1 2 3 4�4 �3 �2 �10

z1

z

�z1

Is P(z > z1) = P(z < –z1)? Yes

P(0) = 0

P(z > 0) = 0.5 P(z < 0) = 0.5

P z( )−∞ < < +∞ =1

P z( ) .≥ =0 0 5 P z( ) .≤ =0 0 5



Question 4 (b) (iv)P z z P z z( ) ( )> = − ≤1 11

Question 4 (b) (v)


r

8

h

12

4

8

h

12

4

r� �

124 4

34

12 3

=−

=−

∴ − =

hr

hrr h

V r h r rr r

r r

= = − =

− =

− + =

π π π2 2

2 3

3 2

12 3 2412 3 24

4 8 0

( ) f r r rffr

h rh

( )( )( )

= − += − + ≠= − + =

∴ == −

∴ = − =

3 24 81 1 4 8 02 8 16 8 0

212 3

12 6 6


The two highlighted triangles are similar triangles.

Question 3 (a) Question 3 (b) Question 3 (c) Sample 3

Paper 2



Question 6 (25 marks)Question 6 (a)Given: Three parallel lines AB, CD and EF such that C is on AE and D is on BF with |AC| = |CE|.

to prove: |BD| = |DF|.construction: Draw a line GH through D parallel to AE such that G is on AB and H is on EF.

proof: ACDG is a parallelogram ⇒ = =AC GD CE

CEHD is a parallelogram ⇒ =CE DH

∴ =GD DH

Now triangle GDB and triangle FDH are congruent (ASA) because:

∠ = ∠ =BGD FHD X [Alternate angles]

∠ = ∠ =GDB FDH Y [Vertically opposite angles]

GD DH= [Already proved]

∴ =BD DF

A

F

D

E

C

B

G

H

X

Y

Y

X

A

F

D

E

C

B


G: Germinate, NG: Not germinateP(G) = 0.6, P(NG) = 0.4

P(G, G, G, G) = = =( . )( . )( . )( . ) ( . )0 6 0 6 0 6 0 6 0 6 81625

4

P(G, NG, NG, NG) = × =( . )( . )( . )( . ) !!

0 6 0 4 0 4 0 4 43

96625

P(NG, NG, NG, NG) = = =( . )( . )( . )( . ) ( . )0 4 0 4 0 4 0 4 0 4 16625

4

Independence is necessary to do the calculation in (a) as the probability would change depending on whether or not the other seeds germinate. Independence may not be valid as the growing conditions (temperature, moisture and soil type) may vary from seed to seed.

Expected value = xP(x) = 4(0.6) = 2.4


Question 5 (a) (i)

Question 5 (a) (ii)


Question 5 (b)

Question 5 (c)

Question 5 (d)Null hypothesis Ho: P = 0.9Alternative hypothesis H1: P < 0.9

Sample proportion = p = 97120

True population proportion = Sample proportion ± 1.96(Standard error of the proportion)

Standard error of the proportion = -p pn

( )1

True population proportion = ±-

=97120

1 961120

0 797

12097

120.( )

. 338 0 879, .

Confidence interval: 0.738 ↔ 0.879There is evidence to support the gardener’s claim that less than 90% of the seeds germinate because, based on the sample data, any values in the range 73.8% − 87.9% are possible values for the proportion of seeds in the sample that germinate.P = 0.9 is not in the confidence interval. At the 5% significance level, we accept the alternative hypothesis and agree that the gardener’s suspicions are well-founded.



Question 6 (25 marks)Question 6 (a)Given: Three parallel lines AB, CD and EF such that C is on AE and D is on BF with |AC| = |CE|.

to prove: |BD| = |DF|.construction: Draw a line GH through D parallel to AE such that G is on AB and H is on EF.

proof: ACDG is a parallelogram ⇒ = =AC GD CE

CEHD is a parallelogram ⇒ =CE DH

∴ =GD DH

Now triangle GDB and triangle FDH are congruent (ASA) because:

∠ = ∠ =BGD FHD X [Alternate angles]

∠ = ∠ =GDB FDH Y [Vertically opposite angles]

GD DH= [Already proved]

∴ =BD DF

A

F

D

E

C

B

G

H

X

Y

Y

X

A

F

D

E

C

B


G: Germinate, NG: Not germinateP(G) = 0.6, P(NG) = 0.4

P(G, G, G, G) = = =( . )( . )( . )( . ) ( . )0 6 0 6 0 6 0 6 0 6 81625

4

P(G, NG, NG, NG) = × =( . )( . )( . )( . ) !!

0 6 0 4 0 4 0 4 43

96625

P(NG, NG, NG, NG) = = =( . )( . )( . )( . ) ( . )0 4 0 4 0 4 0 4 0 4 16625

4

Independence is necessary to do the calculation in (a) as the probability would change depending on whether or not the other seeds germinate. Independence may not be valid as the growing conditions (temperature, moisture and soil type) may vary from seed to seed.

Expected value = xP(x) = 4(0.6) = 2.4


Question 5 (a) (i)

Question 5 (a) (ii)


Question 5 (b)

Question 5 (c)

Question 5 (d)Null hypothesis Ho: P = 0.9Alternative hypothesis H1: P < 0.9

Sample proportion = p = 97120



( )1

True population proportion = ±-

=97120

1 961120

0 797

12097

120.( )

. 338 0 879, .

Confidence interval: 0.738 ↔ 0.879There is evidence to support the gardener’s claim that less than 90% of the seeds germinate because, based on the sample data, any values in the range 73.8% − 87.9% are possible values for the proportion of seeds in the sample that germinate.P = 0.9 is not in the confidence interval. At the 5% significance level, we accept the alternative hypothesis and agree that the gardener’s suspicions are well-founded.

Sample 3

Paper 2



400

100 200 300 400 500 6000

100

200

300

800

500

600

700

Fore

ign I

ncom

e (

Bil

lions o

f euro

)

Domestic Income (Billions of euro)

700

1000

900

1100

1200

( , )x y



r = 0.6378

Question 7 (b)


Question 6 (b)

x y y xx y x y y x

x xxx y

+ = ⇒ = −

= ⇒ = = =−

∴ = −=

∴ = =

5 5

4 646

23

2 53

3 10 25 10

2 3

( )

,

4

m

n

12

l

x

y

z

w6

z w z wxy

zw

ww

w w w

w

+ = ⇒ = −

=

=−

⇒ = − ⇒ =

∴ =

13 13

23

13 2 39 3 5 39

39

( )Pythagoras

55265

, z =



400

100 200 300 400 500 6000

100

200

300

800

500

600

700

Fore

ign I

ncom

e (

Bil

lions o

f euro

)


700

1000

900

1100

1200

( , )x y



r = 0.6378

Question 7 (b)


Question 6 (b)

x y y xx y x y y x

x xxx y

+ = ⇒ = −

= ⇒ = = =−

∴ = −=

∴ = =

5 5

4 646

23

2 53

3 10 25 10

2 3

( )

,

4

m

n

12

l

x

y

z

w6

z w z wxy

zw

ww

w w w

w

+ = ⇒ = −

=

=−

⇒ = − ⇒ =

∴ =

13 13

23

13 2 39 3 5 39

39

( )Pythagoras

55265

, z =

Sample 3

Paper 2



400

100 200 300 400 500 6000

100

200

300

800

500

600

700

Fore

ign I

ncom

e (

Bil

lions o

f euro

)


700

1000

900

1100

1200

( , )x y

The slope is now negative, there is a negative moderate correlation. The difference is that as domestic yearly income increases the foreign yearly income decreases.

Question 7 (f) (iii)

Question 7 (f) (iv)


r = –0.5085

y yr

x x

y x

y

y

x

= + −

= +−

−

= −

σσ

( )

( . ) ( )

. (

442 0 5085 16054

230

442 1 51 xxy xy x

−= − += − +

230442 1 51 347 3

1 51 789 3

). .

. .

Slope m = −1.5

Question 7 (e) (iii) Question 7 (f) (i)

Question 7 (f) (ii)

CaSio CaLCuLator (fx-85GT PLUS)Press Shift followed by the Number 1Press 5: RegPress 3: rPress = [Write down the answer]

x

y

=+ + + + + + + + +

=

=+ +

150 170 180 200 250 260 280 320 600 26010

267

400 580 7000 380 500 200 400 220 1200 60010

518

267 518

+ + + + + + +=

=( , ) ( , )x y

( , ) ( , ), ( , ) ( , )

.

x y x y

m

= =

=−−

=

267 518 600 10001000 518600 267

1 4

2 2

As the yearly domestic income increases the foreign yearly income increases.

Outlier (600, 1200); Possible answer: A boom for the economy due to a huge tourist influx and a huge increase in computer/pharmaceutical exports due to the exchange rate of the euro.

Question 7 (c)

Question 7 (d)

Question 7 (e)Remove (600, 1200) from the table and input the results again into your calculator. After pressing the AC button do the following:

x

x

==23054σ

y

y

==442160σ


CaSio CaLCuLator (fx-85GT PLUS)Press Shift followed by the Number 1Press 4: VarPress 2: xPress = [Write down the answer]Press Shift followed by the Number 1Press 4: VarPress 3: σ xPress = [Write down the answer]

CaSio CaLCuLator (fx-85GT PLUS)Press Shift followed by the Number 1Press 4: VarPress 3: yPress = [Write down the answer]Press Shift followed by the Number 1Press 4: VarPress 3: σ yPress = [Write down the answer]



400

100 200 300 400 500 6000

100

200

300

800

500

600

700

Fore

ign I

ncom

e (

Bil

lions o

f euro

)


700

1000

900

1100

1200

( , )x y

The slope is now negative, there is a negative moderate correlation. The difference is that as domestic yearly income increases the foreign yearly income decreases.

Question 7 (f) (iii)

Question 7 (f) (iv)


r = –0.5085

y yr

x x

y x

y

y

x

= + −

= +−

−

= −

σσ

( )

( . ) ( )

. (

442 0 5085 16054

230

442 1 51 xxy xy x

−= − += − +

230442 1 51 347 3

1 51 789 3

). .

. .

Slope m = −1.5

Question 7 (e) (iii) Question 7 (f) (i)

Question 7 (f) (ii)

CaSio CaLCuLator (fx-85GT PLUS)Press Shift followed by the Number 1Press 5: RegPress 3: rPress = [Write down the answer]

x

y

=+ + + + + + + + +

=

=+ +

150 170 180 200 250 260 280 320 600 26010

267

400 580 7000 380 500 200 400 220 1200 60010

518

267 518

+ + + + + + +=

=( , ) ( , )x y

( , ) ( , ), ( , ) ( , )

.

x y x y

m

= =

=−−

=

267 518 600 10001000 518600 267

1 4

2 2

As the yearly domestic income increases the foreign yearly income increases.

Outlier (600, 1200); Possible answer: A boom for the economy due to a huge tourist influx and a huge increase in computer/pharmaceutical exports due to the exchange rate of the euro.

Question 7 (c)

Question 7 (d)

Question 7 (e)Remove (600, 1200) from the table and input the results again into your calculator. After pressing the AC button do the following:

x

x

==23054σ

y

y

==442160σ


CaSio CaLCuLator (fx-85GT PLUS)Press Shift followed by the Number 1Press 4: VarPress 2: xPress = [Write down the answer]Press Shift followed by the Number 1Press 4: VarPress 3: σ xPress = [Write down the answer]

CaSio CaLCuLator (fx-85GT PLUS)Press Shift followed by the Number 1Press 4: VarPress 3: yPress = [Write down the answer]Press Shift followed by the Number 1Press 4: VarPress 3: σ yPress = [Write down the answer]

Sample 3

Paper 2



P tt

t t

= + ==

= ⇒ = = =−

100 20 6 11020 6 10

6 6 603

12

1 12

coscos

cos cos ( ) o π

AS

CT

63

73

53

113

187

t

t

=

=

=

π π

π π

π

, [ ]

, [ ]

,

First quadrant

Fourth quadrant

ππ

π π18

518

1118

= ,

P tdPdt

t t

dPdt t

= +

= − × = −

= −=

100 20 6

20 6 6 120 6

1200

cos

sin sin

sinn ( )

sin sin

6 0 0

120 615

120 25

15

=

= −

= −

=

dPdt t π

π π== −

= −

= −

=

114

120 65

120 65

5

mm Hg

dPdt t π

π πsin sin ==

= −

= − =

=

70 5

120 63

120 2 03

.

sin sin

mm Hg

mm HgdPdt t π

ππ

Question 8 (c) (ii)

63

73

53

113

187

t

t

=

=

=

π π

π π

π

, [ ]

, [ ]

,

First quadrant

Fourth quadrant

ππ

π π18

518

1118

= ,

Question 8 (d)

A

S

B

105

10

15

A: People with high blood pressureB: High level of cholesterol

People with high blood pressure and high level of cholesterol = 10

P

P

P

( )

( )

(

High BP

High level of cholesterol

High

= =

= =

1540

38

2540

58

BBP high level of cholesterol

High BP high l

and

or

)

(

= =1040

14

P eevel of cholesterol) = =3040

34

P A B

P A P B P A B

( )

( ) ( ) ( )

or

and

=

+ − = + − =

34

38

58

14

34

(b)

(c)

(d)

(e)

(f)

(a)



P tt

P

= +=

∴ = + =

100 20 66 1

100 20 1 120

cos: cos

( )Maximum value

mm Hg (SP)

90

t (s)

P (mm Hg)

80

100

110

120

�

4

�

12

�

6

�

3

7�

12

�

2

5�

12

2�

3

�

18

5�

18

7�

18

11�

18

Range = [80, 120]

Period =π3

Duration = π3

s

Number of beats per minute = = =60 180 5713π π


P tt

P

= += −

∴ = + − =

100 20 66 1

100 20 1 80

cos: cos

( )Minimum value

mm Hg (DP))

Question 8 (b) (i)

Question 8 (b) (iii) Question 8 (b) (iv)

Draw a line through P = 100 mm Hg and read off the times as shown above.

P = 110 mm Hg ⇒ =t π π π π18

518

718

1118

s s s s, , ,

Question 8 (c) (i)

t (s) 0 π12

π6

π4

π3

512π π

2712π 2

3π

P (mm Hg) 120 100 80 100 120 100 80 100 120

Question 8 (b) (ii)



P tt

t t

= + ==

= ⇒ = = =−

100 20 6 11020 6 10

6 6 603

12

1 12

coscos

cos cos ( ) o π

AS

CT

63

73

53

113

187

t

t

=

=

=

π π

π π

π

, [ ]

, [ ]

,

First quadrant

Fourth quadrant

ππ

π π18

518

1118

= ,

P tdPdt

t t

dPdt t

= +

= − × = −

= −=

100 20 6

20 6 6 120 6

1200

cos

sin sin

sinn ( )

sin sin

6 0 0

120 615

120 25

15

=

= −

= −

=

dPdt t π

π π== −

= −

= −

=

114

120 65

120 65

5

mm Hg

dPdt t π

π πsin sin ==

= −

= − =

=

70 5

120 63

120 2 03

.

sin sin

mm Hg

mm HgdPdt t π

ππ

Question 8 (c) (ii)

63

73

53

113

187

t

t

=

=

=

π π

π π

π

, [ ]

, [ ]

,

First quadrant

Fourth quadrant

ππ

π π18

518

1118

= ,

Question 8 (d)

A

S

B

105

10

15

A: People with high blood pressureB: High level of cholesterol

People with high blood pressure and high level of cholesterol = 10

P

P

P

( )

( )

(

High BP

High level of cholesterol

High

= =

= =

1540

38

2540

58

BBP high level of cholesterol

High BP high l

and

or

)

(

= =1040

14

P eevel of cholesterol) = =3040

34

P A B

P A P B P A B

( )

( ) ( ) ( )

or

and

=

+ − = + − =

34

38

58

14

34

(b)

(c)

(d)

(e)

(f)

(a)



P tt

P

= +=

∴ = + =

100 20 66 1

100 20 1 120

cos: cos

( )Maximum value

mm Hg (SP)

90

t (s)

P (mm Hg)

80

100

110

120

�

4

�

12

�

6

�

3

7�

12

�

2

5�

12

2�

3

�

18

5�

18

7�

18

11�

18

Range = [80, 120]

Period =π3

Duration = π3

s

Number of beats per minute = = =60 180 5713π π


P tt

P

= += −

∴ = + − =

100 20 66 1

100 20 1 80

cos: cos

( )Minimum value

mm Hg (DP))

Question 8 (b) (i)

Question 8 (b) (iii) Question 8 (b) (iv)

Draw a line through P = 100 mm Hg and read off the times as shown above.

P = 110 mm Hg ⇒ =t π π π π18

518

718

1118

s s s s, , ,

Question 8 (c) (i)

t (s) 0 π12

π6

π4

π3

512π π

2712π 2

3π

P (mm Hg) 120 100 80 100 120 100 80 100 120

Question 8 (b) (ii)

Sample 3

Paper 2



( ) ( )

( ) ( )( )

2 3 4 2 04 0

4 4 2 3 2

2

2

2

− + − + =

− <

− − −

k x k xb ac

k kNo real roots:

<<

− + − + <

+ <+ <

016 8 16 24 0

16 016 0

2

2

k k kk kk k( )

k k k( ) ,+ = ⇒ = −16 0 16 0

ans: −16 < k < 0


x x

x xx x x

x xx x

x

+ + = −

+ = − +

+ = + +

− − =+ − =

5 19 1

5 19 15 19 2 1

3 18 03 6 0

2

2

( )

( )( )== −

= − − + = − + = −

= + = + =

3 6

3 3 4 3 2 1

6 6 49 6 7 13

,

:

:

Check solutions:

x

xans: x = −3


answer: (3, −2, 4)

loglog

(log ) log(log )(log )

lo

22

22

2

2 2

12 7

7 12 03 4 0

xx

x xx x

+ =

− + =− − =

∴ gglog

23

24

3 2 84 2 16

x xx x= ⇒ = =

∴ = ⇒ = =

2 3 207 233 3

x y zx y zx y z

− + =+ + =+ − =

.....( ).......( )

.........( )

(

123

11 2 42 3 5) ( ) : ....( )

( ) ( ) : ....( )( )+ + =− + = ×−

+ =

9 4 434 2 20 2

9 4 4

x zx z

x z 338 4 40

3

27 4 434 16 4

6 12 202

− − = −=

+ == ⇒ =

− + =∴ = −

x zx

zz z

yy


Let axb c

byc a

cza b

k

ax k b cby k c acz k a bax by cz

−=

−=

−=

= −= −= −+ + =

( )( )( )

kkb kc kc ka ka kb− + − + − = 0


z i z iz i z iz i

= − − ⇒ + += − + ⇒ + −

∴ + +

3 33 33

( )( )

( )

is a factoris a factor

(( )z iz z iz z i iz i iz zz z

+ − =

+ − + + − + + − =

+ + + =

+ + =

3 03 3 9 3 3 06 9 1 06 10 0

2 2

2

2

z i z i= − − ⇒ = − +3 3

ORRoots:Sum Product

− − − += −

=

− + =

∴ + + =

3 36

100

6 10 0

2

2

i iSP

z Sz Pz z

,

If z is a root of the cubic equation, its conjugate is also a root. This is because the coefficients in the cubic are all real. Therefore, z z2 6 10 0+ + = is a factor of the cubic equation.

az z bz z z azaz z bz az a z

3 2 2

3 2 3 2

22 40 6 10 422 40 6 4

+ + + = + + +

+ + + = + +

( )( )( ) ++ + +

∴ = + ⇒ =+ = ⇒ =

∴ + + + =

( )

(

10 24 4022 6 4 3

10 24 543 22 54 403 2

a za a

a b bz z z z22

43

6 10 3 4 03 3

+ + + == − − − + −

z zz i i

)( ), ,



Question 1 (a) (ii)

Question 1 (b)

z a bi z a bi= + ⇒ = −



( ) ( )

( ) ( )( )

2 3 4 2 04 0

4 4 2 3 2

2

2

2

− + − + =

− <

− − −

k x k xb ac

k kNo real roots:

<<

− + − + <

+ <+ <

016 8 16 24 0

16 016 0

2

2

k k kk kk k( )

k k k( ) ,+ = ⇒ = −16 0 16 0

ans: −16 < k < 0


x x

x xx x x

x xx x

x

+ + = −

+ = − +

+ = + +

− − =+ − =

5 19 1

5 19 15 19 2 1

3 18 03 6 0

2

2

( )

( )( )== −

= − − + = − + = −

= + = + =

3 6

3 3 4 3 2 1

6 6 49 6 7 13

,

:

:

Check solutions:

x

xans: x = −3


answer: (3, −2, 4)

loglog

(log ) log(log )(log )

lo

22

22

2

2 2

12 7

7 12 03 4 0

xx

x xx x

+ =

− + =− − =

∴ gglog

23

24

3 2 84 2 16

x xx x= ⇒ = =

∴ = ⇒ = =

2 3 207 233 3

x y zx y zx y z

− + =+ + =+ − =

.....( ).......( )

.........( )

(

123

11 2 42 3 5) ( ) : ....( )

( ) ( ) : ....( )( )+ + =− + = ×−

+ =

9 4 434 2 20 2

9 4 4

x zx z

x z 338 4 40

3

27 4 434 16 4

6 12 202

− − = −=

+ == ⇒ =

− + =∴ = −

x zx

zz z

yy


Let axb c

byc a

cza b

k

ax k b cby k c acz k a bax by cz

−=

−=

−=

= −= −= −+ + =

( )( )( )

kkb kc kc ka ka kb− + − + − = 0


z i z iz i z iz i

= − − ⇒ + += − + ⇒ + −

∴ + +

3 33 33

( )( )

( )

is a factoris a factor

(( )z iz z iz z i iz i iz zz z

+ − =

+ − + + − + + − =

+ + + =

+ + =

3 03 3 9 3 3 06 9 1 06 10 0

2 2

2

2

z i z i= − − ⇒ = − +3 3

ORRoots:Sum Product

− − − += −

=

− + =

∴ + + =

3 36

100

6 10 0

2

2

i iSP

z Sz Pz z

,

If z is a root of the cubic equation, its conjugate is also a root. This is because the coefficients in the cubic are all real. Therefore, z z2 6 10 0+ + = is a factor of the cubic equation.

az z bz z z azaz z bz az a z

3 2 2

3 2 3 2

22 40 6 10 422 40 6 4

+ + + = + + +

+ + + = + +

( )( )( ) ++ + +

∴ = + ⇒ =+ = ⇒ =

∴ + + + =

( )

(

10 24 4022 6 4 3

10 24 543 22 54 403 2

a za a

a b bz z z z22

43

6 10 3 4 03 3

+ + + == − − − + −

z zz i i

)( ), ,



Question 1 (a) (ii)

Question 1 (b)

z a bi z a bi= + ⇒ = −

Sample 4

Paper 1



injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is an injective function which means that every y value has its own unique matching x value.

injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is a bijective function as there is a perfect one to one correspondence between the x and y values. A bijective function means it is both an injective and surjective function.

Question 5 (c)

Domain = {0, 1, 2, 3, 4, 5,....}Range = {0, 1, 4, 9, 16,.......}Every element in the domain matches to a unique element in the range. There are elements in the range that do not have a matching element from the domain.

Question 5 (d)

Domain = {Positive real numbers}Range = {Positive real numbers}Every element in the domain matches to a unique element in the range. Every element in the range has a unique matching element from the domain.

✓

✓

✓

✓

injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is a surjective function which means that every y value has at least one matching x value. It is not injective as many y values have more than one correspoding x value.

injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is not a function because x values have two y values.

Question 5 (e)

Domain = {All real numbers}Range = {Positive real numbers}Every element in the domain matches to at least one element in the range. Some values in the range match to two elements in the range. For example, 22 and (−2)2 both map on to 4.

x

y

Question 5 (f)

✓


11

11

132

13

19

127

13

13

, , , ,.......,a r

S ar

= =

=−

=−

=∞

a d a a da d a a d a a

d dd d

− +− + + + = = ⇒ =

− +

− + + + =

, ,

, ,( ) ( )

3 27 9

9 9 99 9 9 293

8

2 2 2

11 18 81 81 18 2932 293 2432 50

255

2 2

2

2

2

− + + + + + =

= −

=

== ±

d d d ddddd (Use eitther)

Numbers: 4, 9, 14



injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion


injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is a surjective function which means that every y value has at least one matching x value. It is not injective as many y values have more than one corresponding x value.

x

y

y f x= ( )

x

y

y g x= ( )

Question 5 (a)

Question 5 (b)

✓

✓

✓

✓



injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is an injective function which means that every y value has its own unique matching x value.

injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion


Question 5 (c)

Domain = {0, 1, 2, 3, 4, 5,....}Range = {0, 1, 4, 9, 16,.......}Every element in the domain matches to a unique element in the range. There are elements in the range that do not have a matching element from the domain.

Question 5 (d)

Domain = {Positive real numbers}Range = {Positive real numbers}Every element in the domain matches to a unique element in the range. Every element in the range has a unique matching element from the domain.

✓

✓

✓

✓

injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is a surjective function which means that every y value has at least one matching x value. It is not injective as many y values have more than one correspoding x value.

injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is not a function because x values have two y values.

Question 5 (e)

Domain = {All real numbers}Range = {Positive real numbers}Every element in the domain matches to at least one element in the range. Some values in the range match to two elements in the range. For example, 22 and (−2)2 both map on to 4.

x

y

Question 5 (f)

✓


11

11

132

13

19

127

13

13

, , , ,.......,a r

S ar

= =

=−

=−

=∞

a d a a da d a a d a a

d dd d

− +− + + + = = ⇒ =

− +

− + + + =

, ,

, ,( ) ( )

3 27 9

9 9 99 9 9 293

8

2 2 2

11 18 81 81 18 2932 293 2432 50

255

2 2

2

2

2

− + + + + + =

= −

=

== ±

d d d ddddd (Use eitther)

Numbers: 4, 9, 14



injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion


injeCtive FunCtion

SurjeCtive FunCtion

BijeCtive FunCtion

This is a surjective function which means that every y value has at least one matching x value. It is not injective as many y values have more than one corresponding x value.

x

y

y f x= ( )

x

y

y g x= ( )

Question 5 (a)

Question 5 (b)

✓

✓

✓

✓

Sample 4

Paper 1



Height of player + arm span above head = 1.85 m + 0.8 m = 2.65 my t tt y= − −= =

3 3 13 4 90 3 3

2. .: . m

Height above base of racket = 3.3 − 2.65 = 0.65 m

y t ty t t

t

= − −

= + − =

=− ± − −

3 3 13 4 90 4 9 13 3 3 0

13 13 4 4 9 3 32

2

2

2 2

. .: . .

( . )( . )(( . )

.4 9

0 23= s

x tt x== = =

700 23 70 0 23 16 1. : ( . ) . m

Distance from baseline to far serve line = 11.885 m + 6.4 m = 18.285 mBall is in as 16.1 m < 18.285 m

x tx t t== = ⇒ =

7011 885 11 885 70 0 17. : . . s

y t ty t tt t

= − −

= = − −

+ − =

3 3 13 4 90 914 0 914 3 3 13 4 9

4 9 13 2 386

2

2

2

. .

. : . . .. . 00

13 13 4 4 9 2 3862 4 9

0 1722

t =− ± −

=( . )( . )

( . ). s

x btt b b== = ⇒ =0 172 11 885 0 172 69 1. : . ( . ) .


6.4 mO

Service line

11.885 m

Net

B

0.914 m

( , )x y

A

C

Question 7 (a)

Question 7 (b)

Question 7 (c)

Question 7 (d)

Question 7 (e)y t tt y= − −

= = − − =

3 3 13 4 90 17 3 3 13 0 17 4 9 0 17 0 948

2

2

. .. : . ( . ) . ( . ) .s m

Yes, it will clear the net.

Question 7 (f) (i) Question 7 (f) (ii)



f (x) is continuous as there are no gaps.g(x) is not continuous at there is a gap at x = 0.

y f x x xf x xf x xx

= = − + +′ = − +′ = ⇒ − + =

− +

( )( )( )

76

2 176

313

73

176

73

1760 0

14 117 01714

=∴ =x

y f x x x

y x x

x xx

= = − + +

= ⇒ − + + =

− − =−

( )

(

76

2 176

313

76

2 176

313

2

0 0

7 17 62 07 311 2 0

22 0 0

317

317

)( ),

( , ), ( , )

xxA B

+ == −

∴ −

y

x

y f x= ( )

y g x= ( )

y g x= ( )

A

C

D

B

Question 6 (a)

Question 6 (b)

Question 6 (c)

Question 6 (d)

Question 6 (e)

( , ) ( )?( )

( ) ( ) ( )

1 12

1 1 1

76

2 176

313

76

2 176

313

76

∈

= − + +

= − + +

= −

f xf x x x

f++ + = =

∈

=

= =

176

313

363 12

1 1212

1 121

12

( )

( , ) ( )?

( )

( ) (

True

True

g x

g xx

g ))

( , ) ( )?( )

( ) ( ) ( )

4 3

4 4 4

76

2 176

313

76

2 176

313

563

∈

= − + +

= − + +

= −

f xf x x x

f++ + = =

∈

=

= =

343

313

93 3

4 312

4 124

3

( )

( , ) ( )?

( )

( ) ( )

True

True

g x

g xx

g

A f x g x dx

x xxdx

x

= −

= − + + −

= − +

∫

∫

( ( ) ( ))

[

1

4

76

2 176

3131

4

718

3 1

12

7712

2 313 1

4

718

3 1712

2 313

12

4 4 4 12 4

x x x+ −

= − + + − − −

ln ]

{ ( ) ( ) ( ) ln( )} { 7718

3 1712

2 313

1114

1 1 1 12 112 4 11 1

( ) ( ) ( ) ln( )}ln .

+ + −

= − ≈



Height of player + arm span above head = 1.85 m + 0.8 m = 2.65 my t tt y= − −= =

3 3 13 4 90 3 3

2. .: . m

Height above base of racket = 3.3 − 2.65 = 0.65 m

y t ty t t

t

= − −

= + − =

=− ± − −

3 3 13 4 90 4 9 13 3 3 0

13 13 4 4 9 3 32

2

2

2 2

. .: . .

( . )( . )(( . )

.4 9

0 23= s

x tt x== = =

700 23 70 0 23 16 1. : ( . ) . m

Distance from baseline to far serve line = 11.885 m + 6.4 m = 18.285 mBall is in as 16.1 m < 18.285 m

x tx t t== = ⇒ =

7011 885 11 885 70 0 17. : . . s

y t ty t tt t

= − −

= = − −

+ − =

3 3 13 4 90 914 0 914 3 3 13 4 9

4 9 13 2 386

2

2

2

. .

. : . . .. . 00

13 13 4 4 9 2 3862 4 9

0 1722

t =− ± −

=( . )( . )

( . ). s

x btt b b== = ⇒ =0 172 11 885 0 172 69 1. : . ( . ) .


6.4 mO

Service line

11.885 m

Net

B

0.914 m

( , )x y

A

C

Question 7 (a)

Question 7 (b)

Question 7 (c)

Question 7 (d)

Question 7 (e)y t tt y= − −

= = − − =

3 3 13 4 90 17 3 3 13 0 17 4 9 0 17 0 948

2

2

. .. : . ( . ) . ( . ) .s m

Yes, it will clear the net.

Question 7 (f) (i) Question 7 (f) (ii)



f (x) is continuous as there are no gaps.g(x) is not continuous at there is a gap at x = 0.

y f x x xf x xf x xx

= = − + +′ = − +′ = ⇒ − + =

− +

( )( )( )

76

2 176

313

73

176

73

1760 0

14 117 01714

=∴ =x

y f x x x

y x x

x xx

= = − + +

= ⇒ − + + =

− − =−

( )

(

76

2 176

313

76

2 176

313

2

0 0

7 17 62 07 311 2 0

22 0 0

317

317

)( ),

( , ), ( , )

xxA B

+ == −

∴ −

y

x

y f x= ( )

y g x= ( )

y g x= ( )

A

C

D

B

Question 6 (a)

Question 6 (b)

Question 6 (c)

Question 6 (d)

Question 6 (e)

( , ) ( )?( )

( ) ( ) ( )

1 12

1 1 1

76

2 176

313

76

2 176

313

76

∈

= − + +

= − + +

= −

f xf x x x

f++ + = =

∈

=

= =

176

313

363 12

1 1212

1 121

12

( )

( , ) ( )?

( )

( ) (

True

True

g x

g xx

g ))

( , ) ( )?( )

( ) ( ) ( )

4 3

4 4 4

76

2 176

313

76

2 176

313

563

∈

= − + +

= − + +

= −

f xf x x x

f++ + = =

∈

=

= =

343

313

93 3

4 312

4 124

3

( )

( , ) ( )?

( )

( ) ( )

True

True

g x

g xx

g

A f x g x dx

x xxdx

x

= −

= − + + −

= − +

∫

∫

( ( ) ( ))

[

1

4

76

2 176

3131

4

718

3 1

12

7712

2 313 1

4

718

3 1712

2 313

12

4 4 4 12 4

x x x+ −

= − + + − − −

ln ]

{ ( ) ( ) ( ) ln( )} { 7718

3 1712

2 313

1114

1 1 1 12 112 4 11 1

( ) ( ) ( ) ln( )}ln .

+ + −

= − ≈

Sample 4

Paper 1



y mx cy t b a

c aa a

m

= += +

=

= ⇒ = =

log log log

log

. log .

10 10 10

10

102 62 6 10 398

==

=−

log( , . ), ( , . )

. .

10

3 2 875 6 3 1553 155 2 875

6

b

m

Points on graph:

−−=

= ⇒ = =3

0 093

0 093 10 1 24100 093

.

. log ..b b

398

y ab

t

t

t

t

=

= ×

=

=

2200 398 1 242200398

1 24

2200398

110 10

.

.

log log ..

log

log ..

24

22003981 24

7 9 810

10

∴ =

= ≈t years

Question 8 (c)

Question 8 (d)

Question 8 (e)


t (Number of years after 2006) 1 2 3 4 5 6y (Cost ) 500 610 720 890 1050 1540log10 y 2.70 2.79 2.86 2.95 3.02 3.19

y aby aby a by a t

t

t

t

=

=

= += +

log log ( )

log log loglog log

10 10

10 10 10

10 10 llog10 b

1 2 3 4 5 6 t (years)0

2.00

1.00

2.875

log10

y

4.00

3.00

3.155

2.600


Question 8 (b) (i)

Question 8 (b) (ii)



y mx cy t b a

c aa a

m

= += +

=

= ⇒ = =

log log log

log

. log .

10 10 10

10

102 62 6 10 398

==

=−

log( , . ), ( , . )

. .

10

3 2 875 6 3 1553 155 2 875

6

b

m

Points on graph:

−−=

= ⇒ = =3

0 093

0 093 10 1 24100 093

.

. log ..b b

398

y ab

t

t

t

t

=

= ×

=

=

2200 398 1 242200398

1 24

2200398

110 10

.

.

log log ..

log

log ..

24

22003981 24

7 9 810

10

∴ =

= ≈t years

Question 8 (c)

Question 8 (d)

Question 8 (e)


t (Number of years after 2006) 1 2 3 4 5 6y (Cost ) 500 610 720 890 1050 1540log10 y 2.70 2.79 2.86 2.95 3.02 3.19

y aby aby a by a t

t

t

t

=

=

= += +

log log ( )

log log loglog log

10 10

10 10 10

10 10 llog10 b

1 2 3 4 5 6 t (years)0

2.00

1.00

2.875

log10

y

4.00

3.00

3.155

2.600


Question 8 (b) (i)

Question 8 (b) (ii)

Sample 4

Paper 1



x y mx y m+ − = ⇒ = −+ − = ⇒ = −

= +− − −+ − −

=

−

3 0 12 1 0 2

1 21 1 2

1

2

tan ( )( )( )

θ11 2

1 213

18 41 13

++

=

∴ = =−θ tan ( ) . o

θ = tan 3�1

(4, 1)�

m m1=

m2= 2

θ θ= ⇒ =− − = ⇒ =

= ±−+

∴ + = −

−tan tan

( ) ( )

1

2

3 32 8 0 2

3 21 2

3 1 2 2

x y mmm

m m33 6 25 5

1

+ = −= −= −

m mmm

Equations of line:

l m x y kl k k

l x y

l m x

1

1

1

217

1 04 1 4 1 0 3

3 0

:( , )

:

:

= − ⇒ + + =− ∈ ⇒ − + = ⇒ = −+ − =

= − ⇒ + 77 04 1 4 7 0 3

7 3 02

2

y kl k k

l x y

+ =

− ∈ ⇒ − + = ⇒ =+ + =

( , ):

sample paper 4: paper 2Question 1 (25 marks)

tanθ = ± −+

m mm m

1 2

1 21

∴ + = − −+ = − += −= −

3 1 2 23 6 27 1

17

( ) ( )m mm m

mm

Question 1 (b)

Question 1 (a)


Question 9 (50 marks)Question 9 (a) (i)h = 1 + x

r

1

O

A B

h

x

1

x r

r x

2 2 2

2

1

1

+ =

= −

V r h

x x

x x x

=

= − +

= + − +

13

2

13

2

13

2 3

1 1

1

p

p

p

( )( )

( )

V x x xdVdx

x x

x xx x

= + − −

= − − =

+ − =− + =

13

2 3

13

2

2

1

1 2 3 0

3 2 1 03 1 1

p

p

( )

( )

( )( ) 00

1

1

13

13

13

13

2 13

3 3281

x

V

= −

∴ = + − − =

,

( ( ) ( ) ( ) )Max. p p

xh x

r x

=

= + =

= − = − = =

13

43

2 13

2 89

23

1

1 1 2( )

O

A B

x

θ

1

r = 23 2

h = 43

Question 9 (a) (ii)

Question 9 (b)


Cone: cm s

Sphere: cm s

dVdtdVdt

V h hdVdh

= −

= +

= −

=

−

−

0 2

0 2

3 1

3 1

2 13

3

.

.

p p

22

2

0 2 2

2

2

0 3

p p

p p

h h

dVdt

dVdh

dhdt

h h dhdt

dVdt h

−

= × = −

= =

=

( )

. [.

pp p

p p

( . ) ( . ) ]

.[ ( . ) ( . ) ]

.

0 3 0 3

0 22 0 3 0 3

0 125

2

21

−

∴ =−

= −

dhdt

dhdt

cms

1

h

Question 9 (e)

tan

tan

θ

θ

= = =

=

≈

−

rh

23

43

1

2 12

12

35o



x y mx y m+ − = ⇒ = −+ − = ⇒ = −

= +− − −+ − −

=

−

3 0 12 1 0 2

1 21 1 2

1

2

tan ( )( )( )

θ11 2

1 213

18 41 13

++

=

∴ = =−θ tan ( ) . o

θ = tan 3�1

(4, 1)�

m m1=

m2= 2

θ θ= ⇒ =− − = ⇒ =

= ±−+

∴ + = −

−tan tan

( ) ( )

1

2

3 32 8 0 2

3 21 2

3 1 2 2

x y mmm

m m33 6 25 5

1

+ = −= −= −

m mmm

Equations of line:

l m x y kl k k

l x y

l m x

1

1

1

217

1 04 1 4 1 0 3

3 0

:( , )

:

:

= − ⇒ + + =− ∈ ⇒ − + = ⇒ = −+ − =

= − ⇒ + 77 04 1 4 7 0 3

7 3 02

2

y kl k k

l x y

+ =

− ∈ ⇒ − + = ⇒ =+ + =

( , ):

sample paper 4: paper 2Question 1 (25 marks)

tanθ = ± −+

m mm m

1 2

1 21

∴ + = − −+ = − += −= −

3 1 2 23 6 27 1

17

( ) ( )m mm m

mm

Question 1 (b)

Question 1 (a)


Question 9 (50 marks)Question 9 (a) (i)h = 1 + x

r

1

O

A B

h

x

1

x r

r x

2 2 2

2

1

1

+ =

= −

V r h

x x

x x x

=

= − +

= + − +

13

2

13

2

13

2 3

1 1

1

p

p

p

( )( )

( )

V x x xdVdx

x x

x xx x

= + − −

= − − =

+ − =− + =

13

2 3

13

2

2

1

1 2 3 0

3 2 1 03 1 1

p

p

( )

( )

( )( ) 00

1

1

13

13

13

13

2 13

3 3281

x

V

= −

∴ = + − − =

,

( ( ) ( ) ( ) )Max. p p

xh x

r x

=

= + =

= − = − = =

13

43

2 13

2 89

23

1

1 1 2( )

O

A B

x

θ

1

r = 23 2

h = 43

Question 9 (a) (ii)

Question 9 (b)


Cone: cm s

Sphere: cm s

dVdtdVdt

V h hdVdh

= −

= +

= −

=

−

−

0 2

0 2

3 1

3 1

2 13

3

.

.

p p

22

2

0 2 2

2

2

0 3

p p

p p

h h

dVdt

dVdh

dhdt

h h dhdt

dVdt h

−

= × = −

= =

=

( )

. [.

pp p

p p

( . ) ( . ) ]

.[ ( . ) ( . ) ]

.

0 3 0 3

0 22 0 3 0 3

0 125

2

21

−

∴ =−

= −

dhdt

dhdt

cms

1

h

Question 9 (e)

tan

tan

θ

θ

= = =

=

≈

−

rh

23

43

1

2 12

12

35o Sample 4

Paper 2



AB

CD

q

p

k

h

θ

AB

CD

qp

k

h

h

CD k

θ 180o −θ

The adjacent interior angles in a parallelogram add up to 180o.

cos( ) cos cos sin sin( ) cos ( )sin

cos

180 180 1801 0

o o o− = −= − −= −

θ θ θθ θ

θ

p h k kh

q h k khq h k kh

2 2 2

2 2 2

2 2 2

2

2 1802

= + −

= + − −

= + +

cos ....( )

cos( )

θ

θ

1

o

ccos ....( )θ 2

∴ + = +p q h k2 2 2 22 2


Question 3 (b)

Question 3 (c)

Add equations (1) and (2):


s

r =5

2(1, 2)

( 2, 6)�

r = 5

s x y x y

r

:

( , ),

2 2

2 2

2 4 20 0

1 2 1 2 20 25 5

+ − − − =

= + + = =Centre

Radius of new circle: r = 52

Centre of new circle is midpoint of (1, 2) and (−2, 6).

Centre = − + +

= −

2 12

6 22

412, ( , )

Equation of new circle: ( ) ( ) ( )x y

x x y y

x y x y

+ + − =

+ + + − + =

+ + − + =

12

2 2 52

2

2 14

2 254

2 2

4

8 16

8 10 0



h k rx h y k r− + − =2 2 2



AB

CD

q

p

k

h

θ

AB

CD

qp

k

h

h

CD k

θ 180o −θ

The adjacent interior angles in a parallelogram add up to 180o.

cos( ) cos cos sin sin( ) cos ( )sin

cos

180 180 1801 0

o o o− = −= − −= −

θ θ θθ θ

θ

p h k kh

q h k khq h k kh

2 2 2

2 2 2

2 2 2

2

2 1802

= + −

= + − −

= + +

cos ....( )

cos( )

θ

θ

1

o

ccos ....( )θ 2

∴ + = +p q h k2 2 2 22 2


Question 3 (b)

Question 3 (c)

Add equations (1) and (2):


s

r =5

2(1, 2)

( 2, 6)�

r = 5

s x y x y

r

:

( , ),

2 2

2 2

2 4 20 0

1 2 1 2 20 25 5

+ − − − =

= + + = =Centre

Radius of new circle: r = 52

Centre of new circle is midpoint of (1, 2) and (−2, 6).

Centre = − + +

= −

2 12

6 22

412, ( , )

Equation of new circle: ( ) ( ) ( )x y

x x y y

x y x y

+ + − =

+ + + − + =

+ + − + =

12

2 2 52

2

2 14

2 254

2 2

4

8 16

8 10 0



h k rx h y k r− + − =2 2 2

Sample 4

Paper 2



x 1 2 3 4 5P(x) 0.1 0.2 0.3 0.3 0.1

P P P( ) [ ( ) ( )]. . .

At least 3 1 1 21 0 1 0 2 0 7

= − += − − =

xxP xP

= =+ + + +

+ + +∑∑

( ) ( . ) ( . ) ( . ) ( . ) ( . ). . .

9 0 1 2 0 2 3 0 3 4 0 3 5 0 10 1 0 2 0 3 0.. .

.3 0 1

3 1+

=

Gender For Against TotalMale 58 85 143Female 84 73 157Total 142 158 300

P( )Male Against treatyand = =85300

1760

P( )Female For treatyor =+

=58 157

3004360


Question 5 (a) (i)

Question 5 (a) (ii)

Question 5 (b) (ii)

Question 5 (b) (i)


OrP(3) + P(4) + P(5) = 0.3 + 0.3 + 0.1 = 0.7

P = =58

1422971


Height

µ = 175LQ = 170 UQ = 180

(i) Median M = 175 cm(ii) Lower quartile (LQ) = 170 cm(iii) Interquartile range = 180 − 170 = 10 cm

P xP z Z z

x z x

( ) .( ) . .

: .

≤ =≤ = ⇒ =

= =−

⇒ =−

180 0 750 75 0 675

180 0 675 180 175µσ σσ

σ∴ =−

=180 175

0 6757 4

.. cm

x z

x z

P x

= =−

= −

= =−

=

≤ ≤

170 170 1757 4

0 675

180 180 1757 4

0 675

170 1

:.

.

:.

.

( 8800 675 0 675

0 675 0 6750 675

)( . . )( . ) ( . )( . ) {

= − ≤ ≤= ≤ − ≤ −= ≤ −

P zP z P zP z 11 0 6752 0 675 12 0 75 11 5 10 5

− ≤= ≤ −= −= −=

P zP z

( . )}( . )

( . )..


Question 4 (a)

75% of students have a height less than 180 cm.Question 4 (b)

Question 4 (c)

Question 4 (d)



x 1 2 3 4 5P(x) 0.1 0.2 0.3 0.3 0.1

P P P( ) [ ( ) ( )]. . .

At least 3 1 1 21 0 1 0 2 0 7

= − += − − =

xxP xP

= =+ + + +

+ + +∑∑

( ) ( . ) ( . ) ( . ) ( . ) ( . ). . .

9 0 1 2 0 2 3 0 3 4 0 3 5 0 10 1 0 2 0 3 0.. .

.3 0 1

3 1+

=

Gender For Against TotalMale 58 85 143Female 84 73 157Total 142 158 300

P( )Male Against treatyand = =85300

1760

P( )Female For treatyor =+

=58 157

3004360


Question 5 (a) (i)

Question 5 (a) (ii)

Question 5 (b) (ii)

Question 5 (b) (i)


OrP(3) + P(4) + P(5) = 0.3 + 0.3 + 0.1 = 0.7

P = =58

1422971


Height

µ = 175LQ = 170 UQ = 180

(i) Median M = 175 cm(ii) Lower quartile (LQ) = 170 cm(iii) Interquartile range = 180 − 170 = 10 cm

P xP z Z z

x z x

( ) .( ) . .

: .

≤ =≤ = ⇒ =

= =−

⇒ =−

180 0 750 75 0 675

180 0 675 180 175µσ σσ

σ∴ =−

=180 175

0 6757 4

.. cm

x z

x z

P x

= =−

= −

= =−

=

≤ ≤

170 170 1757 4

0 675

180 180 1757 4

0 675

170 1

:.

.

:.

.

( 8800 675 0 675

0 675 0 6750 675

)( . . )( . ) ( . )( . ) {

= − ≤ ≤= ≤ − ≤ −= ≤ −

P zP z P zP z 11 0 6752 0 675 12 0 75 11 5 10 5

− ≤= ≤ −= −= −=

P zP z

( . )}( . )

( . )..


Question 4 (a)

75% of students have a height less than 180 cm.Question 4 (b)

Question 4 (c)

Question 4 (d)

Sample 4

Paper 2



Outcome Not a Prime Sum Prime Sum

P 712

512

Net income to Bob –3 3

xP(x) − 74

54

(i) A random variable is a function that associates a unique numerical value with every outcome of an experiment. It may vary from trial to trial as the experiment is repeated.(ii) DiSCrete: Ex. A coin is tossed five times. The random variable is the number of heads. Its values can be 0, 1, 2, 3, 4, 5. ContinuouS: Ex. The temperature in a house during the day can take any positive or negative value within a certain range.(iii) Expected value µ = =∑ xP x( ) Mean of a probability distribution.

(i) A prime number is a whole positive integer (excluding 1) divisible by itself and 1 only.

(ii)

4

5

6

1

3

2

Die A

Die

B

1

2

3

3 4 5 6

4 5 6 7 8

2 3 4 5 6 7

7 8 9 10 11 12

6 7 8 9 10 11

5 6 7 8 9 10

4 5 6 7 8 9

Primes in table: 2, 3, 5, 7, 11Number of primes = 15

P( )Sum that is a prime Number of primesNumber of numbers

= =15536

512

=

P( )Sum that is a prime Number of non-primesNumber of n

not =uumbers

= =2136

712


Question 7 (b)

(iii)

(iv)

Question 7 (c)

E xP x= = − + = −∑ ( ) 74

54

12

E = –50 c, on average Bob loses 50 c per game.

Expected losses: 30 1512× − = −( )

Question 7 (d)(i)

(ii)


′ = −

→ − −

′ =

→ −

′ =

A

B

C

12

32

6 32

2 92

92

92

1

, ,

, ,

332

0 0 0

6 92

32

92

2712

12

, ( , )

( )

→

= −

− −

−

= −A −− = − =

274

1354

1358

12

AreaArea

′ ′ ′=

= =

=

A B CABC

k

1358

152

94

32

22

A B CA x y x y

( , ), ( , ), ( , )

( )( ) ( )( )

0 0 1 4 4 1

1 1 4 4

15

12 1 2 2 1

12

12

15

= −

= −

= −

= 22

kOAOA

= =′=

Image LengthObject Length

32

A(0, 0)

O( 1, 3)�

A’

3

2

1

A x y( , )1 1

a

b

B x y( , )2 2

C x y( , )

x ax bxa b

y ay bya b

=−−

=−−

2 1 2 1,

a bO x y A x y

A

= =− = =

′ =− −−

3 11 3 0 0

3 0 1 13 1

3 01 1 2 2

,( , ) ( , ); ( , ) ( , )

( ) ( ) , ( ) −−−

= −

= =− = =

1 33 1

12

32

3 11 3 1 41 1

( ) ,

,( , ) ( , ); ( , )a bO x y B (( , )

( ) ( ) , ( ) ( ) ,

,

x y

B

a

2 2

3 1 1 13 1

3 4 1 33 1

2 92

3

′ =− −−

−−

=

= bbO x y C x y

C

=− = =

′ =− −−

−

11 3 4 1

3 4 1 13 1

3 1 11 1 2 2( , ) ( , ); ( , ) ( , )

( ) ( ) , ( ) (333 1

132

0) ,−

=


Question 6 (b)

Question 6 (c)

external Division



Outcome Not a Prime Sum Prime Sum

P 712

512

Net income to Bob –3 3

xP(x) − 74

54

(i) A random variable is a function that associates a unique numerical value with every outcome of an experiment. It may vary from trial to trial as the experiment is repeated.(ii) DiSCrete: Ex. A coin is tossed five times. The random variable is the number of heads. Its values can be 0, 1, 2, 3, 4, 5. ContinuouS: Ex. The temperature in a house during the day can take any positive or negative value within a certain range.(iii) Expected value µ = =∑ xP x( ) Mean of a probability distribution.

(i) A prime number is a whole positive integer (excluding 1) divisible by itself and 1 only.

(ii)

4

5

6

1

3

2

Die A

Die

B

1

2

3

3 4 5 6

4 5 6 7 8

2 3 4 5 6 7

7 8 9 10 11 12

6 7 8 9 10 11

5 6 7 8 9 10

4 5 6 7 8 9

Primes in table: 2, 3, 5, 7, 11Number of primes = 15

P( )Sum that is a prime Number of primesNumber of numbers

= =15536

512

=

P( )Sum that is a prime Number of non-primesNumber of n

not =uumbers

= =2136

712


Question 7 (b)

(iii)

(iv)

Question 7 (c)

E xP x= = − + = −∑ ( ) 74

54

12

E = –50 c, on average Bob loses 50 c per game.

Expected losses: 30 1512× − = −( )

Question 7 (d)(i)

(ii)


′ = −

→ − −

′ =

→ −

′ =

A

B

C

12

32

6 32

2 92

92

92

1

, ,

, ,

332

0 0 0

6 92

32

92

2712

12

, ( , )

( )

→

= −

− −

−

= −A −− = − =

274

1354

1358

12

AreaArea

′ ′ ′=

= =

=

A B CABC

k

1358

152

94

32

22

A B CA x y x y

( , ), ( , ), ( , )

( )( ) ( )( )

0 0 1 4 4 1

1 1 4 4

15

12 1 2 2 1

12

12

15

= −

= −

= −

= 22

kOAOA

= =′=

Image LengthObject Length

32

A(0, 0)

O( 1, 3)�

A’

3

2

1

A x y( , )1 1

a

b

B x y( , )2 2

C x y( , )

x ax bxa b

y ay bya b

=−−

=−−

2 1 2 1,

a bO x y A x y

A

= =− = =

′ =− −−

3 11 3 0 0

3 0 1 13 1

3 01 1 2 2

,( , ) ( , ); ( , ) ( , )

( ) ( ) , ( ) −−−

= −

= =− = =

1 33 1

12

32

3 11 3 1 41 1

( ) ,

,( , ) ( , ); ( , )a bO x y B (( , )

( ) ( ) , ( ) ( ) ,

,

x y

B

a

2 2

3 1 1 13 1

3 4 1 33 1

2 92

3

′ =− −−

−−

=

= bbO x y C x y

C

=− = =

′ =− −−

−

11 3 4 1

3 4 1 13 1

3 1 11 1 2 2( , ) ( , ); ( , ) ( , )

( ) ( ) , ( ) (333 1

132

0) ,−

=


Question 6 (b)

Question 6 (c)

external Division

Sample 4

Paper 2



sin( ) sin

sin( ) sin

sin sin

∠=

∠ =

∠ = −

CAD

CAD

CAD

4010054

40 10054

40 11

o

o

00054

46 8o

o

= .

AC

AC

2 2 230 40 2 30 40 100

54

= + −

∴ =

( )( ) cos o

cm

sin( )

sin .

sin . .

∠ =

∴ =

= =

CADBD

BD

BD

12

30

46 860

60 46 8 43 7

o

o m

Question 9 (c)

(i)

(ii)

(iii)

(iv) Area cmo= =( )( )sin30 40 100 1182 2

A

C

Question 9 (d)

Draw a line segment [AC] of length 10 cm.Using a compass draw an arc of radius 10 cm with C as centre.Using a set square place a rule perpendicular to AC and move the ruler into position such that it measures a chord of distance of 10 cm. Draw a line along the ruler in this position.


− × + =

− + ==

= =

321 5 6

5 27275

5 40

712

512

12x

xx

x .

Question 7 (e)

A probability of 10.2% means that a fair die will give six or more 4s 10.2% of the time in 20 throws of a die. So there is nothing unusual about the die used in the casino. If the probability of at least six 4s when a fair die was tossed 20 times was less that 5% then this probability is so small that it hasn’t happened by chance and so the die is biased.

PP

P C C C

( )( )

{ ( ) ( ) ( ) ( )

44

1

16

56

200

16

0 56

20 201

16

1 56

19 202

=

=

= − + +

Not

(( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )16

2 56

18 203

16

3 56

17 204

16

4 56

16 205

16

5 56+ + +C C C 115

1 0 8980 10210 2

}.

.. %

= −==


Question 8 (b)

(i) A rhombus is a simple quadilateral whose sides are equal.(ii) The angles in a quadrilateral add up to 360o. If each angle is equal, each angle will be 90o and hence a square is the shape formed.

Area of Area of Ar

∆ = =

∆ = =

ADC p q pqABC p q pq

12

12

14

12

12

14

( )( )( )( )

eea of ABCD pq pq pq= + =14

14

12

Area of Area of Area of

∆ =

∆ =

=

ADC abABC abABCD ab

1212

sinsin

s

θ

θiinθ

3030

40

p

C

A

BDq

O

40

100o

100o


(i) Perimeter P = 2a + 2bQuestion 9 (b)

(ii)



sin( ) sin

sin( ) sin

sin sin

∠=

∠ =

∠ = −

CAD

CAD

CAD

4010054

40 10054

40 11

o

o

00054

46 8o

o

= .

AC

AC

2 2 230 40 2 30 40 100

54

= + −

∴ =

( )( ) cos o

cm

sin( )

sin .

sin . .

∠ =

∴ =

= =

CADBD

BD

BD

12

30

46 860

60 46 8 43 7

o

o m

Question 9 (c)

(i)

(ii)

(iii)

(iv) Area cmo= =( )( )sin30 40 100 1182 2

A

C

Question 9 (d)

Draw a line segment [AC] of length 10 cm.Using a compass draw an arc of radius 10 cm with C as centre.Using a set square place a rule perpendicular to AC and move the ruler into position such that it measures a chord of distance of 10 cm. Draw a line along the ruler in this position.


− × + =

− + ==

= =

321 5 6

5 27275

5 40

712

512

12x

xx

x .

Question 7 (e)

A probability of 10.2% means that a fair die will give six or more 4s 10.2% of the time in 20 throws of a die. So there is nothing unusual about the die used in the casino. If the probability of at least six 4s when a fair die was tossed 20 times was less that 5% then this probability is so small that it hasn’t happened by chance and so the die is biased.

PP

P C C C

( )( )

{ ( ) ( ) ( ) ( )

44

1

16

56

200

16

0 56

20 201

16

1 56

19 202

=

=

= − + +

Not

(( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )16

2 56

18 203

16

3 56

17 204

16

4 56

16 205

16

5 56+ + +C C C 115

1 0 8980 10210 2

}.

.. %

= −==


Question 8 (b)

(i) A rhombus is a simple quadilateral whose sides are equal.(ii) The angles in a quadrilateral add up to 360o. If each angle is equal, each angle will be 90o and hence a square is the shape formed.

Area of Area of Ar

∆ = =

∆ = =

ADC p q pqABC p q pq

12

12

14

12

12

14

( )( )( )( )

eea of ABCD pq pq pq= + =14

14

12

Area of Area of Area of

∆ =

∆ =

=

ADC abABC abABCD ab

1212

sinsin

s

θ

θiinθ

3030

40

p

C

A

BDq

O

40

100o

100o


(i) Perimeter P = 2a + 2bQuestion 9 (b)

(ii)

Sample 4

Paper 2



tan tan( )tan tan

tan tan

15 45 3045 30

1 45 30

1 13

1 13

1

o o o

o o

o o

= −

=−

+

=−

+

=−−

+

× =−+

=−+

×−−

=− +

13

1 13

33

3 13 1

3 13 1

3 13 1

3 2 3( )( )

( )( )

113 1

4 2 32

2 3

−

=−

= −

Question 9 (f)

sin sin ( )θ θ= = ⇒ = =−d

d2 1 1

212

30oC

A

BD

d

2

d

2

d

d

E

15o

60o

75o

75o

θ 30o30

o

60o

∠ = ∠ = ∠ = ∠ =ADC ABC BAD BCD75 75 150 60o o o o, , ,

Question 9 (e) (ii)


A

C

D B

Draw the kite ABCD.

C

A

BD

d

2

d

2

d

dd

x

h

E

Consider the right-angled triangle BEC:

x d h d d d d= − = − = −

=

−

32

1 32

2 32

d h d

d h d

h d d d

h d d

2 22

2 22

2 22 2

2

2

4

43

43

43

2

= +

= +

∴ = − =

= =

Consider the right-angled triangle BEA:

tan( ) ( )

tan ( )

∠ =

−

= −

∴ ∠ = − =−

ABDd

d

ABD

2 32

2

2 3

2 3 151 o

Question 9 (e) (i)



tan tan( )tan tan

tan tan

15 45 3045 30

1 45 30

1 13

1 13

1

o o o

o o

o o

= −

=−

+

=−

+

=−−

+

× =−+

=−+

×−−

=− +

13

1 13

33

3 13 1

3 13 1

3 13 1

3 2 3( )( )

( )( )

113 1

4 2 32

2 3

−

=−

= −

Question 9 (f)

sin sin ( )θ θ= = ⇒ = =−d

d2 1 1

212

30oC

A

BD

d

2

d

2

d

d

E

15o

60o

75o

75o

θ 30o30

o

60o

∠ = ∠ = ∠ = ∠ =ADC ABC BAD BCD75 75 150 60o o o o, , ,

Question 9 (e) (ii)


A

C

D B

Draw the kite ABCD.

C

A

BD

d

2

d

2

d

dd

x

h

E

Consider the right-angled triangle BEC:

x d h d d d d= − = − = −

=

−

32

1 32

2 32

d h d

d h d

h d d d

h d d

2 22

2 22

2 22 2

2

2

4

43

43

43

2

= +

= +

∴ = − =

= =

Consider the right-angled triangle BEA:

tan( ) ( )

tan ( )

∠ =

−

= −

∴ ∠ = − =−

ABDd

d

ABD

2 32

2

2 3

2 3 151 o

Question 9 (e) (i)

Sample 4

Paper 2



x yx y y x

x xx x xx

2 2

2 2

2 2

2

2 23 7 3 7

2 2 3 72 2 9 42 492 1

− == + ⇒ = −

− = −

− = − +

− =

( )( )88 84 98

0 17 84 1000 17 50 2

22 3 2

2

2

5017

x xx xx x

xx y

− +

= − += − −

∴ =

= =

( )( ),

: ( )): ( )

: ( , ), (

− = − = −= = − = − =

−

7 6 7 13 7 7

2 1

5017

5017

15017

3117

5017

x yAns ,, )31

17

x kx x x x axx kx x x a x a

3 2 2

3 2 3 2

22 56 4 1422 56 4 4

− − + = + + +

− − + = + + + +

( )( )( ) ( 114 56

22 4 14 4 36 94 9 4 5 5

)x

a a ak a k k

+

∴− = + ⇒ = − ⇒ = −∴− = + ⇒ − = − + = − ⇒ =

x x x x x xx x x

x

3 2 25 22 56 4 9 14 04 2 7 0

4 2 7

− − + = + − + =+ − − =

∴ = −

( )( )( )( )( )

, ,


Question 2 (b)

[Replace y in the quadratic equation.]

[A cubic is equal to a linear by a quadratic.]


Im

Re

1

2

3

4

0�1�2�3�4

�1

�2

�3

�4

1 2 3 4

z i z=4

iz

i z3

i z2

z i m

iz i i i i i m

i z i i

= + → =

= + = + = − + → = −

= − + = − −

3 4

3 4 3 4 4 3

1 3 4 3 4

43

2 34

2

( )

( ) →→ =

= − + = − − = − → = −

= + = + → =

m

i z i i i i i m

i z i i m

43

3 2 34

4 43

3 4 3 4 4 3

1 3 4 3 4

( )

( )

Multiplying a complex number z by: Causes:i an anti-clockwise rotation of 90o to zi

2 an anti-clockwise rotation of 180o to zi


4 an anti-clockwise rotation of 360o to z



Question 1 (b)

Question 1 (c)



x yx y y x

x xx x xx

2 2

2 2

2 2

2

2 23 7 3 7

2 2 3 72 2 9 42 492 1

− == + ⇒ = −

− = −

− = − +

− =

( )( )88 84 98

0 17 84 1000 17 50 2

22 3 2

2

2

5017

x xx xx x

xx y

− +

= − += − −

∴ =

= =

( )( ),

: ( )): ( )

: ( , ), (

− = − = −= = − = − =

−

7 6 7 13 7 7

2 1

5017

5017

15017

3117

5017

x yAns ,, )31

17

x kx x x x axx kx x x a x a

3 2 2

3 2 3 2

22 56 4 1422 56 4 4

− − + = + + +

− − + = + + + +

( )( )( ) ( 114 56

22 4 14 4 36 94 9 4 5 5

)x

a a ak a k k

+

∴− = + ⇒ = − ⇒ = −∴− = + ⇒ − = − + = − ⇒ =

x x x x x xx x x

x

3 2 25 22 56 4 9 14 04 2 7 0

4 2 7

− − + = + − + =+ − − =

∴ = −

( )( )( )( )( )

, ,


Question 2 (b)

[Replace y in the quadratic equation.]

[A cubic is equal to a linear by a quadratic.]


Im

Re

1

2

3

4

0�1�2�3�4

�1

�2

�3

�4

1 2 3 4

z i z=4

iz

i z3

i z2

z i m

iz i i i i i m

i z i i

= + → =

= + = + = − + → = −

= − + = − −

3 4

3 4 3 4 4 3

1 3 4 3 4

43

2 34

2

( )

( ) →→ =

= − + = − − = − → = −

= + = + → =

m

i z i i i i i m

i z i i m

43

3 2 34

4 43

3 4 3 4 4 3

1 3 4 3 4

( )

( )

Multiplying a complex number z by: Causes:i an anti-clockwise rotation of 90o to zi



4 an anti-clockwise rotation of 360o to z



Question 1 (b)

Question 1 (c)

Sample 5

Paper 1



S S a rr

b rr

ar

br r

a br

n n

n n

2

2 2

2

11

11

1 1 1

1

1

= ⇒−−

=−−

−=

+ −

=+

+

( ) ( )

( ) ( )( )

( rr ba

r ba

) =

∴ = −1


Question 4 (b)



1. Prove true for n = 7:

Therefore, assuming true for n = k means it is true for n = k + 1. So true for n = 7 and true for n = k means it is true for n = k + 1. This implies it is true for all n n≥ ∈7, .�

7 50403 2187

7 3

7

7

!

!

=

=

∴ >

3. Prove true for n = k + 1: Prove (k + 1)! > 3k + 1

Proof:


2. Assume true for n = k: Assume k! > 3k

b br brb r

S b rn

n

, , ,..........,

(

2 4

2

21First term Common ratio= =

=− ))

, , ,..........,

(

1 2

2

2

−

= =

=

r

a ar ara r

S an

First term Common ratio111

2−−rr

n )

( )! ( ) !( )

( ) !

k k kk

k k

k

k

k

+ = +

> +

> ×

∴ + > +

1 11 3

3 31 3 1

[From step 2 where k! > 3k][(k + 1) > 3 because k ≥ 7 ⇒ (k + 1) ≥ 8]


f x b bb b

f a b bb b

b b b bb

x x

x x

a a

a a

a a a a

a

( )

( )

=+−

= ⇒+−

=

+ = −

=

−

−

−

−

− −

−

3 3

3 34 2242

2 2

bbbb

a

a

a

a

=

∴ =

−

f a b bb b

bb

bb

a a

a a

aa

aa

( )2

1

122

53

2 2

2 2

22

22

1212

5232

=+−

=+

−=

+−

= =−

−

log log

log log

log log

(log ) log

(log

b b

b b

b b

b b

b

x x

x x

x x

x x

=

=

=

=

12

12

14

2

xx xx xx x bx x b

b

b

b b

b

b

) loglog (log )

log

log

2

0

4

4 04 0

0 1

4

− =− =

∴ = ⇒ = =

∴ = ⇒ =

44

4

1 171622 0

+ =

== ±

∴ = >

bbb bas


Question 3 (a) (ii)

Question 3 (b)

[Square both sides.]

[The roots add to 17.]



S S a rr

b rr

ar

br r

a br

n n

n n

2

2 2

2

11

11

1 1 1

1

1

= ⇒−−

=−−

−=

+ −

=+

+

( ) ( )

( ) ( )( )

( rr ba

r ba

) =

∴ = −1


Question 4 (b)



1. Prove true for n = 7:

Therefore, assuming true for n = k means it is true for n = k + 1. So true for n = 7 and true for n = k means it is true for n = k + 1. This implies it is true for all n n≥ ∈7, .�

7 50403 2187

7 3

7

7

!

!

=

=

∴ >

3. Prove true for n = k + 1: Prove (k + 1)! > 3k + 1

Proof:


2. Assume true for n = k: Assume k! > 3k

b br brb r

S b rn

n

, , ,..........,

(

2 4

2

21First term Common ratio= =

=− ))

, , ,..........,

(

1 2

2

2

−

= =

=

r

a ar ara r

S an

First term Common ratio111

2−−rr

n )

( )! ( ) !( )

( ) !

k k kk

k k

k

k

k

+ = +

> +

> ×

∴ + > +

1 11 3

3 31 3 1

[From step 2 where k! > 3k][(k + 1) > 3 because k ≥ 7 ⇒ (k + 1) ≥ 8]


f x b bb b

f a b bb b

b b b bb

x x

x x

a a

a a

a a a a

a

( )

( )

=+−

= ⇒+−

=

+ = −

=

−

−

−

−

− −

−

3 3

3 34 2242

2 2

bbbb

a

a

a

a

=

∴ =

−

f a b bb b

bb

bb

a a

a a

aa

aa

( )2

1

122

53

2 2

2 2

22

22

1212

5232

=+−

=+

−=

+−

= =−

−

log log

log log

log log

(log ) log

(log

b b

b b

b b

b b

b

x x

x x

x x

x x

=

=

=

=

12

12

14

2

xx xx xx x bx x b

b

b

b b

b

b

) loglog (log )

log

log

2

0

4

4 04 0

0 1

4

− =− =

∴ = ⇒ = =

∴ = ⇒ =

44

4

1 171622 0

+ =

== ±

∴ = >

bbb bas


Question 3 (a) (ii)

Question 3 (b)

[Square both sides.]

[The roots add to 17.]

Sample 5

Paper 1




y

y

x=

= = ≈

3

3 3 1 712 .

A

x

1.7

1

0�1�2 21

y

B

2

f x( ) = 3x

x = 12

y xdydx x

= +

=+

ln( )11

1

xx

xx x

++

=+ ++

= ++

31

1 21

1 21

( )( )

xx

dx

xdx

x x

++

= ++

= + += + − +

∫

∫

31

1 21

2 13 2 4 1 2

1

3

1

3

13[ ln( )]

( ln ) ( lln )ln ln(ln ln )ln

23 2 4 1 2 22 2 4 22 2 2

= + − −= + −= +

Average value = +−

= +2 2 2

3 11 2ln ln

The function is not continuous. At x = −1 the function is not defined.

Question 5 (e)

Question 6 (b)

Question 6 (c)

Question 6 (d)



A

x

2

1

0�1�2 21

y

B

f x af x f a

aa

x( )( , ) ( ) ( )

=

− ∈ ⇒ − = =

= ⇒ =

−1 11 1

33

13

1 13

f xfB

x( )( )

( , )

=

= =∴

30 3 1

0 1

0

The function is not bijective as every y value does not have a corresponding x value. For example, y = − 1

2 has no x value.

A

y x=2

1

0�1�2 21

y

B

f x�1

( )

f x( )

x

y a dydx

a ax x= ⇒ = × ln

y dydx

x x= ⇒ = × >3 3 3 0ln

This is always increasing as ln 3 > 0 and 3x > 0 for all values of x.

Question 5 (b) (i)

Question 5 (b) (ii)

Question 5 (c) [Graph to right]

Question 5 (d)

dydxd ydx

d ydx

x

x

= ×

= ×

=

(ln )

(ln )

3 3

3 32

22

2

2Points of inflection: 00

3 3 03 0

2(ln )( )× =

=

x

x No solutions

There are no points of inflection.




y

y

x=

= = ≈

3

3 3 1 712 .

A

x

1.7

1

0�1�2 21

y

B

2

f x( ) = 3x

x = 12

y xdydx x

= +

=+

ln( )11

1

xx

xx x

++

=+ ++

= ++

31

1 21

1 21

( )( )

xx

dx

xdx

x x

++

= ++

= + += + − +

∫

∫

31

1 21

2 13 2 4 1 2

1

3

1

3

13[ ln( )]

( ln ) ( lln )ln ln(ln ln )ln

23 2 4 1 2 22 2 4 22 2 2

= + − −= + −= +

Average value = +−

= +2 2 2

3 11 2ln ln

The function is not continuous. At x = −1 the function is not defined.

Question 5 (e)

Question 6 (b)

Question 6 (c)

Question 6 (d)



A

x

2

1

0�1�2 21

y

B

f x af x f a

aa

x( )( , ) ( ) ( )

=

− ∈ ⇒ − = =

= ⇒ =

−1 11 1

33

13

1 13

f xfB

x( )( )

( , )

=

= =∴

30 3 1

0 1

0

The function is not bijective as every y value does not have a corresponding x value. For example, y = − 1

2 has no x value.

A

y x=2

1

0�1�2 21

y

B

f x�1

( )

f x( )

x

y a dydx

a ax x= ⇒ = × ln

y dydx

x x= ⇒ = × >3 3 3 0ln

This is always increasing as ln 3 > 0 and 3x > 0 for all values of x.

Question 5 (b) (i)

Question 5 (b) (ii)

Question 5 (c) [Graph to right]

Question 5 (d)

dydxd ydx

d ydx

x

x

= ×

= ×

=

(ln )

(ln )

3 3

3 32

22

2

2Points of inflection: 00

3 3 03 0

2(ln )( )× =

=

x

x No solutions

There are no points of inflection.

Sample 5

Paper 1



p ee

=+

≈3000 0 21 0 2

16050 25 7

0 25 7

( . )( . )

. ( )

. ( )

p ee

ee

e

t

t

t

t=+

=+

=

3000 0 21 0 2

6001 0 2

600

0 25

0 25

0 25

0 25

0

( . )( . ) .

.

.

.

.

.225

0 25

0 25

0 25

0 25

0 251 0 2600

0 2

t

t

t

t

t

te

eee

e.

.

.

.

.. .+=

+ −

lim ,n

nr r→∞

= ∞ >1

lim lim.

lim.

.

.

t t t t

t

pe

e→∞ →∞ − →∞

=+

=

+

6000 2

600

0 2 10 25

0 25

=

+∞

=+

=600

0 2 1600

0 2 03000

. .

pe

e

dpdt

e

tt

t

=+

= +

= − +

−− −

− −

6000 2

600 0 2

600 0 2

0 250 25 1

0 25

.( . )

( . )

..

. 22 0 25

0 25 0 25 2

1

0 25

1500 2

1

( ( . ))

( . )

.

. .

e

e edpdt

t

t t

t

−

−

=

× −

=+

=

5500 2

1220 25 0 25 2e e. .( . )+≈− flowers/year

p = 1600 after seven yearsQuestion 7 (d) (i) Question 7 (d) (ii)

Question 7 (e)

Question 7 (f) (i)

Question 7 (f) (ii)

Question 7 (g)


p keke

t keke

kk

k

t

t=+

= =+

=+

+

30001

0 500 30001

30001

500 1

0 25

0 25

0

0

.

.

:

( ))

.

=+ ==

∴ = =

3000500 500 3000500 2500

5002500

0 2

kk kk

k

p keke

ee

t

t

t

t

=+

=+

30001

1233 3000 0 21 0 2

1233

0 25

0 25

0 25

0 25

.

.

.

.

( . ).

(11 0 2 6001233 246 6 6001233 353

0 25 0 25

0 25 0 25

+ =

+ =

=

. )..

. .

. .

e ee e

t t

t t

441233353 4

1233353 4

0 25

4 12333

0 25

0 25

e

e

t

t

t

t

.

.

.

ln.

.

ln

=

=

∴ =553 4

5.

≈ years


t (years) 0 2 4 6 8 10p 500 744 1057 1418 1789 2127

1000

4

2000

1500

2500

500

0

p

86 102 t (Years)

1600

Question 7 (c) (i)

Question 7 (c) (ii)



p ee

=+

≈3000 0 21 0 2

16050 25 7

0 25 7

( . )( . )

. ( )

. ( )

p ee

ee

e

t

t

t

t=+

=+

=

3000 0 21 0 2

6001 0 2

600

0 25

0 25

0 25

0 25

0

( . )( . ) .

.

.

.

.

.225

0 25

0 25

0 25

0 25

0 251 0 2600

0 2

t

t

t

t

t

te

eee

e.

.

.

.

.. .+=

+ −

lim ,n

nr r→∞

= ∞ >1

lim lim.

lim.

.

.

t t t t

t

pe

e→∞ →∞ − →∞

=+

=

+

6000 2

600

0 2 10 25

0 25

=

+∞

=+

=600

0 2 1600

0 2 03000

. .

pe

e

dpdt

e

tt

t

=+

= +

= − +

−− −

− −

6000 2

600 0 2

600 0 2

0 250 25 1

0 25

.( . )

( . )

..

. 22 0 25

0 25 0 25 2

1

0 25

1500 2

1

( ( . ))

( . )

.

. .

e

e edpdt

t

t t

t

−

−

=

× −

=+

=

5500 2

1220 25 0 25 2e e. .( . )+≈− flowers/year

p = 1600 after seven yearsQuestion 7 (d) (i) Question 7 (d) (ii)

Question 7 (e)

Question 7 (f) (i)

Question 7 (f) (ii)

Question 7 (g)


p keke

t keke

kk

k

t

t=+

= =+

=+

+

30001

0 500 30001

30001

500 1

0 25

0 25

0

0

.

.

:

( ))

.

=+ ==

∴ = =

3000500 500 3000500 2500

5002500

0 2

kk kk

k

p keke

ee

t

t

t

t

=+

=+

30001

1233 3000 0 21 0 2

1233

0 25

0 25

0 25

0 25

.

.

.

.

( . ).

(11 0 2 6001233 246 6 6001233 353

0 25 0 25

0 25 0 25

+ =

+ =

=

. )..

. .

. .

e ee e

t t

t t

441233353 4

1233353 4

0 25

4 12333

0 25

0 25

e

e

t

t

t

t

.

.

.

ln.

.

ln

=

=

∴ =553 4

5.

≈ years


t (years) 0 2 4 6 8 10p 500 744 1057 1418 1789 2127

1000

4

2000

1500

2500

500

0

p

86 102 t (Years)

1600

Question 7 (c) (i)

Question 7 (c) (ii)

Sample 5

Paper 1



( . ) .. . %

1 055 1 0044717 10 0044717 0 44717

112 = = +

= =i

i

3002 68 6 244 13 72 438 72. . .× − × =


Question 8 (c) (i)

Question 8 (c) (ii)


A P i ii

t

t=+

+ −

= ×−

=

( )(( ) )

. ( . )( . )

.

11 1

15 000 0 055 1 0551 055 1

3002 66

6 88

A P i ii

t

t=+

+ −

= ×

( )(( ) )

. ( . )( .

11 1

15 000 0 0044717 1 00447171 004471

72

77 1244 1372 −

=)

.


First repayment: A

1 055., Second repayment:

A( . )

,1 055 2 Third repayment:

A( . )1 055 3

Question 8 (a) (ii)

P A A A A A A= + + + + +

1 055 1 055 1 055 1 055 1 055 1 0552 3 4 5 6. ( . ) ( . ) ( . ) ( . ) ( . )

P A A A A A A= + + + + +

1 055 1 055 1 055 1 055 1 055 1 0552 3 4 5 6. ( . ) ( . ) ( . ) ( . ) ( . )

aa A r

S a rr

A

= =

=−−

=−

−

1 0551

1 055

11

1 0551 1

1 055

1 11

6

6 6

.,

.

( ) . .

.0055

15 000 1 0551 1

1 055

1 11 055

15 000 1 11 0556

∴ =−

−⇒

−

A. .

.

.

−

=

∴ =× −

−

1 11 055

1 055

1 055 15 000 1 11 055

1 11

6..

..

.

A

A

0055

3002 68

6

= .

P Ai

Ai

Ai

Ai t

=+

++

++

+ ++( ) ( ) ( )

..................( )1 1 1 11 2 3

P Ai

Ai

Ai

Ai

a A

t=+

++

++

+ ++

=+

( ) ( ) ( )..................

( )

(

1 1 1 1

1

1 2 3

iir

i

S P a rr

Ai i

i

t

t t

),

( )

( ) ( ) ( )

( )

=+

= =−−

=+

−+

−+

=

11

11

11 1

1

1 11

AA ii

i ii

A ii

t

t

t( )( )

( )( )

( )( )

1 11

1 1 11

1 11

+ −+

++ −+

=

+ −+ tt t

t

t

t

iA ii i

A P i ii

=

+ −+

=+

+ −

(( ) )( )

( )(( ) )

1 11

11 1

\



Question 8 (b) (i)

Question 8 (b) (ii)

[This is a geometric series.]




( . ) .. . %

1 055 1 0044717 10 0044717 0 44717

112 = = +

= =i

i

3002 68 6 244 13 72 438 72. . .× − × =


Question 8 (c) (i)

Question 8 (c) (ii)


A P i ii

t

t=+

+ −

= ×−

=

( )(( ) )

. ( . )( . )

.

11 1

15 000 0 055 1 0551 055 1

3002 66

6 88

A P i ii

t

t=+

+ −

= ×

( )(( ) )

. ( . )( .

11 1

15 000 0 0044717 1 00447171 004471

72

77 1244 1372 −

=)

.


First repayment: A

1 055., Second repayment:

A( . )

,1 055 2 Third repayment:

A( . )1 055 3

Question 8 (a) (ii)

P A A A A A A= + + + + +

1 055 1 055 1 055 1 055 1 055 1 0552 3 4 5 6. ( . ) ( . ) ( . ) ( . ) ( . )

P A A A A A A= + + + + +

1 055 1 055 1 055 1 055 1 055 1 0552 3 4 5 6. ( . ) ( . ) ( . ) ( . ) ( . )

aa A r

S a rr

A

= =

=−−

=−

−

1 0551

1 055

11

1 0551 1

1 055

1 11

6

6 6

.,

.

( ) . .

.0055

15 000 1 0551 1

1 055

1 11 055

15 000 1 11 0556

∴ =−

−⇒

−

A. .

.

.

−

=

∴ =× −

−

1 11 055

1 055

1 055 15 000 1 11 055

1 11

6..

..

.

A

A

0055

3002 68

6

= .

P Ai

Ai

Ai

Ai t

=+

++

++

+ ++( ) ( ) ( )

..................( )1 1 1 11 2 3

P Ai

Ai

Ai

Ai

a A

t=+

++

++

+ ++

=+

( ) ( ) ( )..................

( )

(

1 1 1 1

1

1 2 3

iir

i

S P a rr

Ai i

i

t

t t

),

( )

( ) ( ) ( )

( )

=+

= =−−

=+

−+

−+

=

11

11

11 1

1

1 11

AA ii

i ii

A ii

t

t

t( )( )

( )( )

( )( )

1 11

1 1 11

1 11

+ −+

++ −+

=

+ −+ tt t

t

t

t

iA ii i

A P i ii

=

+ −+

=+

+ −

(( ) )( )

( )(( ) )

1 11

11 1

\



Question 8 (b) (i)

Question 8 (b) (ii)



Sample 5

Paper 1



AC DC CAC x yDC x y

x xy yC

∩ =− − =+ − =

− = ⇒ =− − = ⇒ = −

∴ −

{ }::

( ,

4 02 5 03 9 0 3

3 4 0 13 11)

E is the midpoint of A and C

A C

E

( , ), ( , )

, ( , )

5 1 3 15 3

21 1

24 0

−

=+ −

=

AC BDA C

AC

BD

⊥−

=− −−

=−−

=

= −

( , ), ( , )5 1 3 11 1

3 522

1Slope of

Slope of 115 3

21 1

24 0

1 4 00 1 4

1 1

E

DB m x yy xy

=+ −

=

= − =− = − −

, ( , )

: , ( , ) ( , )( )

== − ++ − =x

x y4

4 0

DC DB DDC x yDB x y

x xy yD

D

∩ =+ − =

+ − =− = ⇒ =

+ − = ⇒ =∴

{ }::

( , )

2 5 04 01 0 1

1 4 0 31 3

(( , ) ( , ) ( , )1 3 4 0 7 3→ → −E B

BA (5, 1)

D C

E

sample paper 5: paper 2Question 1 (25 marks)Question 1 (a)

Question 1 (b) Question 1 (c)[Diagonals of rhombus are perpendicular.]



dNdt

et

ÊËÁ

ˆ¯̃

= ==10

0 006 100 8 0 85. .. ( ) billions of barrels per year

Boundary conditions: At t = 0, N = 0.

dNdt

e

dN e dt

N e c e

t

t

t

=

=

= + =

Ú Ú

0 8

0 8

0 80 006

4003

0 006

0 006

0 006 0

.

.

..

.

.

. ..006t c+

N e c

e c c

N e

t

t

= +

\ = + fi = -

= - =

4003

0 4003

4003

4003

4003

4003

0 006

0

0 006

.

. (ee t0 006 1. )-

t N e= = - =20 4003

1 170 006 20: ( ). ( ) billion

N e

e

t

t

t

= - =

= ¥ + =

=

55 4003

1 55

3 55400

1 11380

10 006

11

0 006

0 006

: ( )

.ln

.

.

3380

57 56 58ÊËÁ

ˆ¯̃

= ª. years

dNdt

et

ÊËÁ

ˆ¯̃

= ==20


Question 9 (a) (i)

Question 9 (a) (ii)

Question 9 (b)

Question 9 (c)

1 0 159

4 714 10 4 714 100 159

2 965 1

3

10 310

barrel m

m barrels

=

¥ = ¥

= ¥

.

. ..

. 00296 5 10296 5

11

9

barrelsbarrels

billion barrels= ¥=

.

.

N e

e

t

t

t

= − =

=×

+ =

296 5 4003

1 296 5

3 296 5400

1 2579800

0 006

0 006

. : ( ) .

.

.

.

==

≈

10 006

2579800

195.

ln years

Question 9 (d) Question 9 (e)



AC DC CAC x yDC x y

x xy yC

∩ =− − =+ − =

− = ⇒ =− − = ⇒ = −

∴ −

{ }::

( ,

4 02 5 03 9 0 3

3 4 0 13 11)

E is the midpoint of A and C

A C

E

( , ), ( , )

, ( , )

5 1 3 15 3

21 1

24 0

−

=+ −

=

AC BDA C

AC

BD

⊥−

=− −−

=−−

=

= −

( , ), ( , )5 1 3 11 1

3 522

1Slope of

Slope of 115 3

21 1

24 0

1 4 00 1 4

1 1

E

DB m x yy xy

=+ −

=

= − =− = − −

, ( , )

: , ( , ) ( , )( )

== − ++ − =x

x y4

4 0

DC DB DDC x yDB x y

x xy yD

D

∩ =+ − =

+ − =− = ⇒ =

+ − = ⇒ =∴

{ }::

( , )

2 5 04 01 0 1

1 4 0 31 3

(( , ) ( , ) ( , )1 3 4 0 7 3→ → −E B

BA (5, 1)

D C

E

sample paper 5: paper 2Question 1 (25 marks)Question 1 (a)

Question 1 (b) Question 1 (c)[Diagonals of rhombus are perpendicular.]



dNdt

et

ÊËÁ

ˆ¯̃

= ==10


Boundary conditions: At t = 0, N = 0.

dNdt

e

dN e dt

N e c e

t

t

t

=

=

= + =

Ú Ú

0 8

0 8

0 80 006

4003

0 006

0 006

0 006 0

.

.

..

.

.

. ..006t c+

N e c

e c c

N e

t

t

= +

\ = + fi = -

= - =

4003

0 4003

4003

4003

4003

4003

0 006

0

0 006

.

. (ee t0 006 1. )-

t N e= = - =20 4003

1 170 006 20: ( ). ( ) billion

N e

e

t

t

t

= - =

= ¥ + =

=

55 4003

1 55

3 55400

1 11380

10 006

11

0 006

0 006

: ( )

.ln

.

.

3380

57 56 58ÊËÁ

ˆ¯̃

= ª. years

dNdt

et

ÊËÁ

ˆ¯̃

= ==20


Question 9 (a) (i)

Question 9 (a) (ii)

Question 9 (b)

Question 9 (c)

1 0 159

4 714 10 4 714 100 159

2 965 1

3

10 310

barrel m

m barrels

=

¥ = ¥

= ¥

.

. ..

. 00296 5 10296 5

11

9

barrelsbarrels

billion barrels= ¥=

.

.

N e

e

t

t

t

= − =

=×

+ =

296 5 4003

1 296 5

3 296 5400

1 2579800

0 006

0 006

. : ( ) .

.

.

.

==

≈

10 006

2579800

195.

ln years


Sample 5

Paper 2



tantan

6060 3

oo= ⇒ = =

hAC

AC h h

Consider the right-angled triangle DBC.

tantan

3030 1

3

3oo= ⇒ = =

=hDC

DC h h h

Consider the right-angled triangle ADC.

DC AC AD

h h

h h

h

h

2 2 2

22

2

22

2

33

24

33

576

83

576

576 38

= +

=

+

= +

=

∴ =×

( )

==14 7. m


Question 3 (b)60

o

h

B

CA

30o

h

B

C

A

24 m

60o

D


PO(1, 2)

5

Q(2, 1)�

10

5− k

s x y x y k

O r k k

:

( , ),

2 2

2 2

2 4 0

1 2 1 2 5

+ − − + =

= + − = −Centre

OPQ is a right-angled triangle.

5 5 1025 5 10

20

2 2 2+ − =+ − =

∴ =

( ) ( )kk

k

V( 3, 1)� �

R(0, 0)

U( 2, 6)�

O

r

U V R

RU

( , ), ( , ), ( , )( )

− − −

=− −−

= − = −

2 6 3 1 0 00 2

0 626

13

Slope of

Slope oof RV

RU RV

=− −− −

= =

− × = − ⇒ ⊥

0 30 1

31

3

13

3 1

( )( )

Each angle in a semicircle is a right angle. Therefore [UV] is the diameter of the circle containing points U, V and R.


Question 2 (b)

V( 3, 1)� �

W x y( , )

U( 2, 6)�

O

r

Or

Slope of Slope ofUW UVyx

yx

y y x

⊥

∴−+

×++

= −

− − = − +

( )( )

( )( )

(

62

13

1

5 6 52 2 xxx y x y

+

+ + − =

65 5 02 2

)

U V

UV O

( , ), ( , )

,

(

− − −

=− − −

= −

2 6 3 12 32

6 12

Midpoint of Centre

552

52

12

12

2 2

12

12

3 2 1 6

1 49 50

, ) ( , )

( ( )) ( )

=

= = − − − + − −

= + =

h k

r UV

Equatioon of circle : ( ) ( )

( ) ( ) ( )

x h y k r

x y

x x

− + − =

+ + − =

+

2 2 2

52

2 52

2 12

2

2

50

5 ++ + − + =

+ + − =

+

254

2 254

504

2 2

2 2

5

5 5 0

y y

x y x yOr

x yEquation of circle : ++ + + =

∈ + + + + = ⇒ =− − = − ⇒

2 2 00 0 2 2 0 02 2

52

52

gx fy cR x y gx fy c cg f

( , )( , ) ( , ) gg f

x y x y

= = −

∴ + + − =

52

52

2 2 5 5 0

,



tantan

6060 3

oo= ⇒ = =

hAC

AC h h

Consider the right-angled triangle DBC.

tantan

3030 1

3

3oo= ⇒ = =

=hDC

DC h h h

Consider the right-angled triangle ADC.

DC AC AD

h h

h h

h

h

2 2 2

22

2

22

2

33

24

33

576

83

576

576 38

= +

=

+

= +

=

∴ =×

( )

==14 7. m


Question 3 (b)60

o

h

B

CA

30o

h

B

C

A

24 m

60o

D


PO(1, 2)

5

Q(2, 1)�

10

5− k

s x y x y k

O r k k

:

( , ),

2 2

2 2

2 4 0

1 2 1 2 5

+ − − + =

= + − = −Centre

OPQ is a right-angled triangle.

5 5 1025 5 10

20

2 2 2+ − =+ − =

∴ =

( ) ( )kk

k

V( 3, 1)� �

R(0, 0)

U( 2, 6)�

O

r

U V R

RU

( , ), ( , ), ( , )( )

− − −

=− −−

= − = −

2 6 3 1 0 00 2

0 626

13

Slope of

Slope oof RV

RU RV

=− −− −

= =

− × = − ⇒ ⊥

0 30 1

31

3

13

3 1

( )( )

Each angle in a semicircle is a right angle. Therefore [UV] is the diameter of the circle containing points U, V and R.


Question 2 (b)

V( 3, 1)� �

W x y( , )

U( 2, 6)�

O

r

Or

Slope of Slope ofUW UVyx

yx

y y x

⊥

∴−+

×++

= −

− − = − +

( )( )

( )( )

(

62

13

1

5 6 52 2 xxx y x y

+

+ + − =

65 5 02 2

)

U V

UV O

( , ), ( , )

,

(

− − −

=− − −

= −

2 6 3 12 32

6 12

Midpoint of Centre

552

52

12

12

2 2

12

12

3 2 1 6

1 49 50

, ) ( , )

( ( )) ( )

=

= = − − − + − −

= + =

h k

r UV

Equatioon of circle : ( ) ( )

( ) ( ) ( )

x h y k r

x y

x x

− + − =

+ + − =

+

2 2 2

52

2 52

2 12

2

2

50

5 ++ + − + =

+ + − =

+

254

2 254

504

2 2

2 2

5

5 5 0

y y

x y x yOr

x yEquation of circle : ++ + + =

∈ + + + + = ⇒ =− − = − ⇒

2 2 00 0 2 2 0 02 2

52

52

gx fy cR x y gx fy c cg f

( , )( , ) ( , ) gg f

x y x y

= = −

∴ + + − =

52

52

2 2 5 5 0

,

Sample 5

Paper 2



σ µµ σµ σ

= =+ = + =− = − =

18 5 55 255 2 18 5 73 755 2 18 5 36 7

. , .. . .. . .

68% of students scored within 1 standard deviation of the mean.

New mean = 58.2New standard deviation = 18.5 (The spread of scores remain unchanged.)

Question 4 (c) (i)


Question 4 (c) (ii)

(i) Two events are said to be independent if one event does not affect the outcome of the other.

(ii) If A and B are independent events P A B P A P B( ) ( ) ( )∩ = ×

(iii) If A and B are independent events P(B | A) = P(B)

(iv) For all events P A B P A P B A( ) ( ) ( | )∩ = ×

0.30.2 0.4

A B

P A B P A P B P A B( ) ( ) ( ) ( ). . ..

∪ = + − ∩= + −=

0 7 0 5 0 30 9

P A B P A BP B

( | ) ( )( )

.

.=

∩= =

0 30 7

37

The events are not independent. P A B P A( | ) ( )

.ππ3

7 0 5


The events A and B are such that P(A) = 0.5, P(B) = 0.7 and P A B( ) . .∩ = 0 3Question 5 (b)

(i)

(ii)

(iii)


Score f x fx CF

0–10 20 5 100 20

10–20 60 15 900 80

20–30 140 25 3500 220

30–40 200 35 7000 420

40–50 400 45 18 000 820

50–60 500 55 27 500 1320

60–70 430 65 27 950 1750

70–80 260 75 19 500 2010

80–90 140 85 11 900 2150

90–100 60 95 5700 2210

2210 122 050

Number of students = 2210 Mean µ = =122 050

221055 2.



Question 4 (b)

The median and mean are nearly the same as the distribution is close to normal.

Num

ber

of

stu

dents

200

100

400

300

600

500

Test Score

10 20 30 40 50 60 70 80 90 1000

The 1105th. student has the median score. This student lies in the 50−60 class interval.

Median =−

× + =( ) .1105 820

50010 50 55 7



σ µµ σµ σ

= =+ = + =− = − =

18 5 55 255 2 18 5 73 755 2 18 5 36 7

. , .. . .. . .

68% of students scored within 1 standard deviation of the mean.

New mean = 58.2New standard deviation = 18.5 (The spread of scores remain unchanged.)

Question 4 (c) (i)


Question 4 (c) (ii)

(i) Two events are said to be independent if one event does not affect the outcome of the other.

(ii) If A and B are independent events P A B P A P B( ) ( ) ( )∩ = ×

(iii) If A and B are independent events P(B | A) = P(B)

(iv) For all events P A B P A P B A( ) ( ) ( | )∩ = ×

0.30.2 0.4

A B

P A B P A P B P A B( ) ( ) ( ) ( ). . ..

∪ = + − ∩= + −=

0 7 0 5 0 30 9

P A B P A BP B

( | ) ( )( )

.

.=

∩= =

0 30 7

37

The events are not independent. P A B P A( | ) ( )

.ππ3

7 0 5


The events A and B are such that P(A) = 0.5, P(B) = 0.7 and P A B( ) . .∩ = 0 3Question 5 (b)

(i)

(ii)

(iii)


Score f x fx CF

0–10 20 5 100 20

10–20 60 15 900 80

20–30 140 25 3500 220

30–40 200 35 7000 420

40–50 400 45 18 000 820

50–60 500 55 27 500 1320

60–70 430 65 27 950 1750

70–80 260 75 19 500 2010

80–90 140 85 11 900 2150

90–100 60 95 5700 2210

2210 122 050

Number of students = 2210 Mean µ = =122 050

221055 2.



Question 4 (b)

The median and mean are nearly the same as the distribution is close to normal.

Num

ber

of

stu

dents

200

100

400

300

600

500

Test Score

10 20 30 40 50 60 70 80 90 1000

The 1105th. student has the median score. This student lies in the 50−60 class interval.

Median =−

× + =( ) .1105 820

50010 50 55 7

Sample 5

Paper 2



No. Maybe there are fewer heavier cars insured with this company or because the premium on these vehicles is so high claims may not be made when minor damage is incurred.

Num

ber

of

cla

ims i

n a

year

10

3000 50004000 600020001000

100

90

80

70

60

40

30

20

Weight of car (kg)

0

50

There is a strong negative correlation between the number of claims per year and the weight of the car.

( , ), ( , )

.

2000 53 5000 1414 53

5000 20000 01m =

−−

= −

Question 7 (e) (i)

Question 7 (e) (ii)


0 8 0 15 500 0 04 4500 0 01 9500 2000 8 0 15 7. . ( ) . ( ) . ( ). .

M M M MM M+ − + − + − =+ − 55 0 04 180 0 01 95 200

350 200550

+ − + − =− =

∴ =

. .M MM

M

Question 7 (d)

Profit per car () M M – 500 M – 4500 M – 9500Probability 0.8 0.15 0.04 0.01


A

B8

E

1.5

4.5

FD C

kEABA

= =+

= =1 5 4 5

1 56

1 54. .

. .

AF

AF

kAFAC AC

AC

2 2 28 6

10

4 10

104

2 5

= +

∴ =

= ⇒ =

∴ = = .

Area of triangle ABC: A = × × =12 1 5 2 1 5. .


Question 6 (b)

Question 6 (c)

BC = − =2 5 1 5 22 2. .

12

12

2 2

1 5

2 5 1 53

2 51 2

1 5 1 2 0 9

× =

× × =

∴ = =

= − =

AC BD

BD

BD

AD

.

. .

..

. . .

This is the average payout per claim per year for the company.

P x( ) . . . .> = + + =450 0 15 0 04 0 01 0 2

Question 7 (75 marks)Question 7 (a)Amount of claim x (/year) 0 1000 5000 10 000Probability 0.8 0.15 0.04 0.01

Question 7 (b)

Question 7 (c)

E xP x= = + + + =∑ ( ) ( . ) ( . ) ( . ) ( . )0 0 8 1000 0 15 5000 0 04 10 000 0 01 450



No. Maybe there are fewer heavier cars insured with this company or because the premium on these vehicles is so high claims may not be made when minor damage is incurred.

Num

ber

of

cla

ims i

n a

year

10

3000 50004000 600020001000

100

90

80

70

60

40

30

20

Weight of car (kg)

0

50

There is a strong negative correlation between the number of claims per year and the weight of the car.

( , ), ( , )

.

2000 53 5000 1414 53

5000 20000 01m =

−−

= −

Question 7 (e) (i)

Question 7 (e) (ii)


0 8 0 15 500 0 04 4500 0 01 9500 2000 8 0 15 7. . ( ) . ( ) . ( ). .

M M M MM M+ − + − + − =+ − 55 0 04 180 0 01 95 200

350 200550

+ − + − =− =

∴ =

. .M MM

M

Question 7 (d)

Profit per car () M M – 500 M – 4500 M – 9500Probability 0.8 0.15 0.04 0.01


A

B8

E

1.5

4.5

FD C

kEABA

= =+

= =1 5 4 5

1 56

1 54. .

. .

AF

AF

kAFAC AC

AC

2 2 28 6

10

4 10

104

2 5

= +

∴ =

= ⇒ =

∴ = = .

Area of triangle ABC: A = × × =12 1 5 2 1 5. .


Question 6 (b)

Question 6 (c)

BC = − =2 5 1 5 22 2. .

12

12

2 2

1 5

2 5 1 53

2 51 2

1 5 1 2 0 9

× =

× × =

∴ = =

= − =

AC BD

BD

BD

AD

.

. .

..

. . .

This is the average payout per claim per year for the company.

P x( ) . . . .> = + + =450 0 15 0 04 0 01 0 2

Question 7 (75 marks)Question 7 (a)Amount of claim x (/year) 0 1000 5000 10 000Probability 0.8 0.15 0.04 0.01

Question 7 (b)

Question 7 (c)

E xP x= = + + + =∑ ( ) ( . ) ( . ) ( . ) ( . )0 0 8 1000 0 15 5000 0 04 10 000 0 01 450

Sample 5

Paper 2



B

7.8 cm

CA 8 cm

5 cm

O

7.8 cm

CA 8 cm

5 cm

B

Tell students to draw the base of the triangle up near the centre of the rectangle in order to leave room below to draw the perpendicular bisectors.


Question 8 (b)


According to this data, the insurance company could claim that its lowest risk is a person aged 69 years driving a car of weight 5800 kg and the highest risk is a person aged 21 years driving a car of weight 1300 kg.

Num

ber

of

cla

ims i

n a

year

10

30 5040 702010

100

90

80

70

60

50

40

30

20

Age in years

0 60

Question 7 (f) (i)

Question 7 (f) (ii)

Outlier (18, 20), learners usually have an experienced driver with them.

Strong negative correlation.

Question 7 (g)



B

7.8 cm

CA 8 cm

5 cm

O

7.8 cm

CA 8 cm

5 cm

B

Tell students to draw the base of the triangle up near the centre of the rectangle in order to leave room below to draw the perpendicular bisectors.


Question 8 (b)


According to this data, the insurance company could claim that its lowest risk is a person aged 69 years driving a car of weight 5800 kg and the highest risk is a person aged 21 years driving a car of weight 1300 kg.

Num

ber

of

cla

ims i

n a

year

10

30 5040 702010

100

90

80

70

60

50

40

30

20

Age in years

0 60

Question 7 (f) (i)

Question 7 (f) (ii)

Outlier (18, 20), learners usually have an experienced driver with them.

Strong negative correlation.

Question 7 (g)

Sample 5

Paper 2



θ

25

7

24

The angle AOC at the centre standing on the arc AC is twice the angle ABC at the circle.

40 2 21600 2 2 21600 2 1 216

2 2 2

2 2

2

= + −

= −

= −

r r r rr rr

( )( ) coscos

( cos )

θ

θ

θ

000 2 1800800 2

400

2 2 2

2 2 2

2 2

= − +

= +

=

=

rr

r

r

( cos sin )(sin sin )

sin

θ θ

θ θ

θ

22224

2520 25

2420 83

∴ =×

=r . km

Question 8 (d)

cos ( cos )

cos cos

2 12

2

1 2

2 2 1

q q

q q

= +

\ = -

40 2 21600 2 2 21600 2 1 216

2 2 2

2 2

2

= + −

= −

= −

r r r rr rr

( )( ) coscos

( cos )

θ

θ

θ

000 2 1 2 11600 4 1

400 1 49625

2 2

2 2

2

= − +

= −

= × −

=

rr

r r

( cos )( cos )

θ

θ

22 576625

400 625576

20 83

∴ =×

=r . km

Alternative method:Apply the Cosine rule to triangle AOC:Apply the Cosine rule to triangle AOC:


O

7.8 cm

CA 8 cm

5 cm

B

r

r r

O

39 km

CA 40 km

25 km

B

r

r r

θ

2θ

Let ∠ =

= + −=

∴ =

ABC θ

θθ

θ

40 25 39 2 25 391950 546

54619

2 2 2 ( )( ) coscos

cos550

725

=

Question 8 (c)



θ

25

7

24

The angle AOC at the centre standing on the arc AC is twice the angle ABC at the circle.

40 2 21600 2 2 21600 2 1 216

2 2 2

2 2

2

= + −

= −

= −

r r r rr rr

( )( ) coscos

( cos )

θ

θ

θ

000 2 1800800 2

400

2 2 2

2 2 2

2 2

= − +

= +

=

=

rr

r

r

( cos sin )(sin sin )

sin

θ θ

θ θ

θ

22224

2520 25

2420 83

∴ =×

=r . km

Question 8 (d)

cos ( cos )

cos cos

2 12

2

1 2

2 2 1

q q

q q

= +

\ = -

40 2 21600 2 2 21600 2 1 216

2 2 2

2 2

2

= + −

= −

= −

r r r rr rr

( )( ) coscos

( cos )

θ

θ

θ

000 2 1 2 11600 4 1

400 1 49625

2 2

2 2

2

= − +

= −

= × −

=

rr

r r

( cos )( cos )

θ

θ

22 576625

400 625576

20 83

∴ =×

=r . km

Alternative method:Apply the Cosine rule to triangle AOC:Apply the Cosine rule to triangle AOC:


O

7.8 cm

CA 8 cm

5 cm

B

r

r r

O

39 km

CA 40 km

25 km

B

r

r r

θ

2θ

Let ∠ =

= + −=

∴ =

ABC θ

θθ

θ

40 25 39 2 25 391950 546

54619

2 2 2 ( )( ) coscos

cos550

725

=

Question 8 (c)

Sample 5

Paper 2



4 1 1 1

1 4 1 1 1 0 1 0 011 1

1

T in T i i T

nn n= + − +

= = + − + = + = ⇒ =

[ ( ) ][ ]

: [ ( ) ][ ] [ ][ ]nn T i Tn T= = + − + = − = = ⇒ =

= = +

2 4 1 1 1 2 1 1 2 0 0 03 4 1

22 2

2

3

: [ ( ) ][ ] [ ][ ] [ ][ ]: [ (( ) ][ ] [ ][ ] [ ][ ]

: [ ( ) ][

− + = − − = − = ⇒ =

= = + −

1 1 1 1 1 0 1 0 0

4 4 1 1

3 33

44

i i i Tn T 11 1 1 1 1 2 2 4 1

0 0 0 125 0

44

100

+ = + + = = ⇒ =

∴ = × +

i T

S

] [ ][ ] [ ][ ]

, , , ,........225 0 25 0 25 1 25× + × + × =



Question 1 (b)



1. Prove true for n = − = − =1 7 4 7 4 31 1:


3. Prove true for n k b bk k= + − = ∈+ +1 7 4 31 1: ,Prove � Proof:


2. Assume true for n k a a ak k k k= − = ∈ ∴ = +: , . .7 4 3 7 3 4�

7 4 7 7 47 3 4 421 7 4 421 4 7

1 1 1

1

1

k k k k

k k

k k

k

aaa

+ + +

+

+

− = −

= + −

= + × −

= + −

( )( )

( 4421 4 33 7 43

1)( )

( )= +

= +=

aa

b

k

k


s1

s2

A

1.5

1.5

1.5

2.75

2.75

h

B

C

D

E

θ

2.75

R

H

BCAB

DEAD h hh h

= ⇒+

=+

+ = ++

1 51 5

2 755 75

1 5 5 75 2 75 1 58 625

..

..

. ( . ) . ( . )

. 11 5 4 125 2 754 5 1 25

4 51 25

3 6

1 51 5 3 6

. . .. .

..

.

sin .. .

h hh

h

= +=

∴ = =

=+

=

cm

θ11 55 1

1 55 1

17 11..

sin ..

.⇒ =

=

−θ o

Height cm

Radius

H

R RH

R

= + + =

= ⇒ =

3 6 2 1 5 2 2 75 12 1

12 1

. ( . ) ( . ) .

: tan . tθ aan . .17 1 3 72o cm=

V R H= = =13

2 13

2 33 72 12 1 175π π ( . ) ( . ) cm

s1

s2

A

1.5

1.5

1.5

2.75

2.75

h

B

C

D

E

θ

2.75

R

H

rr

sin ..

( . ) sin .. sin . sin ..

17 13 6

3 6 17 13 6 17 1 17 13 6

o

o

o o

=−

− =

− =

rr

r rr r

ssin . sin .. sin . ( sin . )

. sin .

17 1 17 13 6 17 1 1 17 1

3 6 17

o o

o o

= +

= +

∴ =

r rr

r 111 17 1

0 82o

o cm( sin . )

.+

=

Volume of empty space with two spheres:V1

43

3 43

3 3175 2 75 1 5 73 75= − − =π π( . ) ( . ) . cm

Volume of empty space with three spheres:V2

43

3 373 75 0 82 71 44= − =. ( . ) .π cm

% . ..

% . %decrease in empty space = −× =

73 75 71 4473 75

100 3 13


Question 9 (b)

Question 9 (c)

Question 9 (d)



4 1 1 1

1 4 1 1 1 0 1 0 011 1

1

T in T i i T

nn n= + − +

= = + − + = + = ⇒ =

[ ( ) ][ ]

: [ ( ) ][ ] [ ][ ]nn T i Tn T= = + − + = − = = ⇒ =

= = +

2 4 1 1 1 2 1 1 2 0 0 03 4 1

22 2

2

3

: [ ( ) ][ ] [ ][ ] [ ][ ]: [ (( ) ][ ] [ ][ ] [ ][ ]

: [ ( ) ][

− + = − − = − = ⇒ =

= = + −

1 1 1 1 1 0 1 0 0

4 4 1 1

3 33

44

i i i Tn T 11 1 1 1 1 2 2 4 1

0 0 0 125 0

44

100

+ = + + = = ⇒ =

∴ = × +

i T

S

] [ ][ ] [ ][ ]

, , , ,........225 0 25 0 25 1 25× + × + × =



Question 1 (b)



1. Prove true for n = − = − =1 7 4 7 4 31 1:


3. Prove true for n k b bk k= + − = ∈+ +1 7 4 31 1: ,Prove � Proof:


2. Assume true for n k a a ak k k k= − = ∈ ∴ = +: , . .7 4 3 7 3 4�

7 4 7 7 47 3 4 421 7 4 421 4 7

1 1 1

1

1

k k k k

k k

k k

k

aaa

+ + +

+

+

− = −

= + −

= + × −

= + −

( )( )

( 4421 4 33 7 43

1)( )

( )= +

= +=

aa

b

k

k


s1

s2

A

1.5

1.5

1.5

2.75

2.75

h

B

C

D

E

θ

2.75

R

H

BCAB

DEAD h hh h

= ⇒+

=+

+ = ++

1 51 5

2 755 75

1 5 5 75 2 75 1 58 625

..

..

. ( . ) . ( . )

. 11 5 4 125 2 754 5 1 25

4 51 25

3 6

1 51 5 3 6

. . .. .

..

.

sin .. .

h hh

h

= +=

∴ = =

=+

=

cm

θ11 55 1

1 55 1

17 11..

sin ..

.⇒ =

=

−θ o

Height cm

Radius

H

R RH

R

= + + =

= ⇒ =

3 6 2 1 5 2 2 75 12 1

12 1

. ( . ) ( . ) .

: tan . tθ aan . .17 1 3 72o cm=

V R H= = =13

2 13

2 33 72 12 1 175π π ( . ) ( . ) cm

s1

s2

A

1.5

1.5

1.5

2.75

2.75

h

B

C

D

E

θ

2.75

R

H

rr

sin ..

( . ) sin .. sin . sin ..

17 13 6

3 6 17 13 6 17 1 17 13 6

o

o

o o

=−

− =

− =

rr

r rr r

ssin . sin .. sin . ( sin . )

. sin .

17 1 17 13 6 17 1 1 17 1

3 6 17

o o

o o

= +

= +

∴ =

r rr

r 111 17 1

0 82o

o cm( sin . )

.+

=

Volume of empty space with two spheres:V1

43

3 43

3 3175 2 75 1 5 73 75= − − =π π( . ) ( . ) . cm

Volume of empty space with three spheres:V2

43

3 373 75 0 82 71 44= − =. ( . ) .π cm

% . ..

% . %decrease in empty space = −× =

73 75 71 4473 75

100 3 13


Question 9 (b)

Question 9 (c)

Question 9 (d)

Sample 6

Paper 1



T T

TT

n

n

n

n

n

n

n

=

=

=

−

−

−

−

−

2 13

2 13

2 13

2 13

1

1

2

1

1

,

=−n 213


Question 3 (b)

3l d�

120o

3l

3 2l d�

( ) ( ) ( ) ( )( ) cos3 3 3 2 2 3 3 2 1209 9 6

2 2 2

2 2 2

l l d l d l d l dl l ld d

= − + − − − −

= − +

o

++ − + − − − +

= − + + − +

9 12 4 2 9 9 2

9 9 6 9 12

2 2 12

2 2

2 2 2 2

l ld d l ld d

l l ld d l ld

( )( )

44 9 9 20 7 27 180 7 6 3

67

3

3 3

2 2 2

2 2

d l ld dd ld ld l d l

d l l

l

+ − +

= − += − −

=

( )( )

,

, ll l l l l l l− − =

67

3 127

3 157

97

, , ,

T S Sn n n= − −1TTn

n−

=1

Constant

a b c bc A2 2 2 2= + − cos

= −

− +

= −

+

=

−

−

3 3 13

3 3 13

3 13

3 13

3 13

1

1

n n

n n

−

=

=

−

− −

n

n n

1

1 1

1 13

3 13

23

2 13

S

S

T S S

n

n

n

n

n n n

= −

= −

= −

−

−

3 1 13

3 1 131

1

−−

−

= −

− −

1

1

3 1 13

3 1 13

n n


z i i

z i

=

+

= +

+ = + =

+

cos sinπ π3 3

12

32

1 32

32

32

32

2

= + = =

294

34

124

3

z x iy

x y

x y

= ⇒ + =

∴ + =

+ =

5 5

5

25

2 2

2 2

1

Re

Im

2 3 4 5�1�2�3�4�5

�1

�2

�3

�4

�5

1

2

3

4

5

x = 4

x y2

+ = 252

A B i i∩ = + −{ , }4 3 4 3


Question 2 (b)

z zx iy x iyxx

+ =+ + − ==

∴ =

88

2 84

This is a circle with centre (0, 0) and radius 5.

This is a straight line.



T T

TT

n

n

n

n

n

n

n

=

=

=

−

−

−

−

−

2 13

2 13

2 13

2 13

1

1

2

1

1

,

=−n 213


Question 3 (b)

3l d�

120o

3l

3 2l d�

( ) ( ) ( ) ( )( ) cos3 3 3 2 2 3 3 2 1209 9 6

2 2 2

2 2 2

l l d l d l d l dl l ld d

= − + − − − −

= − +

o

++ − + − − − +

= − + + − +

9 12 4 2 9 9 2

9 9 6 9 12

2 2 12

2 2

2 2 2 2

l ld d l ld d

l l ld d l ld

( )( )

44 9 9 20 7 27 180 7 6 3

67

3

3 3

2 2 2

2 2

d l ld dd ld ld l d l

d l l

l

+ − +

= − += − −

=

( )( )

,

, ll l l l l l l− − =

67

3 127

3 157

97

, , ,

T S Sn n n= − −1TTn

n−

=1

Constant

a b c bc A2 2 2 2= + − cos

= −

− +

= −

+

=

−

−

3 3 13

3 3 13

3 13

3 13

3 13

1

1

n n

n n

−

=

=

−

− −

n

n n

1

1 1

1 13

3 13

23

2 13

S

S

T S S

n

n

n

n

n n n

= −

= −

= −

−

−

3 1 13

3 1 131

1

−−

−

= −

− −

1

1

3 1 13

3 1 13

n n


z i i

z i

=

+

= +

+ = + =

+

cos sinπ π3 3

12

32

1 32

32

32

32

2

= + = =

294

34

124

3

z x iy

x y

x y

= ⇒ + =

∴ + =

+ =

5 5

5

25

2 2

2 2

1

Re

Im

2 3 4 5�1�2�3�4�5

�1

�2

�3

�4

�5

1

2

3

4

5

x = 4

x y2

+ = 252

A B i i∩ = + −{ , }4 3 4 3


Question 2 (b)

z zx iy x iyxx

+ =+ + − ==

∴ =

88

2 84

This is a circle with centre (0, 0) and radius 5.

This is a straight line.

Sample 6

Paper 1




Question 5 (a)

y f xx

x x x

dydx

xx

= = − +

= − +

= − +

−

−

( ) ln ln2 12

2 12

2 1

2

1

2

× = − +

= ⇒ − + =

− + =∴ =

= = − +

12

2 1

0 2 1 0

2 02

2 22

1 22

2

2

x x

dydx x x

xx

y f ( ) ln = 0

2 0Local minimum A( , )

dydx x x

x x

d ydx

x xx x

d ydx x

= − + = − +

= − = −

= ⇒

− −

− −

2 1 2

4 1 4 1

0 4

22 1

2

23 2

3 2

2

2 3 −− =

− =∴ =

= − +

= − + = −

1 0

4 04

4 24

1 42

1 2 2

2

12

12

xxx

f ( ) ln ln ln

Point oof inflection B( , ln )4 2 12−

dydx x xdydx

t mx

= − +

= − + = − + =

=

2 1

24

14

18

14

18

2

42

Equation of : == −

− + =− + + = ⇒ = −

18

124 2

8 04 8 2 4 0 8 2 8

, ( , ln )

ln ln

Bx y k

k ktEquation of :: (ln )x y− + − =8 8 2 1 0

Question 5 (d)

Question 5 (c)Question 5 (b)

y

x

y f x= ( )

tA

B

dydx x x

xx

dydx

xx

xx x

= − + =−

> ⇒−

>

− >∴ > ∈

2 1 2

0 2 0

2 02

2 2

2

, �


x y xy

x y

x yy y

y yy

x

: :

( )

= ⇒ =

∴ =

− =

− =

− =∴ =

= =

3 2 32

32

832

8

3 2 1616

3 162

24

V1 = Volume of small bucketV2 = Volume of large bucket

∴ = ⇒ = =V V V

V1 2 2

1238

86

43

( )( )( )( )( )x a x b x cx a x bx cx bcx bx cx bcx ax abx

+ + +

= + + + +

= + + + + + +

2

3 2 2 2 aacx abcx a b c x ab ac bc abc

+

= + + + + + + +3 2( ) ( )

( )( )( )( ) ( )

x x xx x xx x x

+ + +

= + + + + + + +

= + + +

1 2 31 2 3 2 3 6 66 11 6

3 2

3 2

( )( )( )( ) ( )

x x xx x xx x x

− − +

= + − − + + − − +

= + − +

1 5 71 5 7 5 7 35 35

37 35

3 2

3 2


Question 4 (c)





Question 5 (a)

y f xx

x x x

dydx

xx

= = − +

= − +

= − +

−

−

( ) ln ln2 12

2 12

2 1

2

1

2

× = − +

= ⇒ − + =

− + =∴ =

= = − +

12

2 1

0 2 1 0

2 02

2 22

1 22

2

2

x x

dydx x x

xx

y f ( ) ln = 0

2 0Local minimum A( , )

dydx x x

x x

d ydx

x xx x

d ydx x

= − + = − +

= − = −

= ⇒

− −

− −

2 1 2

4 1 4 1

0 4

22 1

2

23 2

3 2

2

2 3 −− =

− =∴ =

= − +

= − + = −

1 0

4 04

4 24

1 42

1 2 2

2

12

12

xxx

f ( ) ln ln ln

Point oof inflection B( , ln )4 2 12−

dydx x xdydx

t mx

= − +

= − + = − + =

=

2 1

24

14

18

14

18

2

42

Equation of : == −

− + =− + + = ⇒ = −

18

124 2

8 04 8 2 4 0 8 2 8

, ( , ln )

ln ln

Bx y k

k ktEquation of :: (ln )x y− + − =8 8 2 1 0

Question 5 (d)

Question 5 (c)Question 5 (b)

y

x

y f x= ( )

tA

B

dydx x x

xx

dydx

xx

xx x

= − + =−

> ⇒−

>

− >∴ > ∈

2 1 2

0 2 0

2 02

2 2

2

, �


x y xy

x y

x yy y

y yy

x

: :

( )

= ⇒ =

∴ =

− =

− =

− =∴ =

= =

3 2 32

32

832

8

3 2 1616

3 162

24

V1 = Volume of small bucketV2 = Volume of large bucket

∴ = ⇒ = =V V V

V1 2 2

1238

86

43

( )( )( )( )( )x a x b x cx a x bx cx bcx bx cx bcx ax abx

+ + +

= + + + +

= + + + + + +

2

3 2 2 2 aacx abcx a b c x ab ac bc abc

+

= + + + + + + +3 2( ) ( )

( )( )( )( ) ( )

x x xx x xx x x

+ + +

= + + + + + + +

= + + +

1 2 31 2 3 2 3 6 66 11 6

3 2

3 2

( )( )( )( ) ( )

x x xx x xx x x

− − +

= + − − + + − − +

= + − +

1 5 71 5 7 5 7 35 35

37 35

3 2

3 2


Question 4 (c)


Sample 6

Paper 1



t (hours) θ ( )o C 0 37

t 27

t + 1 26

FC F=

= − = − =

71 632 71 6 32 225

959

.( ) ( . )

o

o

FC

k t T TT T T e

a

a akt

= = = =

= + −

= + −

−

−

1 8 0 5 96 22

22 96 22

10

0

. , . , ,

( )

( )

h h C Co o

θ

ee− × ≈1 8 0 5 52. . o C

θ

θ

θ

= + −

− = −

=−−

=

−

−

−

T T T eT T T e

eT T

T

e T

a akt

a akt

kta

a

kt

( )

( )( )( )

(

0

0

0

0

1

−−−

=−−

TT

kt T TT

a

a

a

a

)( )

ln

θ

θ0

Ta = 20oC, T0 = 37oC

kt

k t

=−−

=

+ =−−

ln ln

( ) ln

37 2027 20

177

1 37 2026 20

==

+ − =

−

+ − =

ln

( ) ln ln

ln

176

1 176

177

17

k t kt

kt k kt66

717

76

0 154 1

×

=

=

−k ln . h


Question 7 (d) (i)

Question 7 (c)

0 154 177

10 154

177

5 76 5 46

. ln

.ln .

t

t

=

∴ =

= =h h mins

Time of death: 4:44 pm

Question 7 (b)

Question 7 (d) (ii)

Question 7 (d) (iii)


Question 6 (25 marks)Question 6 (a)f x x xf x h x h x h

x h x hx h

f x h

( )( ) ( ) ( )

( )

= + −

+ = + + − +

= + + − − −

+

5 45 45 4 4 2

2

2

2 2

−− = + + − − − − − +

= − −

+ −=

−

f x x h x hx h x xh hx h

f x h f xh

h

( )

( ) ( )

5 4 4 2 5 44 2

4

2 2 2

2

22 4 2

4 2

2

0

hx hh

x h

dydx

f x h f xh

xh

−= − −

=+ −

= −

→lim ( ) ( )

y x xx xx xxx

= + −

= − − −

= − − + −

= − − −

= − −

5 44 54 4 92 9

9 2

2

2

2

2

2

( )( )(( ) )

( )Locall maximum: Local maximum: (2, 9)

y x= =9 2,

y x xdydx

x

= + −

= −

5 4

4 2

2

It is not injective because it is not strictly increasing or decreasing for all x∈�. This means that there are y values that correspond to more than one value of x.

Question 6 (b)

Question 6 (d)

Question 6 (c)



t (hours) θ ( )o C 0 37

t 27

t + 1 26

FC F=

= − = − =

71 632 71 6 32 225

959

.( ) ( . )

o

o

FC

k t T TT T T e

a

a akt

= = = =

= + −

= + −

−

−

1 8 0 5 96 22

22 96 22

10

0

. , . , ,

( )

( )

h h C Co o

θ

ee− × ≈1 8 0 5 52. . o C

θ

θ

θ

= + −

− = −

=−−

=

−

−

−

T T T eT T T e

eT T

T

e T

a akt

a akt

kta

a

kt

( )

( )( )( )

(

0

0

0

0

1

−−−

=−−

TT

kt T TT

a

a

a

a

)( )

ln

θ

θ0

Ta = 20oC, T0 = 37oC

kt

k t

=−−

=

+ =−−

ln ln

( ) ln

37 2027 20

177

1 37 2026 20

==

+ − =

−

+ − =

ln

( ) ln ln

ln

176

1 176

177

17

k t kt

kt k kt66

717

76

0 154 1

×

=

=

−k ln . h


Question 7 (d) (i)

Question 7 (c)

0 154 177

10 154

177

5 76 5 46

. ln

.ln .

t

t

=

∴ =

= =h h mins

Time of death: 4:44 pm

Question 7 (b)

Question 7 (d) (ii)

Question 7 (d) (iii)


Question 6 (25 marks)Question 6 (a)f x x xf x h x h x h

x h x hx h

f x h

( )( ) ( ) ( )

( )

= + −

+ = + + − +

= + + − − −

+

5 45 45 4 4 2

2

2

2 2

−− = + + − − − − − +

= − −

+ −=

−

f x x h x hx h x xh hx h

f x h f xh

h

( )

( ) ( )

5 4 4 2 5 44 2

4

2 2 2

2

22 4 2

4 2

2

0

hx hh

x h

dydx

f x h f xh

xh

−= − −

=+ −

= −

→lim ( ) ( )

y x xx xx xxx

= + −

= − − −

= − − + −

= − − −

= − −

5 44 54 4 92 9

9 2

2

2

2

2

2

( )( )(( ) )

( )Locall maximum: Local maximum: (2, 9)

y x= =9 2,

y x xdydx

x

= + −

= −

5 4

4 2

2

It is not injective because it is not strictly increasing or decreasing for all x∈�. This means that there are y values that correspond to more than one value of x.

Question 6 (b)

Question 6 (d)

Question 6 (c)

Sample 6

Paper 1



y (m)

x (m)

80604020 100�40 0�20�100 �60�80

60

80

20

40

200

100

180

140

160

120

h =32 m, y1 = 0 m, y2 = 126 m, y3 = 178 m, y4 = 192 m, y5 = 178 m, y6 = 126 m, y7 = 0 m

A h y y y y y yn n= + + + + + +

= + +

−22

322

0 0 2 1

1 2 3 4 1[ ( ................. )]

[ ( 226 178 192 178 126

25600 2

+ + + +

=

)]

m

e x x

e x x x x

e e x

x

x

x x

= + +

= + − +−

= − +

∴ + = +

−

−

12

12

12

2

2

2 2

2

( ) ( )

Question 8 (c)

Question 8 (d) (i)

e e x

e e x x

y e

x x

x x

+ = +

∴ + = + = +

= −

−

−

2

2 21521

231 19 57

2

139

22

139

139

13

( )

. { 99139

231 19 57 21521

231 39 14 0 013191

2

2

x xe

x

x

+

= − +

= − −

=

− }

.

. ..886 0 013 2− . x

Question 8 (d) (ii)


x (m) –96 –64 –32 0 32 64 96

y (m) 0 126 178 192 178 126 0

y (m)

x (m)

80604020 100�40 0�20�100 �60�80

60

80

20

40

200

100

180

140

160

120

Maximum height = 192 mx = 50 m, y = 154 m

1 0 30481

0 30481

10 3048

192 192 192 629 92126

ft m

ft m

ft m m ft

=

=

× = ⇒ =

.

.

..

%% . % . %error = −

× =

630 629 92126630

100 0 0125


Question 8 (a) (ii)

Question 8 (b) (i)

Question 8 (b) (ii)



y (m)

x (m)

80604020 100�40 0�20�100 �60�80

60

80

20

40

200

100

180

140

160

120

h =32 m, y1 = 0 m, y2 = 126 m, y3 = 178 m, y4 = 192 m, y5 = 178 m, y6 = 126 m, y7 = 0 m

A h y y y y y yn n= + + + + + +

= + +

−22

322

0 0 2 1

1 2 3 4 1[ ( ................. )]

[ ( 226 178 192 178 126

25600 2

+ + + +

=

)]

m

e x x

e x x x x

e e x

x

x

x x

= + +

= + − +−

= − +

∴ + = +

−

−

12

12

12

2

2

2 2

2

( ) ( )

Question 8 (c)

Question 8 (d) (i)

e e x

e e x x

y e

x x

x x

+ = +

∴ + = + = +

= −

−

−

2

2 21521

231 19 57

2

139

22

139

139

13

( )

. { 99139

231 19 57 21521

231 39 14 0 013191

2

2

x xe

x

x

+

= − +

= − −

=

− }

.

. ..886 0 013 2− . x

Question 8 (d) (ii)


x (m) –96 –64 –32 0 32 64 96

y (m) 0 126 178 192 178 126 0

y (m)

x (m)

80604020 100�40 0�20�100 �60�80

60

80

20

40

200

100

180

140

160

120

Maximum height = 192 mx = 50 m, y = 154 m

1 0 30481

0 30481

10 3048

192 192 192 629 92126

ft m

ft m

ft m m ft

=

=

× = ⇒ =

.

.

..

%% . % . %error = −

× =

630 629 92126630

100 0 0125


Question 8 (a) (ii)

Question 8 (b) (i)

Question 8 (b) (ii) Sample 6

Paper 1



V t dt

t

=

=−

∫3400 02

100

3400 02

100100

0

0 02

0

0 0

.sin

.cos( )

.

.

p

pp

22

3402

100 0 02 100 0

3402

2 0

= − −

= − −

pp p

pp

(cos( ( . ) cos( ( ))

(cos( cos() ))) = 0

Question 9 (d)

Question 9 (e)

Question 9 (f)

Period sV 2 2200

1100

0 01= = =pp

.

V tdVdt

t t

dVd

=

= × =

340 100

340 100 100 34 000 100

sin( )

cos( ) cos( )

p

p p p p

ttdVdt

t

t

= =

=

−

=

0

1

0 005

34 000 100 0 34 000p p pcos( ( ))

.

Vs

== = =

−

=

34 000 100 0 005 34 000 012

1

0

p p p pcos( ( . )) cos( ) Vs

dVdt t ..

cos( ( . )) cos( )01

134 000 100 0 01 34 000 34 000= = = − −p p p p p Vs

V tV t

t

=

== × −

=

340 100340 100

115 600 1 20057

2 2 2

12

sin( )sin ( )

( cos )

pp

p8800 1 200( cos )− pt

V t dt

t t

2

0

0 0110 01

57 800 1 200

5 780 000 200200

= −

= −

∫.( cos )

sin( )

.p

pp

= −

=

0

0 01

5 780 000 0 01 200 0 01200

5 780 000 0

.

. sin( . ))pp(

.. sin01 2200

57 800 2

−

=

pp

V


y xy x

x

x

= −

= ⇒ − =

=

=

191 86 0 0130 191 86 0 013 0

191 86 0 013

191 86

2

2

2

. .. .. .

.00 013

243.

≈ m

y ax

y xa

a

a

= −

= = =

=

∴ = =

191 86

0 191 86 96

191 86 9216

191 869216

0 02

2.

: .

.

. . 008



t 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

100pt 0 12 p p 3

2 p 2p 52 p 3p 7

2 p 4p

sin( )100pt 0 1 0 -1 0 1 0 -1 0

340 100sin( )pt 0 340 0 -340 0 340 0 -340 0

V t

P

R

=

= = =

= −

340 1002

1001

500 02

340 340

sin( )

.

[ ,

ppp

Period s s

Range ]]

Maximum voltage is 340 V.

0.01 0.02 0.03 0.04

100

200

300

400

�100

�200

�300

�400

V (V)

t (s)

500

�500

0


Question 9 (c)



V t dt

t

=

=−

∫3400 02

100

3400 02

100100

0

0 02

0

0 0

.sin

.cos( )

.

.

p

pp

22

3402

100 0 02 100 0

3402

2 0

= − −

= − −

pp p

pp

(cos( ( . ) cos( ( ))

(cos( cos() ))) = 0

Question 9 (d)

Question 9 (e)

Question 9 (f)

Period sV 2 2200

1100

0 01= = =pp

.

V tdVdt

t t

dVd

=

= × =

340 100

340 100 100 34 000 100

sin( )

cos( ) cos( )

p

p p p p

ttdVdt

t

t

= =

=

−

=

0

1

0 005

34 000 100 0 34 000p p pcos( ( ))

.

Vs

== = =

−

=

34 000 100 0 005 34 000 012

1

0

p p p pcos( ( . )) cos( ) Vs

dVdt t ..

cos( ( . )) cos( )01

134 000 100 0 01 34 000 34 000= = = − −p p p p p Vs

V tV t

t

=

== × −

=

340 100340 100

115 600 1 20057

2 2 2

12

sin( )sin ( )

( cos )

pp

p8800 1 200( cos )− pt

V t dt

t t

2

0

0 0110 01

57 800 1 200

5 780 000 200200

= −

= −

∫.( cos )

sin( )

.p

pp

= −

=

0

0 01

5 780 000 0 01 200 0 01200

5 780 000 0

.

. sin( . ))pp(

.. sin01 2200

57 800 2

−

=

pp

V


y xy x

x

x

= −

= ⇒ − =

=

=

191 86 0 0130 191 86 0 013 0

191 86 0 013

191 86

2

2

2

. .. .. .

.00 013

243.

≈ m

y ax

y xa

a

a

= −

= = =

=

∴ = =

191 86

0 191 86 96

191 86 9216

191 869216

0 02

2.

: .

.

. . 008



t 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

100pt 0 12 p p 3

2 p 2p 52 p 3p 7

2 p 4p

sin( )100pt 0 1 0 -1 0 1 0 -1 0

340 100sin( )pt 0 340 0 -340 0 340 0 -340 0

V t

P

R

=

= = =

= −

340 1002

100150

0 02

340 340

sin( )

.

[ ,

ppp

Period s s

Range ]]

Maximum voltage is 340 V.

0.01 0.02 0.03 0.04

100

200

300

400

�100

�200

�300

�400

V (V)

t (s)

500

�500

0


Question 9 (c)

Sample 6

Paper 1



3 + 4 + 1 = 0x y

O( 1, 4)�

10

3 + 4 + = 0x y k

r = 5r = 5

43

+=

0x

yk

�

The length of a side of the square is 10 units. A line parallel to 3x + 4y + 1 = 0 has the form 3x + 4y + k = 0.Pick a point on the line 3x + 4y + 1 = 0. (1, -1) is on this line.

d = 10, (x1, y1) = (1, -1)

103 1 4 1

3 4

103 4

550 1

50 149 51

2 2=

+ − +

+

=− +

= −

± = −∴ = −

( ) ( )

,

k

k

kk

k

Equations: 3x + 4y - 49 = 0, 3x + 4y + 51 = 0

3 + 4 + 1 = 0x y

43

+=

0x

yk

�

O( 1, 4)�

r = 5

s x y x y

O r

:

( , ), ( )

2 2

2 2

2 8 8 0

1 4 1 4 8 25 5

+ + − − =

− = − + + = =Centre

Equation of tangents: 4x - 3y + k = 0

dax by c

a b=

+ +

+1 1

2 2

4 1 3 4

4 35

4 1225

5

16 2516 25

9 41

2 2

( ) ( )

( )

,

− − +

+ −=

− − +=

− + =

− + = ±∴ = −

k

k

kk

k

t1: 4x – 3y + 41 = 0, t2: 4x – 3y – 9 = 0


Question 2 (b)

dax by c

a b=

+ +

+1 1

2 2


(x, 0) is a point on the x-axis.

dax by c

a b=

+ +

+1 1

2 2Distance d1 of (x, 0) to line k: d

x x1 2 2

3 4 0 5

3 4

3 55

=− +

+=

+( )

Distance d2 of (x, 0) to line l: dx x

2 2 2

5 12 0 1

5 12

5 113

=− −

+=

−( )

d dx x

x xx x xx

1 2

3 55

5 113

13 3 5 5 5 139 65 25 5 14 70

= ⇒+

=−

+ = ± −+ = − ⇒ = −

∴ =

( ) ( )

−−

+ = − + ⇒ = −∴ = −

5

39 65 25 5 64 601516

orx x xx

m2 2= −

m t1 10=

m = − 34

tanθ = −+

m mm m

1 2

1 21

tan ( )( )

tan ( )( )(

θ

θ

=− −

+ −=

+−

=− − −+ − −

t

t

tt

1034

1034

34

34

14 3040 3

21 2 ))

( ) ( )

=− ++

=−

=

∴+−

= ±

+ = −+ =

8 34 6

510

12

4 3040 3

12

2 4 30 1 40 38 60 40

tt

t tt −−= − ⇒ = −

+ = − −+ = − += −

311 20

2 4 30 1 40 38 60 40 35 100

2011

tt t

ort t

t tt

( ) ( )

⇒⇒ = −t 20



θθ

Question 1 (b)

anSwerS: ( , ), ( , )- -5 0 01516



3 + 4 + 1 = 0x y

O( 1, 4)�

10

3 + 4 + = 0x y k

r = 5r = 5

43

+=

0x

yk

�

The length of a side of the square is 10 units. A line parallel to 3x + 4y + 1 = 0 has the form 3x + 4y + k = 0.Pick a point on the line 3x + 4y + 1 = 0. (1, -1) is on this line.

d = 10, (x1, y1) = (1, -1)

103 1 4 1

3 4

103 4

550 1

50 149 51

2 2=

+ − +

+

=− +

= −

± = −∴ = −

( ) ( )

,

k

k

kk

k

Equations: 3x + 4y - 49 = 0, 3x + 4y + 51 = 0

3 + 4 + 1 = 0x y

43

+=

0x

yk

�

O( 1, 4)�

r = 5

s x y x y

O r

:

( , ), ( )

2 2

2 2

2 8 8 0

1 4 1 4 8 25 5

+ + − − =

− = − + + = =Centre

Equation of tangents: 4x - 3y + k = 0

dax by c

a b=

+ +

+1 1

2 2

4 1 3 4

4 35

4 1225

5

16 2516 25

9 41

2 2

( ) ( )

( )

,

− − +

+ −=

− − +=

− + =

− + = ±∴ = −

k

k

kk

k

t1: 4x – 3y + 41 = 0, t2: 4x – 3y – 9 = 0


Question 2 (b)

dax by c

a b=

+ +

+1 1

2 2


(x, 0) is a point on the x-axis.

dax by c

a b=

+ +

+1 1

2 2Distance d1 of (x, 0) to line k: d

x x1 2 2

3 4 0 5

3 4

3 55

=− +

+=

+( )

Distance d2 of (x, 0) to line l: dx x

2 2 2

5 12 0 1

5 12

5 113

=− −

+=

−( )

d dx x

x xx x xx

1 2

3 55

5 113

13 3 5 5 5 139 65 25 5 14 70

= ⇒+

=−

+ = ± −+ = − ⇒ = −

∴ =

( ) ( )

−−

+ = − + ⇒ = −∴ = −

5

39 65 25 5 64 601516

orx x xx

m2 2= −

m t1 10=

m = − 34

tanθ = −+

m mm m

1 2

1 21

tan ( )( )

tan ( )( )(

θ

θ

=− −

+ −=

+−

=− − −+ − −

t

t

tt

1034

1034

34

34

14 3040 3

21 2 ))

( ) ( )

=− ++

=−

=

∴+−

= ±

+ = −+ =

8 34 6

510

12

4 3040 3

12

2 4 30 1 40 38 60 40

tt

t tt −−= − ⇒ = −

+ = − −+ = − += −

311 20

2 4 30 1 40 38 60 40 35 100

2011

tt t

ort t

t tt

( ) ( )

⇒⇒ = −t 20



θθ

Question 1 (b)

anSwerS: ( , ), ( , )- -5 0 01516

Sample 6

Paper 2



68%

µ�σ��σ µ

�σ��σ µ�σ��σ

97.5%95%

µ σµ σµ µ

σσ

σ

− =+ =

= ⇒ =

− ==

∴ =

2 24 82 36 4

2 61 2 30 6

30 6 2 24 85 8 2

.

.

. .

. ..

years

22 9. years

µσ

µσ

==≥ =

= =−

=−

= −

30 62 9

30

30 30 30 62 9

0 21

..

( ) ?

: ..

.

yearsyears

P x

x z x

PP x P zP zP zP z

( ) ( . )( . )( . )

{ ( .

≥ = ≥ −= − ≤ −= − ≥= − − ≤

30 0 211 0 211 0 211 1 0 21))}

( . )}.. %

= ≤==

P z 0 210 583258 32



In any normal distribution with mean x and standard deviation s.1. 68% of the data falls within 1s of the mean x.2. 95% of the data falls within 2s of the mean x.3. 99.7% of the data falls within 3s of the mean x.


a b c bc A2 2 2 2= + − cos ,

b a c ac B2 2 2 2= + − cos ,

c a b ab C2 2 2 2= + − cos

A

B C

a

bc

|BD| = |DC|A

B CD

7

3.5

4

x x��

� θθ

Apply the Cosine rule to triangle ABD: 4 3 5 2 3 52 2 2= + −. ( . )( ) cosx x θ

Apply the Cosine rule to triangle ADC: 7 3 5 2 3 5 1802 2 2= + − −. ( . )( ) cos( )x x o θ

16 3 5 749 3 5 765 2 3 5 2

2 2

2 2

2 2

2 814

9

= + −

= + +

= +

= ⇒ =

. cos. cos( . )

x xx x

xx x

θ

θ

22

9∴ =BC


Question 3 (b)

Question 3 (c)



68%

µ�σ��σ µ

�σ��σ µ�σ��σ

97.5%95%

µ σµ σµ µ

σσ

σ

− =+ =

= ⇒ =

− ==

∴ =

2 24 82 36 4

2 61 2 30 6

30 6 2 24 85 8 2

.

.

. .

. ..

years

22 9. years

µσ

µσ

==≥ =

= =−

=−

= −

30 62 9

30

30 30 30 62 9

0 21

..

( ) ?

: ..

.

yearsyears

P x

x z x

PP x P zP zP zP z

( ) ( . )( . )( . )

{ ( .

≥ = ≥ −= − ≤ −= − ≥= − − ≤

30 0 211 0 211 0 211 1 0 21))}

( . )}.. %

= ≤==

P z 0 210 583258 32



In any normal distribution with mean x and standard deviation s.1. 68% of the data falls within 1s of the mean x.2. 95% of the data falls within 2s of the mean x.3. 99.7% of the data falls within 3s of the mean x.


a b c bc A2 2 2 2= + − cos ,

b a c ac B2 2 2 2= + − cos ,

c a b ab C2 2 2 2= + − cos

A

B C

a

bc

|BD| = |DC|A

B CD

7

3.5

4

x x��

� θθ

Apply the Cosine rule to triangle ABD: 4 3 5 2 3 52 2 2= + −. ( . )( ) cosx x θ

Apply the Cosine rule to triangle ADC: 7 3 5 2 3 5 1802 2 2= + − −. ( . )( ) cos( )x x o θ

16 3 5 749 3 5 765 2 3 5 2

2 2

2 2

2 2

2 814

9

= + −

= + +

= +

= ⇒ =

. cos. cos( . )

x xx x

xx x

θ

θ

22

9∴ =BC


Question 3 (b)

Question 3 (c)

Sample 6

Paper 2



s

B

AC

D

E

β

90o

90

o

�

β

90o

β

90 o

�

β

90 o

�

�β

β

Consider triangle ADE. Let ∠ =DAE β .[AE] is a diameter. Therefore, ∠ =ADE 90o. (The angle at the circle standing on a diameter is a right angle).

∴∠ = − − = −DEA 180 90 90o o oβ β

∠ + ∠ =AED ABD 180o (Opposite angles of a cyclic quadrilateral add up to 180o.)

∴∠ = − − = +ABD 180 90 90o o o( )β β

∠ = ∠ =DAE ADB β (Alternate angles)

∴∠ = − − + = −DAB 180 90 90 2o o oβ β β( )

∴∠ = − − − =BAC 90 90 2o oβ β β( )

∴∠ = ∠ = ∠ =BAC DAE ADB β



y(T

housands)

1 2

3.8

4 5

1

2

3

4

5

6

7

8

9

10

x (cm)

( , )x y

3

4.35

x

y

=+ + + + + + + +

=

=+ + +

4 4 3 0 5 2 5 0 2 1 0 0 1 1 3 29

2 7

5100 7900 900 4000

. . . . . . . .

++ + + + +=

4700 7300 9500 8700 26009

5633

( . , ), ( , )

.

2 7 5633 0 90009000 5633

0 2 71247m =

−−

= −

For every increase of 1 cm in rainfall, there is a decrease in 1247 tourists.

m x yy x

x y

= − =− = − −

+ − =

1247 0 90009000 1247 0

1247 9000 0

1 1, ( , ) ( , )( )

(i) Number of tourists = 4350

x yy= + − =

∴ =3 8 1247 3 8 9000 0

4261. : ( . )


Rainfall (x cm) 4.4 3.0 5.2 5.0 2.1 0 0 1.1 3.2Number of tourists (y thousands) 5.1 7.9 0.9 4.0 4.7 7.3 9.5 8.7 2.6

Question 5 (b)

Question 5 (c) [See graph]


Question 5 (f) Question 5 (g)

(ii)

The student’s result is smaller by 89 or is about 2% smaller ( % %).89

4350 100 2× ≈



s

B

AC

D

E

β

90o

90

o

�

β

90o

β

90 o

�

β

90 o

�

�β

β

Consider triangle ADE. Let ∠ =DAE β .[AE] is a diameter. Therefore, ∠ =ADE 90o. (The angle at the circle standing on a diameter is a right angle).

∴∠ = − − = −DEA 180 90 90o o oβ β

∠ + ∠ =AED ABD 180o (Opposite angles of a cyclic quadrilateral add up to 180o.)

∴∠ = − − = +ABD 180 90 90o o o( )β β

∠ = ∠ =DAE ADB β (Alternate angles)

∴∠ = − − + = −DAB 180 90 90 2o o oβ β β( )

∴∠ = − − − =BAC 90 90 2o oβ β β( )

∴∠ = ∠ = ∠ =BAC DAE ADB β



y(T

housands)

1 2

3.8

4 5

1

2

3

4

5

6

7

8

9

10

x (cm)

( , )x y

3

4.35

x

y

=+ + + + + + + +

=

=+ + +

4 4 3 0 5 2 5 0 2 1 0 0 1 1 3 29

2 7

5100 7900 900 4000

. . . . . . . .

++ + + + +=

4700 7300 9500 8700 26009

5633

( . , ), ( , )

.

2 7 5633 0 90009000 5633

0 2 71247m =

−−

= −

For every increase of 1 cm in rainfall, there is a decrease in 1247 tourists.

m x yy x

x y

= − =− = − −

+ − =

1247 0 90009000 1247 0

1247 9000 0

1 1, ( , ) ( , )( )

(i) Number of tourists = 4350

x yy= + − =

∴ =3 8 1247 3 8 9000 0

4261. : ( . )


Rainfall (x cm) 4.4 3.0 5.2 5.0 2.1 0 0 1.1 3.2Number of tourists (y thousands) 5.1 7.9 0.9 4.0 4.7 7.3 9.5 8.7 2.6

Question 5 (b)

Question 5 (c) [See graph]


Question 5 (f) Question 5 (g)

(ii)

The student’s result is smaller by 89 or is about 2% smaller ( % %).89

4350 100 2× ≈

Sample 6

Paper 2



6 1 18 1 38 38 1 18 1 18 1 2 124 38 38 38 3876 52

× + × + = − + × + × + ×+ = − +=

=

x xx x

x

x

( )

55276

1319

77 5 4 80 21319

=

∴ = + × =Median . .

Mean Median

difference

µ ==

=−

× =

81 280 2

81 2 80 281 2

100 1 23

..

% . ..

% . %

(ii) It is approximately normal because the mean is approximately equal to the median.

σ = 4 75.

Question 7 (c)

µ σ

µσ

= =≤ = ⇒ =

=−

⇒ =−

∴ =

81 2 4 750 9 1 28

1 28 81 24 75

87

. , .( ) . .

. ..

.

P z Z z

z x x

x 33%

µ σ

µσ

= =< =

= =−

=−

=

< =

81 2 4 7585

85 85 81 24 75

0 8

0 8

. , .( ) ?

: ..

.

( . )

P x

x z x

P z 00 7881 78 8. . %=

(iii)

(i)

Question 7 (d)(i) (ii)

Conditions for a Bernoulli Trial:Condition: There are only two possible outcomes (success or failure) in each trial.Condition: There is a fixed number of trials n.Condition: The probability of success p is fixed from trial to trial.Condition: The trials are independent.Condition: The binomial random variable is the number of successes in n trials.

(iii) He has two successes and two failures on the first four throws and he scores on the last.

Question 7 (e)

Question 7 (f)(i)

(ii)

PPP C

( ) .( ) .( ) ( .

SuccessFailureScores all five

==

=

0 9270 073

0 955 227 0 073 0 685 68 55 0) ( . ) . . %= =

P C( ) ( . ) ( . ) . . %Scores three out of five = = =53

3 20 927 0 073 0 042 4 2

P C( ) ( . )Scores two out of first four and scores last = 42 0 927 22 20 073 0 927 0 025 2 5( . ) ( . ) . . %= =


Percentage success Frequency f Mid-interval Value x fx

70–74 6 72 432

74–78 18 76 1368

78–82 38 80 3040

82–86 18 84 1512

86–90 18 88 1584

90–94 2 92 184

100 8120

µ = = =∑∑fxf

8120100

81 2.

Mean percentage success rate = 81.2%


40

35

30

25

20

15

10

94

5

86 9078 8270 74

Num

ber

of

pla

yers

Percentage success rate at the free-throw line

Percentage success Cumulative frequency< 74 6< 78 24< 82 62< 86 80< 90 98< 94 100

Median

Question 7 (b)



6 1 18 1 38 38 1 18 1 18 1 2 124 38 38 38 3876 52

× + × + = − + × + × + ×+ = − +=

=

x xx x

x

x

( )

55276

1319

77 5 4 80 21319

=

∴ = + × =Median . .

Mean Median

difference

µ ==

=−

× =

81 280 2

81 2 80 281 2

100 1 23

..

% . ..

% . %

(ii) It is approximately normal because the mean is approximately equal to the median.

σ = 4 75.

Question 7 (c)

µ σ

µσ

= =≤ = ⇒ =

=−

⇒ =−

∴ =

81 2 4 750 9 1 28

1 28 81 24 75

87

. , .( ) . .

. ..

.

P z Z z

z x x

x 33%

µ σ

µσ

= =< =

= =−

=−

=

< =

81 2 4 7585

85 85 81 24 75

0 8

0 8

. , .( ) ?

: ..

.

( . )

P x

x z x

P z 00 7881 78 8. . %=

(iii)

(i)

Question 7 (d)(i) (ii)

Conditions for a Bernoulli Trial:Condition: There are only two possible outcomes (success or failure) in each trial.Condition: There is a fixed number of trials n.Condition: The probability of success p is fixed from trial to trial.Condition: The trials are independent.Condition: The binomial random variable is the number of successes in n trials.

(iii) He has two successes and two failures on the first four throws and he scores on the last.

Question 7 (e)

Question 7 (f)(i)

(ii)

PPP C

( ) .( ) .( ) ( .

SuccessFailureScores all five

==

=

0 9270 073

0 955 227 0 073 0 685 68 55 0) ( . ) . . %= =

P C( ) ( . ) ( . ) . . %Scores three out of five = = =53

3 20 927 0 073 0 042 4 2

P C( ) ( . )Scores two out of first four and scores last = 42 0 927 22 20 073 0 927 0 025 2 5( . ) ( . ) . . %= =


Percentage success Frequency f Mid-interval Value x fx

70–74 6 72 432

74–78 18 76 1368

78–82 38 80 3040

82–86 18 84 1512

86–90 18 88 1584

90–94 2 92 184

100 8120

µ = = =∑∑fxf

8120100

81 2.

Mean percentage success rate = 81.2%


40

35

30

25

20

15

10

94

5

86 9078 8270 74

Num

ber

of

pla

yers

Percentage success rate at the free-throw line

Percentage success Cumulative frequency< 74 6< 78 24< 82 62< 86 80< 90 98< 94 100

Median

Question 7 (b)

Sample 6

Paper 2



OC

OC OD DF FE ECa b

a b

=

= + + +

∴ = + + ++ =

10

10 3 52

cm

A

B

a

C

O

D

E

5

b

F

5

33

*

*

(ii) Triangles OAF and CBF are similar because:

∠ = ∠

∠ = ∠

∴ ∠ = ∠

OAF FBC

OFA BFC

AOF B

( )

( )

90o angles

Vertically opposite

CCF

BCOA

FCOF

ba

a ba b

b a

a b

a a a

= ⇒ =++

+ = +=

∴ =

+ =

+ = ⇒ +

53

53

5 15 3 155 3

532

53

2 3 5aa

aa

b

=

=∴ =

∴ = =

6

8 60 755 0 75

31 25

.( . ) .

cm

cm

AB AF FB

AOF OF OA AF

AF

AF

FB

= +

∆ = +

∴ = +

= − =

∆

:

.

. .

2 2 2

2 2 2

2 2

3 75 3

3 75 3 2 25 cm

CC FC BC FB

FB

FB

AB

:

.

. .

. .

2 2 2

2 2 2

2 2

6 25 5

6 25 5 3 75

2 25 3 75

= +

∴ = +

= − =

∴ = +

cm

== 6 cm

Question 9 (70 marks)Question 9 (a)(i)

(iii)

(iv)





( )1

Confidence interval = [sample parameter - 1.96(standard error), sample parameter + 1.96(standard error)]

For about 95% of all samples the confidence interval covers the population values of the parameter and for the other 5% it does not.For a 99% confidence level, the confidence interval must be wider so that more samples cover the population parameter.


Sample proportion

Population proportion

p

n

= =

=

=

200300

23

300

23

±± - =1 96 1300

0 613 0 7223

23. ( ) . , .

Confidence interval = [0.613, 0.72]For about 95% of all samples, the population proportion of all fifth years who believe the Junior Certificate should be scrapped is between 61.3% and 72%.

m s= ± ¥xn

1 96.Standard deviation of population s = 10 cmMean of the sample x = 173 cmNumber of the sample n = 50Population mean m = ?

m = ± ¥ =173 170 23 175 771.96 1050

. , .

For about 95% of all samples, the population mean height of university students is between 170.23 cm and 175.77 cm.

Question 8 (b)

Question 8 (c)



OC

OC OD DF FE ECa b

a b

=

= + + +

∴ = + + ++ =

10

10 3 52

cm

A

B

a

C

O

D

E

5

b

F

5

33

*

*

(ii) Triangles OAF and CBF are similar because:

∠ = ∠

∠ = ∠

∴ ∠ = ∠

OAF FBC

OFA BFC

AOF B

( )

( )

90o angles

Vertically opposite

CCF

BCOA

FCOF

ba

a ba b

b a

a b

a a a

= ⇒ =++

+ = +=

∴ =

+ =

+ = ⇒ +

53

53

5 15 3 155 3

532

53

2 3 5aa

aa

b

=

=∴ =

∴ = =

6

8 60 755 0 75

31 25

.( . ) .

cm

cm

AB AF FB

AOF OF OA AF

AF

AF

FB

= +

∆ = +

∴ = +

= − =

∆

:

.

. .

2 2 2

2 2 2

2 2

3 75 3

3 75 3 2 25 cm

CC FC BC FB

FB

FB

AB

:

.

. .

. .

2 2 2

2 2 2

2 2

6 25 5

6 25 5 3 75

2 25 3 75

= +

∴ = +

= − =

∴ = +

cm

== 6 cm

Question 9 (70 marks)Question 9 (a)(i)

(iii)

(iv)





( )1


For about 95% of all samples the confidence interval covers the population values of the parameter and for the other 5% it does not.For a 99% confidence level, the confidence interval must be wider so that more samples cover the population parameter.


Sample proportion

Population proportion

p

n

= =

=

=

200300

23

300

23

±± - =1 96 1300

0 613 0 7223

23. ( ) . , .

Confidence interval = [0.613, 0.72]For about 95% of all samples, the population proportion of all fifth years who believe the Junior Certificate should be scrapped is between 61.3% and 72%.

m s= ± ¥xn

1 96.Standard deviation of population s = 10 cmMean of the sample x = 173 cmNumber of the sample n = 50Population mean m = ?

m = ± ¥ =173 170 23 175 771.96 1050

. , .

For about 95% of all samples, the population mean height of university students is between 170.23 cm and 175.77 cm.

Question 8 (b)

Question 8 (c)

Sample 6

Paper 2



First 10 natural squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

1 2 2 3 56

1 4 306

5 5

2 2+ =× ×

+ =

=

1 2 3 3 4 76

1 4 9 2 714 14

2 2 2+ + =× ×

+ + = ×=

1 2 3 4 4 5 96

1 4 9 16 1806

30 30

2 2 2 2+ + + =× ×

+ + + =

=

1 2 3 4 10 10 11 216

5 11 7 3852 2 2 2 2+ + + + + =× ×

= × × =.......

1 2 3 1 2 16

2 2 2 2+ + + + =+ +....... ( )( )n n n n S100

100 101 2016

338 350= =( )( )



Consecutive numbers: n − 1, nConsecutive squares: n n2 21, ( )+n n n n n n n n n2 2 2 2 2 21 2 1 2 1 2 1− − = − − + = − + − = −( ) ( )2n is always an even number2n − 1 is always an odd number

Question 1 (a) (ii)


Question 1 (b) (iv)



A

B(3, 4)

C

O

D

E

5

F

5

3

0.75

3

1.25

x

2.25

3.75

c1

c2

t

θ

�

�

�

�

�

�

θ

θ

y

x P

E(0, 5)F(0, 6.25)D(0, 7)O(0, 10)

c x yc x y

12 2

22 2

10 925

: ( ):

+ − =

+ =

tan.

θ = =5

3 7543

∠ = − − = −

∴ ∠ = − − =

FCB

BCP

180 90 90

90 90

o o o

o o

θ θ

θ θ

( )

( )

θ3

45

sin

cos

( , )

θ

θ

= = ⇒ =

= = ⇒ =

∴

y y

x x

B

545

4

535

3

3 4

Slope of BC = 43

Slope of tangent t = − 34

Equation of t: m x y= − =34 1 1 3 4, ( , ) ( , )

y xy xx y

− = − −

− = − ++ − =

4 34 16 3 93 4 25 0

34 ( )

c x y

t x y x y

y y

12 2

22

10 9

3 4 25 0 25 43

25 43

10

: ( )

:

( )

+ − =

+ − = ⇒ =−

∴−

+ − ==

− ++ − + − =

− + + − +

9

625 200 169

20 100 9 0

625 200 16 9 180 900

22

2 2

y y y y

y y y y −− =

− + =− − =

∴ =

=−

= −

81 025 380 1444 05 38 5 38 0

25 43

2

385

385

y yy yy

x

( )( )

( ) 995

95

385∴ −A( , )

A B

AB

( , ), ( , )

( ) ( )

−

= + + − =

95

385

95

2 385

2

3 4

3 4 6

Question 9 (b)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)



First 10 natural squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

1 2 2 3 56

1 4 306

5 5

2 2+ =× ×

+ =

=

1 2 3 3 4 76

1 4 9 2 714 14

2 2 2+ + =× ×

+ + = ×=

1 2 3 4 4 5 96

1 4 9 16 1806

30 30

2 2 2 2+ + + =× ×

+ + + =

=

1 2 3 4 10 10 11 216

5 11 7 3852 2 2 2 2+ + + + + =× ×

= × × =.......

1 2 3 1 2 16

2 2 2 2+ + + + =+ +....... ( )( )n n n n S100

100 101 2016

338 350= =( )( )



Consecutive numbers: n − 1, nConsecutive squares: n n2 21, ( )+n n n n n n n n n2 2 2 2 2 21 2 1 2 1 2 1− − = − − + = − + − = −( ) ( )2n is always an even number2n − 1 is always an odd number

Question 1 (a) (ii)


Question 1 (b) (iv)



A

B(3, 4)

C

O

D

E

5

F

5

3

0.75

3

1.25

x

2.25

3.75

c1

c2

t

θ

�

�

�

�

�

�

θ

θ

y

x P

E(0, 5)F(0, 6.25)D(0, 7)O(0, 10)

c x yc x y

12 2

22 2

10 925

: ( ):

+ − =

+ =

tan.

θ = =5

3 7543

∠ = − − = −

∴ ∠ = − − =

FCB

BCP

180 90 90

90 90

o o o

o o

θ θ

θ θ

( )

( )

θ3

45

sin

cos

( , )

θ

θ

= = ⇒ =

= = ⇒ =

∴

y y

x x

B

545

4

535

3

3 4

Slope of BC = 43

Slope of tangent t = − 34

Equation of t: m x y= − =34 1 1 3 4, ( , ) ( , )

y xy xx y

− = − −

− = − ++ − =

4 34 16 3 93 4 25 0

34 ( )

c x y

t x y x y

y y

12 2

22

10 9

3 4 25 0 25 43

25 43

10

: ( )

:

( )

+ − =

+ − = ⇒ =−

∴−

+ − ==

− ++ − + − =

− + + − +

9

625 200 169

20 100 9 0

625 200 16 9 180 900

22

2 2

y y y y

y y y y −− =

− + =− − =

∴ =

=−

= −

81 025 380 1444 05 38 5 38 0

25 43

2

385

385

y yy yy

x

( )( )

( ) 995

95

385∴ −A( , )

A B

AB

( , ), ( , )

( ) ( )

−

= + + − =

95

385

95

2 385

2

3 4

3 4 6

Question 9 (b)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Sample 7

Paper 1



zzz z z

z i

3

3

2

2

11 01 1 0

1 1 4 1 12 1

1 32

1 32

=

− =

− + + =

=− ± −

=− ± −

=− ±

( )( )

( )( )( )

zz i i= − + − −1 12

32

12

32

, ,

1 21 3

21 3

1 31 3

2 2 31 32 2 3

41 3

2

w i

iii

iii

=− +

=− +

×− −− −

=− −

−

=− −

=− −

( )( )( )

ii w2

2=


Question 3 (b)

w i

w i i i

=− +

= − + = − −

12

32

14

32

34

12

32

22

Roots

Sum :

Product

: , ,ww

i i

S i i

P

1 12

32

12

32

1 32

1 32

1

=− + − −

− ++− −

= −

:: ww

z Sz Pz zz z

× =

− + =

− − + =

∴ + + =

1 1

01 1 0

1 0

2

2

2

( )


S pn qnS p n q n pn pn p qn qS S T T p

n

n

n n n n

= +

= − + − = − + + −

− = ⇒ =−

−

2

12 2

1

1 1 2( ) ( )

nn qn pn pn p qn qpn p q

T p n p qT T p nn

n n

2 2

1

1

22

2 12

+ − + − − += − +

= + − +− = +

+

+

( )( 11 2

2 2 22

) − + − + −= + − + − + −= =

=

p q pn p qpn p p q pn p qp

a T

Common Difference

11

1

22

⇒ = − + = += − =+

a p p q p qd T T pn n

S rr

rr

50 3 3

3

32

2450 25 2 1 49 24500 49 98

2

3 9

= ⇒ + =+ =

=

∴ = =

[ log log ]log

log



a ar ara ar ara a r a r

d a

, ,log , log , log

log , log log , log loglog

2

2

2= + += + llog log log

log[ log ( ) log ]

r a ra aS a n rn

n

− === + −2 2 1



zzz z z

z i

3

3

2

2

11 01 1 0

1 1 4 1 12 1

1 32

1 32

=

− =

− + + =

=− ± −

=− ± −

=− ±

( )( )

( )( )( )

zz i i= − + − −1 12

32

12

32

, ,

1 21 3

21 3

1 31 3

2 2 31 32 2 3

41 3

2

w i

iii

iii

=− +

=− +

×− −− −

=− −

−

=− −

=− −

( )( )( )

ii w2

2=


Question 3 (b)

w i

w i i i

=− +

= − + = − −

12

32

14

32

34

12

32

22

Roots

Sum :

Product

: , ,ww

i i

S i i

P

1 12

32

12

32

1 32

1 32

1

=− + − −

− ++− −

= −

:: ww

z Sz Pz zz z

× =

− + =

− − + =

∴ + + =

1 1

01 1 0

1 0

2

2

2

( )


S pn qnS p n q n pn pn p qn qS S T T p

n

n

n n n n

= +

= − + − = − + + −

− = ⇒ =−

−

2

12 2

1

1 1 2( ) ( )

nn qn pn pn p qn qpn p q

T p n p qT T p nn

n n

2 2

1

1

22

2 12

+ − + − − += − +

= + − +− = +

+

+

( )( 11 2

2 2 22

) − + − + −= + − + − + −= =

=

p q pn p qpn p p q pn p qp

a T

Common Difference

11

1

22

⇒ = − + = += − =+

a p p q p qd T T pn n

S rr

rr

50 3 3

3

32

2450 25 2 1 49 24500 49 98

2

3 9

= ⇒ + =+ =

=

∴ = =

[ log log ]log

log



a ar ara ar ara a r a r

d a

, ,log , log , log

log , log log , log loglog

2

2

2= + += + llog log log

log[ log ( ) log ]

r a ra aS a n rn

n

− === + −2 2 1

Sample 7

Paper 1




f x’( )

x

1 3

6

′ = = + +

∈ ′ ⇒ + + =∴ =

∈ ′ ⇒

f x y ax bx cf x a b c

c

f x

( )( , ) ( ) ( ) ( )

( , ) ( )

2

20 6 0 0 66

1 0 aa ba b

f x a ba

( ) ( )....( )

( , ) ( ) ( ) ( )

1 1 6 06

3 0 3 3 6 09

2

2

+ + =∴ + = −

∈ ′ ⇒ + + =∴

1

++ = −+ = −

− = ⇒ =+ = − ⇒ = −

∴

3 63 2

2 4 22 6 8

ba b

a ab b

....( )

( ) ( ) :( ) ....( )

2

2 11

′′ = = − +f x y x x( ) 2 8 62

Local minimum at x = 3 as dydx

= 0 and the

slope is positive.

Local maximum at x = 1 as dydx

= 0 and the

slope is negative.

f x’( )

x

1 3

6

D

′ = = − +′′ = −′′ = ⇒ − =

∴ =

f x y x xf x xf x xx

( )( )( )

2 8 64 80 4 8 0

2

2

dydx

x x

y x x x cx y cy f x x x x

= − +

= − + +

= = ⇒ =

∴ = = − +

2 8 6

4 60 0 0

4 6

2

23

3 2

23

3 2

,( )

f x x x x

f

( )

( ) ( ) ( ) ( )

= − +

= − + = − + =

23

3 2

23

3 2

4 6

3 3 4 3 6 3 18 36 18 0Local minimmum:

Local maximum:

( , )

( ) ( ) ( ) ( )( ,

3 0

1 1 4 1 6 1 4 61

23

3 2 23

83f = − + = − + =

883

23

3 2 163

432 2 4 2 6 2 16 12

)

( ) ( ) ( ) ( )f = − + = − + =

Point of inflection:: ( , )2 43


Question 5 (c)



Using the lining up method, a cubic equals a quadratic by a linear.

x px qx r x px q x tx px qx r x tx px ptx q

3 2 2

3 2 3 2 2

3 33 3

+ + + = − + +

+ + + = + − − +

( )( )xx qt

x px qx r x t p x q pt x qt+

+ + + = + − + − +3 2 3 23 3 ( ) ( )

Lining up the coefficients gives you three equations. Replace t from equation (1) in the other equations.

34p t pp t= −=

.....( )1 33 42 4

2

2

2

q q ptq q p pq pq p

= −= −

= −

= −

.....( )( )

..( )

2

i

r qtr q pr pq

===

.....( )( )

34

4

Result (i) is proved under equation (2).To prove result (ii) replace q under equation (3).r pq p p p= = − = −4 4 2 82 3( ) ...( )ii

x px qx r x px q x p3 2 23 3 4+ + + = − + +( )( )

⇒ + + + = − − + = − + + =x px qx r x px p x p x p x p x p3 2 2 23 3 2 4 2 4 0( )( ) ( )( )( )∴ = − −x p p p4 2, ,


Question 4 (a)

Question 4 (b)

Question 4 (c)




f x’( )

x

1 3

6

′ = = + +

∈ ′ ⇒ + + =∴ =

∈ ′ ⇒

f x y ax bx cf x a b c

c

f x

( )( , ) ( ) ( ) ( )

( , ) ( )

2

20 6 0 0 66

1 0 aa ba b

f x a ba

( ) ( )....( )

( , ) ( ) ( ) ( )

1 1 6 06

3 0 3 3 6 09

2

2

+ + =∴ + = −

∈ ′ ⇒ + + =∴

1

++ = −+ = −

− = ⇒ =+ = − ⇒ = −

∴

3 63 2

2 4 22 6 8

ba b

a ab b

....( )

( ) ( ) :( ) ....( )

2

2 11

′′ = = − +f x y x x( ) 2 8 62

Local minimum at x = 3 as dydx

= 0 and the

slope is positive.

Local maximum at x = 1 as dydx

= 0 and the

slope is negative.

f x’( )

x

1 3

6

D

′ = = − +′′ = −′′ = ⇒ − =

∴ =

f x y x xf x xf x xx

( )( )( )

2 8 64 80 4 8 0

2

2

dydx

x x

y x x x cx y cy f x x x x

= − +

= − + +

= = ⇒ =

∴ = = − +

2 8 6

4 60 0 0

4 6

2

23

3 2

23

3 2

,( )

f x x x x

f

( )

( ) ( ) ( ) ( )

= − +

= − + = − + =

23

3 2

23

3 2

4 6

3 3 4 3 6 3 18 36 18 0Local minimmum:

Local maximum:

( , )

( ) ( ) ( ) ( )( ,

3 0

1 1 4 1 6 1 4 61

23

3 2 23

83f = − + = − + =

883

23

3 2 163

432 2 4 2 6 2 16 12

)

( ) ( ) ( ) ( )f = − + = − + =

Point of inflection:: ( , )2 43


Question 5 (c)



Using the lining up method, a cubic equals a quadratic by a linear.

x px qx r x px q x tx px qx r x tx px ptx q

3 2 2

3 2 3 2 2

3 33 3

+ + + = − + +

+ + + = + − − +

( )( )xx qt

x px qx r x t p x q pt x qt+

+ + + = + − + − +3 2 3 23 3 ( ) ( )

Lining up the coefficients gives you three equations. Replace t from equation (1) in the other equations.

34p t pp t= −=

.....( )1 33 42 4

2

2

2

q q ptq q p pq pq p

= −= −

= −

= −

.....( )( )

..( )

2

i

r qtr q pr pq

===

.....( )( )

34

4

Result (i) is proved under equation (2).To prove result (ii) replace q under equation (3).r pq p p p= = − = −4 4 2 82 3( ) ...( )ii

x px qx r x px q x p3 2 23 3 4+ + + = − + +( )( )

⇒ + + + = − − + = − + + =x px qx r x px p x p x p x p x p3 2 2 23 3 2 4 2 4 0( )( ) ( )( )( )∴ = − −x p p p4 2, ,


Question 4 (a)

Question 4 (b)

Question 4 (c)

Sample 7

Paper 1



Payment # Fixed Payment Interest Debt Payment Balance

0 60 000

1 14 438.07 3900 10 538.07 49 461.93

2 14 438.07 3215.03 11 223.04 38 238.89

3 14 438.07 2485.53 11 952.54 26 286.35

4 14 438.07 1708.61 12 729.46 13 556.89

5 14 438.07 881.20 13 556.87 0

CalCulation for Year 1Payment Number 1: 14 438.07Interest: 60 000 0 065 3900× =.Debt Payment: 14 438 07 3900 00 10 538 07. . .− =Balance: 60000 10538 07 49461 93− =. .


Question 7 (b)

Question 7 (c) (i)

Question 7 (c) (ii)

P = + + + + + +250 2501 045

2501 045

2501 045

2501 045

2501 045

251 2 3 4 5. . . . .

001 045

2501 045

250 1 11 045

11 045

11 045

11 045

1

6 7

1 2 3 4

. .

. . . .

+

= + + + + +P11 045

11 045

11 045

1 11 045

8

2501 1

1 04

5 6 7. . .

,.

,

.

+ +

= = =

∴ =

−

a r n

P55

1 11 045

1723 18

8

−

=

.

.

A P i ii

t

t=+

+ −( )

( )1

1 1

P Fi t

=+( )1

P = =5000

1 0453515 938( . )

.

Minimum price = 3515.93 + 1723.18 = 5239.11Minimum price bonds can be offered is 5239 to the nearest euro.


P t i

A

= = =

=−

=

60 000 5 0 065

60 000 0 065 1 0651 065 1

15

5

, , .. ( . )( . )

years

44 438 07.


Question 6 (25 marks)y

x

x = 1

1 4

A

yx

x= − +

32 32

42

yx

x

x

= − +

= − + = − + =

32 32

4

4 324

3 42

4 2 6 4 0

2

2: ( )

A = + + + + + + =0 52

34 5 0 2 15 97 9 5 37 3 06 1 36 26 005. { . ( . . . . )} .

A x x dx

x x x

xx x

= − +

=−

− +

= − − +

−

−

∫ ( )32 4

321

34

4

32 34

4

2 321

4

1 2

1

4

2

= − − +

− − − +

= − −

1

4

2 2324

3 44

4 4 321

3 14

4 1

8

( ) ( ) ( ) ( )

112 16 32 424 75

34

994

+ + + −

= = .

% . ..

% %error = −

× ≈

26 005 24 7524 75

100 5

x 1 1.5 2 2.5 3 3.5 4y 34.5 15.97 9.00 5.37 3.06 1.36 0

Question 6 (a)

Question 6 (b)

Question 6 (c)

Question 6 (d)



Payment # Fixed Payment Interest Debt Payment Balance

0 60 000

1 14 438.07 3900 10 538.07 49 461.93

2 14 438.07 3215.03 11 223.04 38 238.89

3 14 438.07 2485.53 11 952.54 26 286.35

4 14 438.07 1708.61 12 729.46 13 556.89

5 14 438.07 881.20 13 556.87 0

CalCulation for Year 1Payment Number 1: 14 438.07Interest: 60 000 0 065 3900× =.Debt Payment: 14 438 07 3900 00 10 538 07. . .− =Balance: 60000 10538 07 49461 93− =. .


Question 7 (b)

Question 7 (c) (i)

Question 7 (c) (ii)

P = + + + + + +250 2501 045

2501 045

2501 045

2501 045

2501 045

251 2 3 4 5. . . . .

001 045

2501 045

250 1 11 045

11 045

11 045

11 045

1

6 7

1 2 3 4

. .

. . . .

+

= + + + + +P11 045

11 045

11 045

1 11 045

8

2501 1

1 04

5 6 7. . .

,.

,

.

+ +

= = =

∴ =

−

a r n

P55

1 11 045

1723 18

8

−

=

.

.

A P i ii

t

t=+

+ −( )

( )1

1 1

P Fi t

=+( )1

P = =5000

1 0453515 938( . )

.

Minimum price = 3515.93 + 1723.18 = 5239.11Minimum price bonds can be offered is 5239 to the nearest euro.


P t i

A

= = =

=−

=

60 000 5 0 065

60 000 0 065 1 0651 065 1

15

5

, , .. ( . )( . )

years

44 438 07.


Question 6 (25 marks)y

x

x = 1

1 4

A

yx

x= − +

32 32

42

yx

x

x

= − +

= − + = − + =

32 32

4

4 324

3 42

4 2 6 4 0

2

2: ( )

A = + + + + + + =0 52

34 5 0 2 15 97 9 5 37 3 06 1 36 26 005. { . ( . . . . )} .

A x x dx

x x x

xx x

= − +

=−

− +

= − − +

−

−

∫ ( )32 4

321

34

4

32 34

4

2 321

4

1 2

1

4

2

= − − +

− − − +

= − −

1

4

2 2324

3 44

4 4 321

3 14

4 1

8

( ) ( ) ( ) ( )

112 16 32 424 75

34

994

+ + + −

= = .

% . ..

% %error = −

× ≈

26 005 24 7524 75

100 5

x 1 1.5 2 2.5 3 3.5 4y 34.5 15.97 9.00 5.37 3.06 1.36 0

Question 6 (a)

Question 6 (b)

Question 6 (c)

Question 6 (d)

Sample 7

Paper 1



pH values: 2 0 2 5 3 02 0 5 6

1 2 5

. , . , . ,......., . ,

( )a d nT a n dn

= = == + − = + (( . ) .0 5 4 5=

pH values: 2.0, 2.5, 3.0, 3.5, 4.0, 4.5.After five hours the pH value is 4.5.

Arithmetic sequence: Geometric sequence:

2 2 5 3 3 510

, . , , . ,....−22 2 5 3

22 5

20 5

8 7

10 10

10 1010

10

10 10

, , ,......

,

.

..

.

− −

−−

−−

− −

= = =

∴ =

a r

22 0 5 1

6 7 0 5 0 5

1010 10

6 7 0 5 0 50 5 0 5 6 7

( )

. . .. . .

.

. . .

− −

− − +=− = − +

= + =

n

n

nn 77 2

7 20 5

14 4

...

.∴ = =n hours

Therefore, the pH value is 8.7 after 13.4 hours.

Question 8 (d) (i)Or

Question 8 (d) (ii)


Period

Range

= = =

= −

2200

1100

0 01

10 10

ppp p

min . min

[ , ]

t (minutes) 0 0.0025 0.005 0.0075 0.01 0.0125 0.015 0.0175 0.02

200pt 0p2

p 32p

2p52p

3p72p

4p

sin( )200pt 0 1 0 −1 0 1 0 −1 0

10 200p psin( )t 0 10p 0 -10p 0 10p 0 -10p 0

Question 9 (b)


pH value [ ]H +

Acidic < 7 > −10 7

Alkaline > 7 < −10 7

Neutral = 7 = −10 7

x yx y x y

+ = ×+ = × ⇒ + =

15 0 40 4 0 25 0 3 15 0 4 0 25 4 5

0

....( )( . ). . . . . . ...( )

12

.. .

. . .. .

4 0 4 60 4 0 25 4 5

0 15 1 5 10

10 15 5

x yx y

y y

x x

+ =+ =

= ⇒ =

+ = ⇒ =

l

l

There is 4.5 l of acid in 15 l of the solution ( . . ).15 0 3 4 5× =When 1 l of water is added, there is now 16 l of solution.

Concentration of acid in solution = × =4 516

100 28 125. % . %

pHpH

= −=

∴ = −

− =

∴ =

+

+

+

− +

log [ ]

log [ ]

log [ ]

[ ]

10

10

107

77

7

10

H

HH

H

Apple juice: moles per litrepH

[ ] .log [ ] log

HH

+

+

=

= − = −

0 00028

10 100

9

0 00028 3 55 7

1 32 10

( . ) . ( )

[ ] .

= <

= ×+ −

Acidic

Ammonia: moles perH litrepH Alkaline= − = − × = >+ −log [ ] log ( . ) . ( )10 10

91 32 10 8 88 7H

Distilled water: pH moles per litre= ⇒ =+ −7 10 7[ ]HpH =

∴ = −

− =

=

∴ = ×

+

+

− +

+

3 223 22

3 22

106

10

103 22

.. log [ ]

. log [ ]

[ ][ ]

.

HH

HH 110 4− moles/litre


Question 8 (a) (ii)

Question 8 (b)

Question 8 (c) (i)

Question 8 (c) (ii)



pH values: 2 0 2 5 3 02 0 5 6

1 2 5

. , . , . ,......., . ,

( )a d nT a n dn

= = == + − = + (( . ) .0 5 4 5=

pH values: 2.0, 2.5, 3.0, 3.5, 4.0, 4.5.After five hours the pH value is 4.5.

Arithmetic sequence: Geometric sequence:

2 2 5 3 3 510

, . , , . ,....−22 2 5 3

22 5

20 5

8 7

10 10

10 1010

10

10 10

, , ,......

,

.

..

.

− −

−−

−−

− −

= = =

∴ =

a r

22 0 5 1

6 7 0 5 0 5

1010 10

6 7 0 5 0 50 5 0 5 6 7

( )

. . .. . .

.

. . .

− −

− − +=− = − +

= + =

n

n

nn 77 2

7 20 5

14 4

...

.∴ = =n hours

Therefore, the pH value is 8.7 after 13.4 hours.

Question 8 (d) (i)Or

Question 8 (d) (ii)


Period

Range

= = =

= −

2200

1100

0 01

10 10

ppp p

min . min

[ , ]

t (minutes) 0 0.0025 0.005 0.0075 0.01 0.0125 0.015 0.0175 0.02

200pt 0p2

p 32p

2p52p

3p72p

4p

sin( )200pt 0 1 0 −1 0 1 0 −1 0

10 200p psin( )t 0 10p 0 -10p 0 10p 0 -10p 0

Question 9 (b)


pH value [ ]H +

Acidic < 7 > −10 7

Alkaline > 7 < −10 7

Neutral = 7 = −10 7

x yx y x y

+ = ×+ = × ⇒ + =

15 0 40 4 0 25 0 3 15 0 4 0 25 4 5

0

....( )( . ). . . . . . ...( )

12

.. .

. . .. .

4 0 4 60 4 0 25 4 5

0 15 1 5 10

10 15 5

x yx y

y y

x x

+ =+ =

= ⇒ =

+ = ⇒ =

l

l

There is 4.5 l of acid in 15 l of the solution ( . . ).15 0 3 4 5× =When 1 l of water is added, there is now 16 l of solution.

Concentration of acid in solution = × =4 516

100 28 125. % . %

pHpH

= −=

∴ = −

− =

∴ =

+

+

+

− +

log [ ]

log [ ]

log [ ]

[ ]

10

10

107

77

7

10

H

HH

H

Apple juice: moles per litrepH

[ ] .log [ ] log

HH

+

+

=

= − = −

0 00028

10 100

9

0 00028 3 55 7

1 32 10

( . ) . ( )

[ ] .

= <

= ×+ −

Acidic

Ammonia: moles perH litrepH Alkaline= − = − × = >+ −log [ ] log ( . ) . ( )10 10

91 32 10 8 88 7H

Distilled water: pH moles per litre= ⇒ =+ −7 10 7[ ]HpH =

∴ = −

− =

=

∴ = ×

+

+

− +

+

3 223 22

3 22

106

10

103 22

.. log [ ]

. log [ ]

[ ][ ]

.

HH

HH 110 4− moles/litre


Question 8 (a) (ii)

Question 8 (b)

Question 8 (c) (i)

Question 8 (c) (ii)

Sample 7

Paper 1



A( 5, 8)�

B(3, 8)�

C( , 2)�

ABC

( , ) ( , )( , ) ( , )( , ) ( , )

( ) (

− →− → −→ −

= − −

5 8 0 03 8 8 166 2 11 6

8 6 1112Area −− = − + = =16 48 176 128 641

212)

A( 5, 8)�

B(3, 8)�C( , 2)�

D( , 18)��

B CA D

( , ) ( , )( , ) ( , )3 8 6 2

5 8 2 18− →

− → −

Equation of BC: B C

m x y

y x

( , ), ( , )( ) , ( , ) ( , )

( ) (

3 8 6 22 8

6 3103

3 8

8 3

1 1

103

−

=− −−

= = −

− − = − ))3 24 10 3010 3 54 0y xx y+ = −− − =

A( 5, 8)�

B(3, 8)�C( , 2)�

D( , 18)��

d

d =− − −

+ −=

10 2 3 18 54

10 31281092 2

( ) ( )

( )



Area = −12 1 2 2 1x y x y

Question 1 (b)

y y m x x− = −1 1( )

B C

CB

( , ), ( , )

( ) ( ( ))

3 8 6 2

6 3 2 8 1092 2

−= ×

= − + − − =

Area Base Height

Base:

AArea = × =109 128109

128

dax by c

a b=

+ +

+1 1

2 2

or

Area of parallelogram ABCD = 2(Area of triangle ABC)


(i) Breathing in: 0 s − 0.0025 s, 0.0075 s − 0.0125 s, 0.0175 s − 0.02 s(ii) Breathing out: 0.0025 s − 0.0075 s, 0.0125 s − 0.0175 s(iii) Maximum values: 0.0025 s, 0.0125 s

Question 9 (c)

dVdt

t

dV t dt

V t

=

=

= −

∫ ∫

10 200

10 200

10200

200

p p

p p

pp

p

sin( )

sin( )

cos( ) ++

= − += = = − +

c

V t ct V c

0 05 2000 2 95 2 95 0 05 200 0

2

. cos( ), . : . . cos( ( ))

pp

.. .

. cos( )

95 0 0530 05 200 3

= − +∴ == − +

cc

V tp

V t

V t d

= − +

=−

− +∫

0 05 200 31

0 01 00 05 200 3

0

0 01

. cos( )

.( . cos( ) )

.

p

pAve. tt

t t= −

= −

100 3 0 5200

200

100 3 0 01 0 05200

0

0 01. sin( )

( . ) . s

.

pp

piin( ( . ))

. sin( )

200 0 01 0

3 0 05200

2

3

p

pp

−

= −

= litres


00.01 0.015 0.02

2�

4�

6�

8�

10�

��2

�4�

�6�

�8�

�10�

M

t (minutes)0.005

R (litres/minute)

M



A( 5, 8)�

B(3, 8)�

C( , 2)�

ABC

( , ) ( , )( , ) ( , )( , ) ( , )

( ) (

− →− → −→ −

= − −

5 8 0 03 8 8 166 2 11 6

8 6 1112Area −− = − + = =16 48 176 128 641

212)

A( 5, 8)�

B(3, 8)�C( , 2)�

D( , 18)��

B CA D

( , ) ( , )( , ) ( , )3 8 6 2

5 8 2 18− →

− → −

Equation of BC: B C

m x y

y x

( , ), ( , )( ) , ( , ) ( , )

( ) (

3 8 6 22 8

6 3103

3 8

8 3

1 1

103

−

=− −−

= = −

− − = − ))3 24 10 3010 3 54 0y xx y+ = −− − =

A( 5, 8)�

B(3, 8)�C( , 2)�

D( , 18)��

d

d =− − −

+ −=

10 2 3 18 54

10 31281092 2

( ) ( )

( )



Area = −12 1 2 2 1x y x y

Question 1 (b)

y y m x x− = −1 1( )

B C

CB

( , ), ( , )

( ) ( ( ))

3 8 6 2

6 3 2 8 1092 2

−= ×

= − + − − =

Area Base Height

Base:

AArea = × =109 128109

128

dax by c

a b=

+ +

+1 1

2 2

or

Area of parallelogram ABCD = 2(Area of triangle ABC)


(i) Breathing in: 0 s − 0.0025 s, 0.0075 s − 0.0125 s, 0.0175 s − 0.02 s(ii) Breathing out: 0.0025 s − 0.0075 s, 0.0125 s − 0.0175 s(iii) Maximum values: 0.0025 s, 0.0125 s

Question 9 (c)

dVdt

t

dV t dt

V t

=

=

= −

∫ ∫

10 200

10 200

10200

200

p p

p p

pp

p

sin( )

sin( )

cos( ) ++

= − += = = − +

c

V t ct V c

0 05 2000 2 95 2 95 0 05 200 0

2

. cos( ), . : . . cos( ( ))

pp

.. .

. cos( )

95 0 0530 05 200 3

= − +∴ == − +

cc

V tp

V t

V t d

= − +

=−

− +∫

0 05 200 31

0 01 00 05 200 3

0

0 01

. cos( )

.( . cos( ) )

.

p

pAve. tt

t t= −

= −

100 3 0 5200

200

100 3 0 01 0 05200

0

0 01. sin( )

( . ) . s

.

pp

piin( ( . ))

. sin( )

200 0 01 0

3 0 05200

2

3

p

pp

−

= −

= litres


00.01 0.015 0.02

2�

4�

6�

8�

10�

��2

�4�

�6�

�8�

�10�

M

t (minutes)0.005

R (litres/minute)

M

Sample 7

Paper 2



135o

45o

A(2, 6)

B

O

y

x

k

Line k makes an angle of 45o with the positive x-axis. The slope of k is the tan of the angle the line makes with the positive x-axis.

tan , ( , ) ( , )( )

:

45 1 2 66 1 26 2

4 0

1 1o = = =

− = −− = −− + =

m x y Ay xy xk x y

P( 1, 2)�

Q x y( , )O(0, 0)

x y y xQ x y x xP

x x

x

+ − = ⇒ = −= −

−

= − − − =

∴

1 0 11

1 22 1 1 7

2

12

( , ) ( , )( , )

( )( )Area

++ − =

+ =

+ = ±

∴ = − >= − = −

∴ −

1 14

1 141 14

13 15 01 1213 12

x

xx

x xy xQ

, ( )

( , )

The midpoint of [PQ] is the centre of the circle. Call it R.

P Q

R

( , ), ( , )

, ( , )

− −

=− + −

= −

1 2 13 121 13

22 12

26 5

Equation of circle:

P R

r PR

( , ), ( , )

( ) ( )

− −

= = + + − − = + =

1 2 6 5

6 1 5 2 49 49 7 22 2

( ) ( )( ) ( )x h y k rx y− + − =

− + + =

2 2 2

2 26 5 98


Question 2 (a)

y y m x x− = −1 1( )

Area = −12 1 2 2 1x y x y

Question 2 (b)

Question 2 (c)

R( , 5)��

Q(13, 12)� O(0, 0)

P( 1, 2)�


A( 5, 8)�

B(3, 8)�C( , 2)�

D( , 18)��

�

�

Slope of AC: m18 25 6

611

=−

− −= −

Slope of BD: m218 8

2 3265

=+

− −= −

tan( )( )

tan

θ

θ

= +−

+

=

− ++ −

∴ =− +−

m mm m

1 2

1 2

611

265

611

265

16

11

1 12265

611

2651

50 5+ −

=( )( )

. o

Question 1 (c)



135o

45o

A(2, 6)

B

O

y

x

k

Line k makes an angle of 45o with the positive x-axis. The slope of k is the tan of the angle the line makes with the positive x-axis.

tan , ( , ) ( , )( )

:

45 1 2 66 1 26 2

4 0

1 1o = = =

− = −− = −− + =

m x y Ay xy xk x y

P( 1, 2)�

Q x y( , )O(0, 0)

x y y xQ x y x xP

x x

x

+ − = ⇒ = −= −

−

= − − − =

∴

1 0 11

1 22 1 1 7

2

12

( , ) ( , )( , )

( )( )Area

++ − =

+ =

+ = ±

∴ = − >= − = −

∴ −

1 14

1 141 14

13 15 01 1213 12

x

xx

x xy xQ

, ( )

( , )

The midpoint of [PQ] is the centre of the circle. Call it R.

P Q

R

( , ), ( , )

, ( , )

− −

=− + −

= −

1 2 13 121 13

22 12

26 5

Equation of circle:

P R

r PR

( , ), ( , )

( ) ( )

− −

= = + + − − = + =

1 2 6 5

6 1 5 2 49 49 7 22 2

( ) ( )( ) ( )x h y k rx y− + − =

− + + =

2 2 2

2 26 5 98


Question 2 (a)

y y m x x− = −1 1( )

Area = −12 1 2 2 1x y x y

Question 2 (b)

Question 2 (c)

R( , 5)��

Q(13, 12)� O(0, 0)

P( 1, 2)�


A( 5, 8)�

B(3, 8)�C( , 2)�

D( , 18)��

�

�

Slope of AC: m18 25 6

611

=−

− −= −

Slope of BD: m218 8

2 3265

=+

− −= −

tan( )( )

tan

θ

θ

= +−

+

=

− ++ −

∴ =− +−

m mm m

1 2

1 2

611

265

611

265

16

11

1 12265

611

2651

50 5+ −

=( )( )

. o

Question 1 (c)

Sample 7

Paper 2



A spinner has nine equal segments numbered 1, 2, 3, 4, 5, 6, 7, 8 and 9 (Nine numbers)Blue: 2, 3, 6, 8, 9 (Five numbers) Red: 1, 4, 5, 7 (Four numbers)E is the event that the pointer lands on an even number. E: 2, 4, 6, 8 (Four numbers)R is the event that the pointer lands on a red colour.

P E R P E P R P E RE R

P E R

( ) ( ) ( ) ( ){ }

( )

∪ = + − ∩∩ =

∪ = + − =

449

49

19

79

R

6

E

2

84

3

1

9

57

P R E P R EP E

( | ) ( )( )

=∩

= = × =1949

19

94

14

P E R P E RP R

( | ) ( )( )

=∩

= = × =1949

19

94

14

(i) Yes, because P(E) = P(R), (ii) No: 1

449π


Question 4 (a)

Question 4 (b)

Question 4 (c)

Question 4 (d)

Question 4 (e)

Question 4 (f)

P E( ) = =Number of even numbers

Number of numbers49

P R( ) = =Number of red colours

Number of numbers49


y x x

x

=

=

sin cos

, , , , , , , , ,

3 2

04 3

23

34

54

43

53

74

π π π ππ

π π π π

x

y

p q r

p

q

r

=

=

=

π

ππ

3

54

( )

( )

( )

Third root

Sixth root

Seventh root

T C

S Ay = 25

x

1

cos x = + =15

AdjacentHypotenuse

The angle x is located in the first and fourth quadrants.

1 55 1 42

2 2 2

2

+ =

= − =∴ =

yy

y

( )

tan ( )

tan tantan

tan

x

x xx

x

= ±

=−

=

2

2 21 2

First and fourth quadrants

22 2 2 21 2

43

2 2 2 21 2

43

2

2

: tan ( )

tan : tan ( )( )

x

x x

=−

= −

= − =−

− −=

T C

S A

T C

S A

sin 3x = 0 cos 2x = 0


Question 3 (b) (i)

Question 3 (b) (ii)

3 0 2 2

23 3

23

03

23

43

53

x n n n

x n n n

x

= + + ∈

= + ∈

=

π π π

π π π

π ππ

π π

, ,

, ,

, , , , ,

�

�

22

2 32

2

434

434

54

74

x n n n

x n n n

x

= + + ∈

= + + ∈

=

ππ

ππ

ππ

ππ

π π π π

, ,

, ,

, , ,

�

�



A spinner has nine equal segments numbered 1, 2, 3, 4, 5, 6, 7, 8 and 9 (Nine numbers)Blue: 2, 3, 6, 8, 9 (Five numbers) Red: 1, 4, 5, 7 (Four numbers)E is the event that the pointer lands on an even number. E: 2, 4, 6, 8 (Four numbers)R is the event that the pointer lands on a red colour.

P E R P E P R P E RE R

P E R

( ) ( ) ( ) ( ){ }

( )

∪ = + − ∩∩ =

∪ = + − =

449

49

19

79

R

6

E

2

84

3

1

9

57

P R E P R EP E

( | ) ( )( )

=∩

= = × =1949

19

94

14

P E R P E RP R

( | ) ( )( )

=∩

= = × =1949

19

94

14

(i) Yes, because P(E) = P(R), (ii) No: 1

449π


Question 4 (a)

Question 4 (b)

Question 4 (c)

Question 4 (d)

Question 4 (e)

Question 4 (f)

P E( ) = =Number of even numbers

Number of numbers49

P R( ) = =Number of red colours

Number of numbers49


y x x

x

=

=

sin cos

, , , , , , , , ,

3 2

04 3

23

34

54

43

53

74

π π π ππ

π π π π

x

y

p q r

p

q

r

=

=

=

π

ππ

3

54

( )

( )

( )

Third root

Sixth root

Seventh root

T C

S Ay = 25

x

1

cos x = + =15

AdjacentHypotenuse

The angle x is located in the first and fourth quadrants.

1 55 1 42

2 2 2

2

+ =

= − =∴ =

yy

y

( )

tan ( )

tan tantan

tan

x

x xx

x

= ±

=−

=

2

2 21 2

First and fourth quadrants

22 2 2 21 2

43

2 2 2 21 2

43

2

2

: tan ( )

tan : tan ( )( )

x

x x

=−

= −

= − =−

− −=

T C

S A

T C

S A

sin 3x = 0 cos 2x = 0


Question 3 (b) (i)

Question 3 (b) (ii)

3 0 2 2

23 3

23

03

23

43

53

x n n n

x n n n

x

= + + ∈

= + ∈

=

π π π

π π π

π ππ

π π

, ,

, ,

, , , , ,

�

�

22

2 32

2

434

434

54

74

x n n n

x n n n

x

= + + ∈

= + + ∈

=

ππ

ππ

ππ

ππ

π π π π

, ,

, ,

, , ,

�

�

Sample 7

Paper 2



B

A

C

D

E

**

∠ = ∠

∠ = ∠

DEA BEC

EAD CBE

( )

(

Vertically opposite

Standing on same arcc)

∴∠ = ∠ADE BCE

AEDE

BEEC

AE EC BE DE

=

∴ =


Question 6 (a) (ii)

Question 6A (b)

r

r2�

G

F

H

I

OJ

2

4

3

FJ JH IJ JGr r

rr

r

=

× = − +

= −

=∴ =

3 4 2 212 416

4

2

2

( )( )

(i) The distribution is normal.

µ σµ σµ µ

σ σ

+ =− =

= ⇒ =

+ = ⇒ =

2 108 62 97 4

2 206 103

103 2 108 6 2 8

..

. .

g

g

µ σ

µσ

= =< =

= =−

=−

= −

<

103 2 8100

100 100 1032 8

1 07

10

g, g.( ) ?

:.

.

(

P x

x z x

P x 00 1 071 07

1 1 071 0 85770 142314 23

) ( . )( . )

( . ).

..

= < −= >= − <= −==

P zP zP z

%%


(ii)

Question 7 (b) Question 7 (c)P xP x ZP z ZP z Zz

( ) .( ) .( ) .( ) .

.

< =< − => =< =

∴ =

100 0 0010 001

0 0010 999

3 08⇒⇒ − = −

− =−

∴ =

zx

x

30 8

3 08 1002 8

108 624

.

..

. g

Weight in grams97.4 g 108.6 g

95%

µ


400030002000

1000

60005000

Cum

ula

tive f

requency

900

600

500

400

300

200

100

800

700

0

IQ Range

1000

Weight (grams)Median10 percentile

th.

Median = 3400 gInterquartile range = 3670 − 3060 = 610 g

n P P= =12 0 15, ( ) . , ( )Need special care Do not need special care ==

= =

0 850 15 0 85 0 292412

22 10

. ,( ) ( . ) ( . ) .P C2 need special care

E = × =100 0 2641 26 41. . occasions


Range of weights below the 10th. percentile = 0−2600 gQuestion 5 (b)

Question 5 (c) (i)

Question 5 (c) (ii)

A.

B. P P( ) (More than two need special care Two or one or none nee= −1 dd special care){ . ( . ) ( . ) ( . )= − + +1 0 2924 0 15 0 85 0 1512

11 11 12

00C C (( . ) }

.0 85

0 2641

12

=



B

A

C

D

E

**

∠ = ∠

∠ = ∠

DEA BEC

EAD CBE

( )

(

Vertically opposite

Standing on same arcc)

∴∠ = ∠ADE BCE

AEDE

BEEC

AE EC BE DE

=

∴ =


Question 6 (a) (ii)

Question 6A (b)

r

r2�

G

F

H

I

OJ

2

4

3

FJ JH IJ JGr r

rr

r

=

× = − +

= −

=∴ =

3 4 2 212 416

4

2

2

( )( )

(i) The distribution is normal.

µ σµ σµ µ

σ σ

+ =− =

= ⇒ =

+ = ⇒ =

2 108 62 97 4

2 206 103

103 2 108 6 2 8

..

. .

g

g

µ σ

µσ

= =< =

= =−

=−

= −

<

103 2 8100

100 100 1032 8

1 07

10

g, g.( ) ?

:.

.

(

P x

x z x

P x 00 1 071 07

1 1 071 0 85770 142314 23

) ( . )( . )

( . ).

..

= < −= >= − <= −==

P zP zP z

%%


(ii)

Question 7 (b) Question 7 (c)P xP x ZP z ZP z Zz

( ) .( ) .( ) .( ) .

.

< =< − => =< =

∴ =

100 0 0010 001

0 0010 999

3 08⇒⇒ − = −

− =−

∴ =

zx

x

30 8

3 08 1002 8

108 624

.

..

. g

Weight in grams97.4 g 108.6 g

95%

µ


400030002000

1000

60005000

Cum

ula

tive f

requency

900

600

500

400

300

200

100

800

700

0

IQ Range

1000

Weight (grams)Median10 percentile

th.

Median = 3400 gInterquartile range = 3670 − 3060 = 610 g

n P P= =12 0 15, ( ) . , ( )Need special care Do not need special care ==

= =

0 850 15 0 85 0 292412

22 10

. ,( ) ( . ) ( . ) .P C2 need special care

E = × =100 0 2641 26 41. . occasions


Range of weights below the 10th. percentile = 0−2600 gQuestion 5 (b)

Question 5 (c) (i)

Question 5 (c) (ii)

A.

B. P P( ) (More than two need special care Two or one or none nee= −1 dd special care){ . ( . ) ( . ) ( . )= − + +1 0 2924 0 15 0 85 0 1512

11 11 12

00C C (( . ) }

.0 85

0 2641

12

=

Sample 7

Paper 2



x

y

M

= = × =

= = × =

∴

10 2 45 10 2 12

10

10 2 45 10 2 12

10

10 10

cos

sin

( , )

o

o

km

km

Bearing of Ships A and B:

The ships travelling from O are moving along tangents to the circle with centre M(10, 10) of radius 2 km. This is analogous to the previous circle. You know the slopes of these two tangents.

dax by c

a b=

+ +

+1 1

2 2

Equation of circle:

x y x yg f

r g f c

2 2

2 2

20 20 196 010 10

100 100

+ − − + == − − =

= + − = +

Centre ( , ) ( , )

−− = =196 4 2

(0, 0)

(10, 10)

r = 2

r = 2

mx y = 0�

Slope of tangent = m.

Question 9 (65 marks)Question 9 (a) (i) Question 9 (a) (ii)

Question 9 (b)

Question 9 (c)

Question 9 (d)

Equation of tangent :

Equation of tang

t mx y kt k

− + =∈ ⇒ =

00 0 0( , )

eent : t mx ymx y x y d

m

m

m

− =− = = =

=−

+

00 10 10 2

210 10

1

2

1 1

2

, ( , ) ( , ),( ) ( )

22

2

1 10 1

1 5 1

+ = −

+ = −

m

m m

m mm m m

m mm mm

2 2

2 2

2

2

1 25 11 25 50 25

24 50 24 012 25 12 04

+ = −

+ = − +

− + =

− + =

( )

( −− − =∴ =

3 3 4 034

43

)( ),

mm

tan

tan ( ) .

α

α

= =

∴ = =−

m 34

1 34 36 87o N of E

tan

tan ( ) .

β

β

= =

∴ = =−

m 43

1 43 53 13o N of E


Question 7 (d)PP

( ) .(Less than advertised weightNot less than advertised

= 0 1423 weight

At least one weighs less than advertised) .

(= 0 8577

P weightOne or more weighs less than advertised weight

)(= P ))

( )( .

= −

= −

11 0 14235

0

PC

None with less than advertised weight)) ( . )

.. %

0 50 85770 53653 6

==

Question 8 (40 marks)Question 8 (a)The parameter in this question is the mean glass thickness, m.Null hypothesis H0: m = 0.954 cmAlternative hypothesis HA: m ≠ 0.954 cm(This is called a two-tailed test as the glass should not be too thick or too thin.)

Question 8 (b)

z x

n

= -ÊËÁ

ˆ¯̃

ms

Mean m = 0.954 cmMean of sample x = 0 96. cmStandard deviation s = 0.13 cmNumber of sample n = 100z = ?

z = -ÊËÁ

ˆ¯̃

=0 96 0 9540 13100

0 46. ..

.

Question 8 (c)p P z= >

= - -= -=

2 0 462 0 5 0 6772 0 52 1 0 67720 6456 0 05

( . ){ . ( . . )}{ . }. .� [0.05 = 5% level of significance]

Therefore, the probability of the null hypothesis is very, very strong. We accept the null hypothesis.

Question 8 (d)The glass company does not have sufficient evidence to conclude it is not meeting the specifications. The difference between the sample mean and the actual mean is not large enough to attribute to anything but sampling error.



x

y

M

= = × =

= = × =

∴

10 2 45 10 2 12

10

10 2 45 10 2 12

10

10 10

cos

sin

( , )

o

o

km

km

Bearing of Ships A and B:

The ships travelling from O are moving along tangents to the circle with centre M(10, 10) of radius 2 km. This is analogous to the previous circle. You know the slopes of these two tangents.

dax by c

a b=

+ +

+1 1

2 2

Equation of circle:

x y x yg f

r g f c

2 2

2 2

20 20 196 010 10

100 100

+ − − + == − − =

= + − = +

Centre ( , ) ( , )

−− = =196 4 2

(0, 0)

(10, 10)

r = 2

r = 2

mx y = 0�

Slope of tangent = m.

Question 9 (65 marks)Question 9 (a) (i) Question 9 (a) (ii)

Question 9 (b)

Question 9 (c)

Question 9 (d)

Equation of tangent :

Equation of tang

t mx y kt k

− + =∈ ⇒ =

00 0 0( , )

eent : t mx ymx y x y d

m

m

m

− =− = = =

=−

+

00 10 10 2

210 10

1

2

1 1

2

, ( , ) ( , ),( ) ( )

22

2

1 10 1

1 5 1

+ = −

+ = −

m

m m

m mm m m

m mm mm

2 2

2 2

2

2

1 25 11 25 50 25

24 50 24 012 25 12 04

+ = −

+ = − +

− + =

− + =

( )

( −− − =∴ =

3 3 4 034

43

)( ),

mm

tan

tan ( ) .

α

α

= =

∴ = =−

m 34

1 34 36 87o N of E

tan

tan ( ) .

β

β

= =

∴ = =−

m 43

1 43 53 13o N of E


Question 7 (d)PP

( ) .(Less than advertised weightNot less than advertised

= 0 1423 weight

At least one weighs less than advertised) .

(= 0 8577

P weightOne or more weighs less than advertised weight

)(= P ))

( )( .

= −

= −

11 0 14235

0

PC

None with less than advertised weight)) ( . )

.. %

0 50 85770 53653 6

==

Question 8 (40 marks)Question 8 (a)The parameter in this question is the mean glass thickness, m.Null hypothesis H0: m = 0.954 cmAlternative hypothesis HA: m ≠ 0.954 cm(This is called a two-tailed test as the glass should not be too thick or too thin.)

Question 8 (b)

z x

n

= -ÊËÁ

ˆ¯̃

ms

Mean m = 0.954 cmMean of sample x = 0 96. cmStandard deviation s = 0.13 cmNumber of sample n = 100z = ?

z = -ÊËÁ

ˆ¯̃

=0 96 0 9540 13100

0 46. ..

.

Question 8 (c)p P z= >

= - -= -=

2 0 462 0 5 0 6772 0 52 1 0 67720 6456 0 05

( . ){ . ( . . )}{ . }. .� [0.05 = 5% level of significance]

Therefore, the probability of the null hypothesis is very, very strong. We accept the null hypothesis.

Question 8 (d)The glass company does not have sufficient evidence to conclude it is not meeting the specifications. The difference between the sample mean and the actual mean is not large enough to attribute to anything but sampling error.

Sample 7

Paper 2



Distances travelled by A and B:Question 9 (e)

Question 9 (f)

θ24

725

tan cosθ θ= ⇒ =724

2425

θ

l

80 km60 km

O

l

l

2 2 2

2 2 2425

60 80 2 60 80

60 80 2 60 80 28

= + −

= + − =

( )( ) cos

( )( )( )

θ

km

Question 9 (g)

v tsB = == × =

30 230 2 60

km/h hkm

,

v st

s v t

v tsA

= ⇒ = ×

= == × =

40 240 2 80

km/h hkm

,

m m143 2

34

43

34

43

34

43

34

4

1 1 1

= =

∴ = +−

+

= +

−+

= +

,

tan( )( )

θ 3334

43

34

2

21212

16 924

724

−

=−

×

=−

=

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